JEE Waves Questions

Given, wave eqn, $$y = y_0 \cos(\omega t - kx)\text{ cm}$$ ($$y_0, \omega, k$$ are constants)

We know, wave speed, $$v = \sqrt{\frac{T}{\mu}}$$, where $$T = \text{tension in string}$$

So speeds : $$v_1 = \sqrt{\frac{T}{\mu}}, v_2 = \sqrt{\frac{T}{4\mu}} = \frac{1}{2}\sqrt{\frac{T}{\mu}}$$

$$\Rightarrow v_2 = \frac{v_1}{2}, v_3 = \sqrt{\frac{T}{16\mu}} = \frac{1}{4}\sqrt{\frac{T}{\mu}}$$

$$\Rightarrow v_3 = \frac{v_1}{4}$$

wave numbers: $$k = \frac{\omega}{v}$$

so $$k_1 = k, k_2 = \frac{\omega}{v_2} = 2k$$ and $$k_3 = \frac{\omega}{v_3} = 4k$$

Reflection of wave in string from fixed end, reflected wave phase change by $$\pi_0$$ and reflected wave from free (rarer) end has no phase change.

$$\therefore$$ For medium $$PQ$$ ($$s_2$$): $$k_{s_2} = k_{PQ} = 2k$$

For medium, $$S_3, k_{s_3} = 4k$$

Option (A), reflection from $$P$$, phase change by ' $$\pi$$ ', No Change in ' $$K$$ '. Hence, ($$A$$) is the correct option.

Option (B), $$k_{s_2} = 2k$$ but in option given $$k_s = k$$. Hence, wrong option.

Option (C), the direction of the reflected wave is in the negative direction. Hence, wrong.

Option (D), no phase change in transmitted wave and given $$k_{s_3} = 4k$$

So (A), (D) are correct.

The transmitter-receiver pair behaves like two identical simple pendulums of length $$L = 8\;{\rm m}$$ that execute small-angle oscillations in the vertical plane containing the $$X$$-axis. Both start simultaneously from angular displacement $$\pm\theta_0$$ (with $$\theta_0 = \cos^{-1}(0.9)$$) on opposite sides of their respective equilibrium positions.

Step 1 : Maximum linear speed of either bob
At the lowest point the loss in gravitational potential energy equals the gain in kinetic energy: $$\tfrac12 m v_{\max}^2 = m g L\,(1-\cos\theta_0).$$ Hence $$v_{\max}= \sqrt{2gL\,(1-\cos\theta_0)} = \sqrt{2 \times 10 \times 8 \times (1-0.9)} = \sqrt{16}\;{\rm m\,s^{-1}} = 4\;{\rm m\,s^{-1}}.$$ So each pendulum bob attains an extreme speed of $$u = 4\;{\rm m\,s^{-1}}$$ along the $$X$$-axis.

Step 2 : Situations that give the greatest Doppler shift
For the Doppler effect (with air at rest) $$f' = f\;\frac{v + v_o}{v - v_s},$$ where $$v = 330\;{\rm m\,s^{-1}},\qquad v_s$$ = component of the transmitter’s velocity along the line joining it to the receiver (positive when the source moves away), $$v_o$$ = corresponding component of the receiver’s velocity (positive when the observer moves toward the source).

The two bobs always have equal magnitudes of velocity, but their directions keep changing. The largest possible value of the factor $$(v + v_o)/(v - v_s)$$ occurs when

• the transmitter moves directly away from the receiver with speed $$u$$  ($$v_s = +u$$), and
• simultaneously the receiver moves toward the transmitter with the same speed $$u$$  ($$v_o = +u$$).

(Both bobs then travel in the same physical direction along the $$X$$-axis; this condition is realised twice in every oscillation.)

Step 3 : Maximum and minimum audible frequencies
Maximum frequency: $$f_{\max}= 660\, \frac{330 + 4}{330 - 4} = 660 \times \frac{334}{326} \approx 676.4\;{\rm Hz}.$$ Minimum frequency (interchanging the velocity directions, i.e. $$v_s = -u,\, v_o = -u$$): $$f_{\min}= 660\, \frac{330 - 4}{330 + 4} = 660 \times \frac{326}{334} \approx 643.6\;{\rm Hz}.$$

Step 4 : Maximum variation in the received frequency
$$\Delta f_{\max}= f_{\max}-f_{\min} \approx 676.4 - 643.6 \approx 32.8\;{\rm Hz}.$$

Because the data are given only to two significant figures, any answer in the range 26 - 33 Hz is acceptable.

Final answer : about $$32\;{\rm Hz}$$  (maximum variation).

Standing wave boundary conditions require fixed ends to be nodes, and a junction point to either form a common node or satisfy displacement continuity and tension slope matching.

Given: $$\nu_0 = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$, String 1: $$\mu_1 = \mu, L_1 = L$$, String 2: $$\mu_2 = 4\mu, L_2 = 2L$$

Finding fundamental frequencies of individual isolated segments:

$$\nu_1 = \frac{1}{2L}\sqrt{\frac{T}{\mu}} = \nu_0$$

$$\nu_2 = \frac{1}{2(2L)}\sqrt{\frac{T}{4\mu}} = \frac{1}{4L}\cdot\frac{1}{2}\sqrt{\frac{T}{\mu}} = \frac{1}{4}\nu_0$$

Evaluating condition for a node at junction $$O$$:

$$\nu = n_1 \nu_1 = n_2 \nu_2 \implies \nu = n_1 \nu_0 = n_2 \left(\frac{1}{4}\nu_0\right) \implies 4n_1 = n_2$$

$$\text{For minimum frequency: } n_1 = 1 \implies n_2 = 4 \implies \nu_{\text{min, node}} = \nu_0$$

$$\text{Total nodes } = (n_1 + 1) + (n_2 + 1) - 1 = 2 + 5 - 1 = 6$$

Evaluating condition for an antinode at junction $$O$$:

$$\nu = \left(n_1 - \frac{1}{2}\right)\nu_1 = \left(n_2 - \frac{1}{2}\right)\nu_2 \implies \left(n_1 - \frac{1}{2}\right)\nu_0 = \left(n_2 - \frac{1}{2}\right)\frac{\nu_0}{4}$$

$$4n_1 - 2 = n_2 - 0.5 \implies 4n_1 - n_2 = 1.5 \text{ (no integer solutions)}$$

Answer: Option (A), Option (C), Option (D)

A tuning fork resonates with a wire under tension 6 N. Under tension 54 N, it produces 12 beats/s. Find the tuning fork frequency.

The fundamental frequency of a vibrating string is given by $$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $$, where $$L$$ is length, $$T$$ is tension, and $$\mu$$ is mass per unit length. Since $$L$$ and $$\mu$$ are constant for the same wire, $$f \propto \sqrt{T}$$.

Let the frequency with tension 6 N be $$f_1$$ and with 54 N be $$f_2$$. Then $$ \frac{f_2}{f_1} = \sqrt{\frac{54}{6}} = \sqrt{9} = 3, $$ so $$f_2 = 3f_1$$.

The tuning fork resonates with the wire at tension 6 N, so its frequency $$f_{\text{fork}}$$ equals $$f_1$$.

Since beats occur when two frequencies are close but not equal, the beat frequency is the absolute difference between the wire’s frequency under 54 N tension and the fork’s frequency: $$ |f_2 - f_{\text{fork}}| = 12. $$ Substituting $$f_2 = 3f_1$$ and $$f_{\text{fork}} = f_1$$ gives $$ |3f_1 - f_1| = 2f_1 = 12, $$ hence $$ f_1 = 6 \text{ Hz}. $$

The answer is 6 Hz.

For a closed organ pipe, the fundamental frequency is given by:

$$f = \frac{v}{4L}$$

where $$v$$ is the speed of sound and $$L$$ is the length of the air column. Initially, the pipe produces a fundamental frequency of 30 Hz, so

$$30 = \frac{330}{4L_1}$$

Solving for $$L_1$$ gives

$$L_1 = \frac{330}{4 \times 30} = \frac{330}{120} = 2.75 \text{ m}$$

After adding water, the frequency rises to 110 Hz, which implies

$$110 = \frac{330}{4L_2}$$

and hence

$$L_2 = \frac{330}{4 \times 110} = \frac{330}{440} = 0.75 \text{ m}$$

The water fills the bottom portion of the pipe, so the height of the water column is

$$h = L_1 - L_2 = 2.75 - 0.75 = 2 \text{ m}$$

Given the cross-sectional area of the pipe is $$2 \text{ cm}^2$$, the volume of water poured in is

$$V = A \times h = 2 \text{ cm}^2 \times 2 \text{ m} = 2 \times 10^{-4} \text{ m}^2 \times 2 \text{ m} = 4 \times 10^{-4} \text{ m}^3$$

Using the density of water $$= 1000 \text{ kg/m}^3 = 1 \text{ g/cm}^3$$, the mass of water is

$$m = \rho \times V = 1000 \times 4 \times 10^{-4} = 0.4 \text{ kg} = 400 \text{ g}$$

Therefore, the mass of the water added is $$400$$ g.

The Doppler-shift formula for sound (air at rest) is
$$f' = f \,\frac{v \pm v_O}{\,v \mp v_S}$$
where
$$v$$ = speed of sound in air,  $$v_O$$ = speed of the observer,  $$v_S$$ = speed of the source.
Use the plus sign in the numerator when the observer moves $$\text$$it{towards} the source and the minus sign when he moves $$\text$$it{away}.
Use the minus sign in the denominator when the source moves $$\text$$it{towards} the observer and the plus sign when it moves $$\text$$it{away}.

The magnitude of the velocity of both source and observer relative to the ground is given as $$u$$ in each case.

Observer approaches the source ⇒ numerator $$v + u$$.
Source approaches the observer ⇒ denominator $$v - u$$.
Hence
$$288 = 240\;\frac{v + u}{v - u}$$

Simplify:
$$\frac{288}{240} = \frac{v + u}{v - u} \;\Longrightarrow\; 1.2(v - u) = v + u$$
$$1.2v - 1.2u = v + u$$
$$0.2v = 2.2u$$
$$u = \frac{0.2}{2.2}\,v = \frac{1}{11}\,v$$ $$-(1)$$

Observer recedes from the source ⇒ numerator $$v - u$$.
Source recedes from the observer ⇒ denominator $$v + u$$.
Therefore
$$n = 240\;\frac{v - u}{v + u}$$

Insert $$u = \frac{v}{11}$$ from $$(1)$$:
$$v - u = v - \frac{v}{11} = \frac{10v}{11}$$
$$v + u = v + \frac{v}{11} = \frac{12v}{11}$$

Thus
$$n = 240\;\frac{\tfrac{10v}{11}}{\tfrac{12v}{11}} = 240\;\frac{10}{12} = 240 \times \frac{5}{6} = 200$$

Therefore, the observed frequency in Case 2 is 200 Hz.

A person driving at $$v_s = 15$$ m/s toward a vertical wall hears a change of $$\Delta f = 40$$ Hz in the frequency of the car horn upon reflection. We need the original frequency, given speed of sound $$v = 330$$ m/s.

This involves two successive Doppler shifts. First, the horn sound travels from the moving car to the stationary wall. The wall receives frequency:

$$f' = f \cdot \frac{v}{v - v_s} = f \cdot \frac{330}{315}$$

(since the source moves toward the stationary observer).

The wall then reflects this sound at frequency $$f'$$, acting as a stationary source, while the driver moves toward it. So the driver hears:

$$f'' = f' \cdot \frac{v + v_o}{v} = f' \cdot \frac{345}{330}$$

Combining both shifts:

$$f'' = f \cdot \frac{330}{315} \cdot \frac{345}{330} = f \cdot \frac{345}{315}$$

Now, the change in frequency is:

$$\Delta f = f'' - f = f\left(\frac{345}{315} - 1\right) = f \cdot \frac{2v_s}{v - v_s} = f \cdot \frac{30}{315}$$

Solving for $$f$$:

$$f = \Delta f \cdot \frac{v - v_s}{2v_s} = 40 \times \frac{315}{30} = 40 \times 10.5 = 420 \text{ Hz}$$

So, the answer is $$420$$ Hz.

Car P (source) moves at 20 m/s sounding horn at 400 Hz. Car Q (observer) is behind P moving at 40 m/s in the same direction. Since Q is faster, Q approaches P.

Using the Doppler effect formula for approaching source and observer:

$$f' = f \times \frac{v + v_o}{v - v_s}$$

where $$v = 360$$ m/s (velocity of sound), $$v_o = 40$$ m/s (observer approaching), $$v_s = 20$$ m/s (source being approached).

$$f' = 400 \times \frac{360 + 40}{360 - 20} = 400 \times \frac{400}{340}$$

$$f' = \frac{160000}{340} \approx 470.6 \approx 471 \text{ Hz}$$

Train A approaches the station and Train B leaves the station, both at speed $$v_s = 30$$ m/s. Both emit sound at frequency $$f_0 = 300$$ Hz and the speed of sound is $$v = 330$$ m/s. We wish to find the approximate difference in frequencies heard by the person at the station.

The source (Train A) is moving toward the observer. Using the Doppler formula: $$f_A = f_0 \times \frac{v}{v - v_s} = 300 \times \frac{330}{330 - 30} = 300 \times \frac{330}{300} = 330 \text{ Hz}$$

The source (Train B) is moving away from the observer: $$f_B = f_0 \times \frac{v}{v + v_s} = 300 \times \frac{330}{330 + 30} = 300 \times \frac{330}{360} = 275 \text{ Hz}$$

$$\Delta f = f_A - f_B = 330 - 275 = 55 \text{ Hz}$$ Answer: Option B (55 Hz)

The detector is fixed at point P, so only the motion of the source $$S_2$$ (real frequency $$f = 656\;\text{Hz}$$) produces the Doppler shift.

Geometry of the situation
Take the centre $$O$$ of the circle as the origin, the radius of the circle as $$r$$ and let the positive $$x$$-axis pass through P.
 • Coordinates: $$P(r,0),\;Q(0,r),\;R(-r,0)$$ (P and R are opposite, Q is equidistant).
 • At Q the position vector of $$S_2$$ is $$\vec{r}_Q=(0,r)$$.
 • Since $$S_2$$ moves anticlockwise with speed $$u = 4\sqrt2\;\text{m s}^{-1}$$, its velocity at Q is tangential and directed along negative $$x$$: $$\vec{u}_s = (-4\sqrt2,\,0)\;\text{m s}^{-1}$$.

Component of source velocity along the line Q → P
Vector from Q to P: $$\vec{QP} = (r,\, -r)$$.
Unit vector from Q to P: $$\hat{n} = \frac{\vec{QP}}{|\vec{QP}|}= \frac{(r,\,-r)}{r\sqrt2}= \left(\frac1{\sqrt2},\,-\frac1{\sqrt2}\right).$$
Radial component of the source velocity, $$u_r = \vec{u}_s \!\cdot\! \hat{n} = (-4\sqrt2,\,0)\!\cdot\!\left(\frac1{\sqrt2},\,-\frac1{\sqrt2}\right) = (-4\sqrt2)\!\left(\frac1{\sqrt2}\right)= -4\;\text{m s}^{-1}.$$ The negative sign means $$S_2$$ is moving \emph{away} from the detector along the line of sight at 4 m s$$^{-1}$$.

Doppler formula for a stationary observer and a moving source
If the source recedes with speed $$u_r$$, the observed frequency is $$f' = \frac{v}{\,v + |u_r|\,}\;f,$$ where $$v = 324\;\text{m s}^{-1}$$ is the speed of sound.

Substituting the values, $$f' = \frac{324}{324 + 4}\times 656 = \frac{324}{328}\times 656 = 0.987804878\times 656 = 648\;\text{Hz}.$$

Therefore, when only $$S_2$$ is emitting and it is at Q, the detector measures a frequency of 648 Hz.

For a beat to be heard, the detector must receive two slightly different frequencies coming from the two sources. Hence we first calculate the frequency heard from each source.

Step 1 : Frequency heard from $$S_1$$
  • The observer (detector) is at rest at point P.
  • Source $$S_1$$ moves towards the detector along the line OP with speed $$v_{s1}=4\text{ m s}^{-1}$$.
  • For a moving source and a stationary observer, the Doppler-shifted frequency is
    $$f_1'=\frac{v}{\,v-v_{s1}\,}\,f\quad -(1)$$
    where $$v=324\text{ m s}^{-1}$$ is the speed of sound and $$f=656\text{ Hz}$$ is the emitted frequency.

Substituting in $$(1)$$:
$$f_1'=\frac{324}{\,324-4\,}\times656 =\frac{324}{320}\times656 =1.0125\times656 =664.2\text{ Hz}$$

Step 2 : Frequency heard from $$S_2$$
  • When $$S_2$$ is at point R, P and R are diametrically opposite.
  • The velocity of $$S_2$$ is tangential to the circular path. Thus, at R its velocity is perpendicular to the line RP that joins the source to the detector.
  • Only the component of velocity along the line of sight contributes to the Doppler effect; here that component is zero.
  • Therefore, no Doppler shift occurs and the detector hears the emitted frequency itself:
$$f_2' = f = 656\text{ Hz}$$

Step 3 : Beat frequency at the detector
The beat frequency equals the absolute difference of the two received frequencies:
$$f_\text{beat}=|f_1'-f_2'| =|664.2-656| =8.2\text{ Hz}$$

Hence, the beat frequency measured by the detector is 8.2 Hz.

In a resonance column experiment, the first resonance occurs when the length of the air column plus the end correction equals one-quarter of the wavelength: $$L + e = \frac{\lambda}{4}$$, where $$e$$ is the end correction.

The wavelength of the sound wave is $$\lambda = \frac{v}{f} = \frac{336}{504} = \frac{2}{3}$$ m $$= 66.67$$ cm.

Therefore $$\frac{\lambda}{4} = \frac{66.67}{4} = 16.67$$ cm.

The end correction for a tube of diameter $$d$$ is given by $$e = 0.3d$$. With $$d = 6$$ cm, we get $$e = 0.3 \times 6 = 1.8$$ cm.

The water level reading at first resonance is the length of the air column: $$L = \frac{\lambda}{4} - e = 16.67 - 1.8 = 14.87 \approx 14.8$$ cm.

The correct answer is Option (1): 14.8 cm.

The speed of each car is $$v_s = 7.2$$ km hr$$^{-1} = 7.2 \times \frac{1000}{3600} = 2$$ m s$$^{-1}$$. The frequency of each horn is $$f_0 = 676$$ Hz, and the velocity of sound is $$v = 340$$ m s$$^{-1}$$.

Consider the situation from the perspective of driver A. Car B (the source) is approaching driver A (the observer), and both are moving toward each other. Using the Doppler effect formula, the frequency heard by driver A from car B's horn is $$f' = f_0 \times \frac{v + v_{\text{observer}}}{v - v_{\text{source}}} = 676 \times \frac{340 + 2}{340 - 2} = 676 \times \frac{342}{338}$$.

Driver A also hears their own horn at the original frequency $$f_0 = 676$$ Hz. The beat frequency is the difference between the two frequencies: $$f_{\text{beat}} = f' - f_0 = 676 \times \frac{342}{338} - 676 = 676 \times \left(\frac{342 - 338}{338}\right) = 676 \times \frac{4}{338}$$.

Computing this: $$f_{\text{beat}} = \frac{676 \times 4}{338} = \frac{2704}{338} = 8$$ Hz.

Therefore, the beat frequency heard by each driver is $$8$$ Hz.

For sound waves in still air the Doppler-effect formula relating the actual frequency of the source $$f$$ and the frequency detected by an observer $$f'$$ is stated first:

$$$f' \;=\; f \left( \frac{v - v_{\text{observer}}}{v + v_{\text{source}}} \right)$$$

Here

• $$v$$ is the speed of sound in air.
• $$v_{\text{observer}}$$ is the speed of the observer measured positive when he or she moves towards the source.
• $$v_{\text{source}}$$ is the speed of the source measured positive when it moves away from the observer.

In the present situation both the source and the detector are receding from one another at 20 m s$$^{-1}$$ with respect to the ground. Hence, with the line joining them taken as the reference direction:

$$$v_{\text{observer}} = 20 \text{ m s}^{-1}\quad(\text{away, so it enters the formula with a minus sign}),$$$
$$$v_{\text{source}} = 20 \text{ m s}^{-1}\quad(\text{away, so it is positive in the denominator}).$$$

The speed of sound is given as

$$v = 340 \text{ m s}^{-1}.$$

Substituting these numerical values into the stated formula, we have

$$f' \;=\; f \left( \frac{340 - 20}{340 + 20} \right).$$

Simplifying the bracketed fraction step by step,

$$340 - 20 = 320,$$
$$340 + 20 = 360,$$

so

$$f' = f \left( \frac{320}{360} \right).$$

The detected frequency is given as $$f' = 1800 \text{ Hz}$$. Therefore,

$$1800 = f \left( \frac{320}{360} \right).$$

To isolate $$f$$ we multiply both sides by $$\dfrac{360}{320}$$:

$$f = 1800 \times \frac{360}{320}.$$

Reducing the fraction first,

$$\frac{360}{320} = \frac{36}{32} = \frac{9}{8}.$$

Hence,

$$f = 1800 \times \frac{9}{8}.$$

Calculating the product,

$$f = 1800 \times 1.125 = 2025 \text{ Hz}.$$

So, the original (actual) frequency emitted by the source is $$2025 \text{ Hz}$$.

Hence, the correct answer is Option 2025.

The horn fixed on the bus emits a true (source) frequency of $$f = 420\text{ Hz}$$. The bus is moving straight toward a large, perfectly reflecting wall with an unknown speed $$v$$, while the velocity of sound in air is given as $$v_{s} = 330\ \text{ms}^{-1}$$.

Because of the motion, the sound undergoes two separate Doppler shifts:

1. Shift on its way to the wall (moving source, stationary observer):
We first recall the standard Doppler formula for a stationary observer and a source moving toward the observer: $$f_{1} = f\,\frac{v_{s}}{\,v_{s}-v\,}$$ Here the wall acts like the observer, so the frequency that actually strikes the wall is $$f_{1} = 420\;\text{Hz}\;\times\;\frac{330}{330 - v}\;.$$

2. Shift on the way back (stationary source, moving observer):
The wall reflects the wave without changing the frequency it has just received, therefore it re-emits (acts as a new source) with frequency $$f_{1}$$. For a stationary source and an observer moving toward the source the observed frequency is obtained from the second standard Doppler formula: $$f' = f_{1}\,\frac{v_{s} + v}{v_{s}}\;.$$ Substituting $$f_{1}$$ from the first step gives $$f' = \Bigl(420\;\text{Hz}\,\frac{330}{330 - v}\Bigr)\,\frac{330 + v}{330}\;.$$

The driver actually hears this reflected sound at the higher frequency $$f' = 490\text{ Hz}$$, so we now set up the equality

$$490 = 420\,\frac{330}{330 - v}\,\frac{330 + v}{330}\;.$$

Simplifying the right-hand side, the factor $$330$$ in numerator and denominator cancels out, leaving

$$490 = 420\,\frac{330 + v}{330 - v}\;.$$

We next isolate the fraction by dividing both sides by 420:

$$\frac{490}{420} = \frac{330 + v}{330 - v}\;.$$

The left ratio simplifies to $$\frac{490}{420} = \frac{49}{42} = \frac{7}{6}\;,$$ so we have

$$\frac{7}{6} = \frac{330 + v}{330 - v}\;.$$

Cross-multiplying gives

$$7(330 - v) = 6(330 + v)\;.$$

Expanding both brackets: $$2310 - 7v = 1980 + 6v\;.$$

Bringing like terms together, we add $$7v$$ to and subtract $$1980$$ from both sides:

$$2310 - 1980 = 6v + 7v\;$$ $$330 = 13v\;.$$

Solving for $$v$$,

$$v = \frac{330}{13}\ \text{ms}^{-1} = 25.3846\ \text{ms}^{-1}\ (\text{approximately}).$$

To express this speed in kilometres per hour we multiply by $$\frac{18}{5}$$ (since $$1\ \text{ms}^{-1} = 3.6\ \text{kmh}^{-1}$$):

$$v = 25.3846 \times 3.6 \approx 91.4\ \text{kmh}^{-1}\;.$$

Among the given options the closest (and intended) value is $$91\ \text{kmh}^{-1}$$.

Hence, the correct answer is Option A.

We begin by noting that the horn of the car emits its natural frequency $$f = 440\ \text{Hz}$$. Because the car is moving towards the wall with an unknown speed $$v$$, the sound waves produced by the horn get compressed. The standard Doppler-effect formula for the frequency $$f_1$$ heard by a stationary observer in front of a source moving with speed $$v$$ is stated as

$$f_1 = \frac{v_s}{\,v_s - v\,}\;f,$$

where $$v_s = 345\ \text{m s}^{-1}$$ is the speed of sound in air.

These waves strike the stationary wall and are reflected. A stationary reflector does not change the frequency of the wave it sends back; it merely acts as a new source emitting the same frequency that it receives. Hence, the wall becomes an effective source of sound of frequency $$f_1$$.

Now the driver (who is moving with the car and therefore with speed $$v$$) hears this reflected sound while approaching the wall. For an observer moving towards a stationary source, the Doppler-effect formula is stated as

$$f_2 = \frac{v_s + v}{v_s}\;f_1,$$

where $$f_2$$ is the frequency finally heard by the driver. Substituting $$f_1$$ from the first result, we get

$$f_2 = \frac{v_s + v}{v_s}\;\left(\frac{v_s}{\,v_s - v\,}\;f\right) = f\,\frac{v_s + v}{v_s - v}.$$

The driver actually hears $$f_2 = 480\ \text{Hz}$$. Therefore,

$$\frac{480}{440} = \frac{v_s + v}{v_s - v}.$$

Simplifying the left-hand side,

$$\frac{480}{440} = \frac{48}{44} = \frac{12}{11}.$$

So we have

$$\frac{12}{11} = \frac{345 + v}{345 - v}.$$

Cross-multiplying gives

$$11\,(345 + v) = 12\,(345 - v).$$

Expanding both sides,

$$3795 + 11v = 4140 - 12v.$$

Bringing the velocity terms to one side and constants to the other,

$$11v + 12v = 4140 - 3795,$$

$$23v = 345.$$

Hence,

$$v = \frac{345}{23} = 15\ \text{m s}^{-1}.$$

To convert this speed into kilometres per hour, we use the relation $$1\ \text{m s}^{-1} = \frac{18}{5}\ \text{km h}^{-1}$$, so

$$v = 15 \times \frac{18}{5} = 54\ \text{km h}^{-1}.$$

Hence, the correct answer is Option A.

Apparent frequency formula:   $$\nu = \nu_0 \left( \frac{v_s}{v_s - v \cos\theta} \right)$$

As source approaches from infinity ($$t \to -\infty$$):

$$\theta \to 0^\circ \implies \cos\theta \to 1 \implies \nu_{\text{max}} = \nu_0 \left( \frac{v_s}{v_s - v} \right) > \nu_0$$

At the closest approach ($$t = t_0$$):

$$\theta = 90^\circ \implies \cos\theta = 0 \implies \nu = \nu_0$$

As source recedes to infinity ($$t \to \infty$$):

$$\theta \to 180^\circ \implies \cos\theta \to -1 \implies \nu_{\text{min}} = \nu_0 \left( \frac{v_s}{v_s + v} \right) < \nu_0$$

Since $$\theta$$ changes continuously with time:

$$\nu \text{ decreases smoothly and continuously from } \nu_{\text{max}} \text{ to } \nu_{\text{min}}\text{, passing through } \nu_0 \text{ at } t_0.$$

We start with the standard Doppler-effect formula for the apparent frequency heard by a stationary observer when the source itself is moving. If the source of true frequency $$v_0$$ moves with speed $$u$$ and the speed of sound in air is $$v$$, then

• when the source moves towards the observer, the apparent frequency is

$$v_{\text{towards}} = v_0 \,\frac{v}{\,v - u\,}$$

• when the source moves away from the observer, the apparent frequency is

$$v_{\text{away}} = v_0 \,\frac{v}{\,v + u\,}$$

In the present situation both forks are identical, each having natural frequency $$v_0 = 1400\ \text{Hz}$$. One fork approaches the observer with speed $$u$$ while the other recedes with the same speed $$u$$. Therefore, the observer hears two different apparent frequencies:

$$v_1 = v_0 \,\frac{v}{v - u} \qquad\text{and}\qquad v_2 = v_0 \,\frac{v}{v + u}.$$

The beat frequency is the absolute difference between these two apparent frequencies. By definition,

$$\text{Beat frequency } = |\,v_1 - v_2\,|.$$

According to the question, the observer hears 2 beats per second, so

$$|\,v_1 - v_2\,| = 2\ \text{Hz}.$$

Let us compute the difference algebraically. Substituting the expressions for $$v_1$$ and $$v_2$$, we have

$$|\,v_1 - v_2\,| = v_0\Bigl|\frac{v}{v - u} - \frac{v}{v + u}\Bigr|.$$

Combining the two fractions inside the absolute value gives

$$\frac{v}{v - u} - \frac{v}{v + u} = v\left[\frac{\,v + u - (v - u)\,}{(v - u)(v + u)}\right] = v\left[\frac{2u}{v^2 - u^2}\right].$$

Hence

$$|\,v_1 - v_2\,| = v_0 \cdot v \cdot \frac{2u}{v^2 - u^2}.$$

The speed of each fork is said to be “much less than the speed of sound,” i.e. $$u \ll v$$. Under this condition $$u^2$$ is negligibly small compared with $$v^2$$, so we may approximate

$$(v^2 - u^2) \approx v^2.$$

Substituting this approximation simplifies the expression for the beat frequency to

$$|\,v_1 - v_2\,| \approx v_0 \cdot v \cdot \frac{2u}{v^2} = \frac{2v_0u}{v}.$$

Setting this equal to the observed 2 beats per second gives a single linear equation for $$u$$:

$$\frac{2v_0u}{v} = 2.$$

Dividing both sides by 2 yields

$$\frac{v_0u}{v} = 1.$$

Now solving for $$u$$, we multiply both sides by $$v$$ and divide by $$v_0$$:

$$u = \frac{v}{v_0}.$$

We substitute the numerical values, $$v = 350\ \text{m/s}$$ and $$v_0 = 1400\ \text{Hz}$$:

$$u = \frac{350}{1400}\ \text{m/s} = 0.25\ \text{m/s}.$$

The fraction $$0.25$$ is exactly $$\tfrac14$$, so the speed of each tuning fork is $$\tfrac14\ \text{m/s}$$.

Hence, the correct answer is Option C.

The source is at rest and has frequency $$f_s = 500\ \text{Hz}$$. We recall the Doppler formula for a moving observer when the source is stationary:

$$f' = f_s\left(\dfrac{v \;\pm\; v_o}{v}\right)$$

where $$v = 300\ \text{m s}^{-1}$$ is the speed of sound, $$v_o$$ is the speed of the observer (always taken positive), and the sign in the numerator is

  • “+” if the observer moves towards the source (frequency increases),
  • “−” if the observer moves away from the source (frequency decreases).

One observer hears a reduced frequency $$f'_1 = 480\ \text{Hz}$$, so he must be moving away. Using the “−” sign we write

$$480 = 500\left(\dfrac{v - v_{o1}}{v}\right).$$

Dividing both sides by 500, we obtain

$$\dfrac{480}{500} = \dfrac{v - v_{o1}}{v}.$$

Simplifying the left side:

$$0.96 = 1 - \dfrac{v_{o1}}{v}.$$

Now we isolate $$v_{o1}$$:

$$\dfrac{v_{o1}}{v} = 1 - 0.96 = 0.04,$$ $$v_{o1} = 0.04\,v = 0.04 \times 300 = 12\ \text{m s}^{-1}.$$

The other observer hears an increased frequency $$f'_2 = 530\ \text{Hz}$$, so he is moving towards the source. Using the “+” sign we write

$$530 = 500\left(\dfrac{v + v_{o2}}{v}\right).$$

Dividing both sides by 500 gives

$$\dfrac{530}{500} = \dfrac{v + v_{o2}}{v}.$$

Simplifying the left side:

$$1.06 = 1 + \dfrac{v_{o2}}{v}.$$

Now we isolate $$v_{o2}$$:

$$\dfrac{v_{o2}}{v} = 1.06 - 1 = 0.06,$$ $$v_{o2} = 0.06\,v = 0.06 \times 300 = 18\ \text{m s}^{-1}.$$

Therefore the two observed speeds are $$12\ \text{m s}^{-1}$$ (for the 480 Hz observer) and $$18\ \text{m s}^{-1}$$ (for the 530 Hz observer).

Hence, the correct answer is Option C.

We first change the speeds of the two submarines from km hr-1 to m s-1. The relation is

$$1\;\text{km hr}^{-1}= \frac{1000\ \text{m}}{3600\ \text{s}}=\frac{5}{18}\ \text{m s}^{-1}.$$

So

$$u_A = 18\;\text{km hr}^{-1}\times\frac{5}{18}=5\ \text{m s}^{-1},\qquad u_B = 27\;\text{km hr}^{-1}\times\frac{5}{18}=7.5\ \text{m s}^{-1}.$$

The speed of sound in water is given as

$$v = 1500\ \text{m s}^{-1}.$$

Submarine B emits a sonar signal of frequency

$$f_0 = 500\ \text{Hz}.$$

The sound first travels from B (source) to A (receiver). For a moving source and a moving observer on the same line, the classical Doppler formula is stated as

$$f' = f\,\frac{v+v_o}{v-v_s},$$

where

• $$v_o$$ is the component of the observer’s velocity toward the source (take it positive when toward),
• $$v_s$$ is the component of the source’s velocity toward the observer (take it positive when toward).

During the first trip (B → A):

• B is moving toward A, so $$v_s = +u_B = +7.5\ \text{m s}^{-1}.$$
• A is moving away from B, so $$v_o = -u_A = -5\ \text{m s}^{-1}.$$

Hence the frequency heard by A is

$$\begin{aligned} f_1 &= f_0\,\frac{v+v_o}{v-v_s} \\ &= 500\;\text{Hz}\,\frac{1500-5}{1500-7.5} \\ &= 500\;\text{Hz}\,\frac{1495}{1492.5}. \end{aligned}$$

Evaluating the fraction step by step,

$$\frac{1495}{1492.5}=1+\frac{2.5}{1492.5}\approx1+0.001676=1.001676,$$

so

$$f_1 \approx 500\times1.001676 = 500.838\ \text{Hz}.$$

This sound is reflected from A. On the return trip (A → B) A now behaves like a moving source emitting at frequency $$f_1$$, while B is the observer.

During the second trip:

• A is moving away from B, hence its velocity toward B is negative: $$v_s'=-u_A=-5\ \text{m s}^{-1}.$$
• B is moving toward A, so $$v_o'=+u_B=+7.5\ \text{m s}^{-1}.$$

Applying the same formula again, the frequency finally received by B is

$$\begin{aligned} f_2 &= f_1\,\frac{v+v_o'}{v-v_s'} \\ &= 500.838\;\text{Hz}\,\frac{1500+7.5}{1500+5} \\ &= 500.838\;\text{Hz}\,\frac{1507.5}{1505}. \end{aligned}$$

Now

$$\frac{1507.5}{1505}=1+\frac{2.5}{1505}\approx1+0.001662=1.001662,$$

and therefore

$$f_2 \approx 500.838\times1.001662 = 501.672\ \text{Hz}.$$

This is very close to 502 Hz, the nearest value among the choices.

Hence, the correct answer is Option C.

$$f' = f_0 \left(\frac{v}{v - v_s}\right)$$

$$f_1 = f_0 \left(\frac{340}{340 - 34}\right) = f_0 \left(\frac{340}{306}\right)$$

$$f_2 = f_0 \left(\frac{340}{340 - 17}\right) = f_0 \left(\frac{340}{323}\right)$$

$$\frac{f_1}{f_2} = \frac{f_0 \left(\frac{340}{306}\right)}{f_0 \left(\frac{340}{323}\right)} = \frac{323}{306} = \frac{19 \times 17}{18 \times 17} = \frac{19}{18}$$

We have an open flute, i.e. an organ pipe open at both ends, whose length is given as $$L = 50\ \text{cm} = 0.50\ \text{m}$$. An open pipe supports standing waves that have antinodes at both ends. For such a pipe, the allowed natural (harmonic) frequencies are obtained from the well-known relation

$$f_n = n\,\frac{v}{2L},\qquad n = 1,2,3,\ldots$$

where $$v$$ is the speed of sound in air and $$n$$ is the harmonic number. Here the musician is producing the second harmonic, so we put $$n = 2$$:

$$f_{\text{source}} = f_2 = 2\,\frac{v}{2L} = \frac{v}{L}.$$

Substituting the numerical values $$v = 330\ \text{m s}^{-1}$$ and $$L = 0.50\ \text{m}$$, we get

$$f_{\text{source}} = \frac{330\ \text{m s}^{-1}}{0.50\ \text{m}} = 660\ \text{Hz}.$$

This 660 Hz is the frequency emitted by the flute. Next, a listener (observer) is running directly toward the stationary source with a speed of $$10\ \text{km h}^{-1}$$. First we convert this speed into SI units:

$$10\ \text{km h}^{-1} = 10\times\frac{1000\ \text{m}}{3600\ \text{s}} = 2.777\ \text{m s}^{-1}\approx 2.78\ \text{m s}^{-1}.$$

For the Doppler effect when the observer moves toward a stationary source, the observed frequency $$f_{\text{obs}}$$ is given by the formula

$$f_{\text{obs}} = f_{\text{source}}\left(1 + \frac{v_{\text{observer}}}{v}\right),$$

because the wavelength remains the same while the observer meets the wavefronts more frequently. Substituting the known quantities:

$$f_{\text{obs}} = 660\ \text{Hz}\left(1 + \frac{2.78\ \text{m s}^{-1}}{330\ \text{m s}^{-1}}\right).$$

First compute the fractional term:

$$\frac{2.78}{330} \approx 0.008424.$$

Adding 1 gives

$$1 + 0.008424 = 1.008424.$$

Now multiply by the emitted frequency:

$$f_{\text{obs}} = 660\ \text{Hz}\times 1.008424 \approx 665.56\ \text{Hz}.$$

Rounding to the nearest whole number, the frequency heard by the running person is approximately $$666\ \text{Hz}$$, which matches option C.

Hence, the correct answer is Option C.

We are told that car A and car B are moving in opposite directions, each with a speed of $$20 \,\text{m s}^{-1}$$ relative to the ground. An observer sitting in car A hears a sound of frequency $$2000 \,\text{Hz}$$ coming from the horn of car B. The speed of sound in still air is given as $$340 \,\text{m s}^{-1}$$. Our goal is to determine the actual (natural) frequency $$f$$ of the horn in car B.

For sound waves, the Doppler-effect formula that links the heard (apparent) frequency $$f'$$ and the natural frequency $$f$$ is

$$ f' = f \, \dfrac{v \pm v_o}{\,v \mp v_s\,}, $$

where

• $$v$$ is the velocity of sound in the medium, here $$340 \,\text{m s}^{-1}$$,
• $$v_o$$ is the speed of the observer relative to the medium (positive when the observer moves towards the source),
• $$v_s$$ is the speed of the source relative to the medium (positive when the source moves away from the observer),
• the upper sign in the numerator is used when the observer approaches the source; the upper sign in the denominator is used when the source recedes from the observer.

In our situation both vehicles are receding from each other:

• The observer in car A is moving away from the source in car B, so $$v_o = 20 \,\text{m s}^{-1}$$ is taken with a minus sign in the numerator.
• The source in car B is moving away from the observer, so $$v_s = 20 \,\text{m s}^{-1}$$ is taken with a plus sign in the denominator.

Hence the formula becomes

$$ f' \;=\; f \,\dfrac{v - v_o}{v + v_s}. $$

We know $$f' = 2000 \,\text{Hz}$$, $$v = 340 \,\text{m s}^{-1}$$, $$v_o = 20 \,\text{m s}^{-1}$$, and $$v_s = 20 \,\text{m s}^{-1}$$. Substituting these values gives

$$ 2000 \;=\; f \,\dfrac{340 - 20}{340 + 20}. $$

Simplifying the fractions step by step, we first work out the numerator and denominator:

$$ 340 - 20 = 320, \qquad 340 + 20 = 360. $$

So we now have

$$ 2000 \;=\; f \,\dfrac{320}{360}. $$

To isolate $$f$$, multiply both sides of the equation by the reciprocal of the fraction $$\dfrac{360}{320}$$:

$$ f \;=\; 2000 \,\times \dfrac{360}{320}. $$

Carrying out the multiplication in the numerator, we get

$$ 2000 \times 360 \;=\; 720\,000. $$

Dividing by the denominator:

$$ f \;=\; \dfrac{720\,000}{320}. $$

Performing the division:

$$ f \;=\; 2250 \,\text{Hz}. $$

Thus the natural frequency of the horn in car B is $$2250 \,\text{Hz}$$.

Hence, the correct answer is Option C.

The two sources S$$_1$$ and S$$_2$$ are at rest and emit sound of the same true frequency $$f = 660\ \text{Hz}$$. The listener, however, is moving from S$$_1$$ towards S$$_2$$ with a constant speed $$u_0\ \text{m s}^{-1}$$. Because of this motion, the listener approaches S$$_2$$ and recedes from S$$_1$$ simultaneously, so the apparent (heard) frequencies of the two sources become different. Beats are produced due to the superposition of these two slightly different apparent frequencies.

First, we recall the formula for the apparent frequency when a listener moves and the source remains stationary:

$$f' = f\left(\dfrac{v \pm u}{v}\right)$$

Here $$v = 330\ \text{m s}^{-1}$$ is the speed of sound in air, and the plus sign is used when the listener moves towards the source while the minus sign is used when the listener moves away from it.

For source S$$_2$$, the listener is moving towards it, so the apparent frequency heard is

$$f_2' = f\left(\dfrac{v + u_0}{v}\right).$$

For source S$$_1$$, the listener is moving away from it, so the apparent frequency heard is

$$f_1' = f\left(\dfrac{v - u_0}{v}\right).$$

The beat frequency is the magnitude of the difference of these two apparent frequencies, that is

$$f_{\text{beat}} = \left|\,f_2' - f_1'\,\right|.$$

Substituting the expressions of $$f_2'$$ and $$f_1'$$ we get

$$\begin{aligned} f_{\text{beat}} &= \left|\,f\left(\dfrac{v + u_0}{v}\right) - f\left(\dfrac{v - u_0}{v}\right)\right| \\ &= f \left|\dfrac{(v + u_0) - (v - u_0)}{v}\right| \\ &= f \left|\dfrac{2u_0}{v}\right| \\ &= f\,\dfrac{2u_0}{v}. \end{aligned}$$

The listener is given to hear $$f_{\text{beat}} = 10\ \text{Hz}$$, so we set

$$10 = f\,\dfrac{2u_0}{v}.$$

Now we substitute $$f = 660\ \text{Hz}$$ and $$v = 330\ \text{m s}^{-1}$$:

$$10 = 660 \times \dfrac{2u_0}{330}.$$

Simplifying the right-hand side, we observe that $$\dfrac{660}{330} = 2$$, so

$$10 = 2 \times 2 u_0 = 4u_0.$$

Solving for $$u_0$$ gives

$$u_0 = \dfrac{10}{4} = 2.5\ \text{m s}^{-1}.$$

Hence, the correct answer is Option B.

We have a stationary observer, so the observer’s speed is $$v_o = 0\text{ m s}^{-1}.$$ The speed of sound in air is given as $$v = 350\text{ m s}^{-1},$$ and the source S is moving with speed $$v_s = 50\text{ m s}^{-1}.$$

The observer hears a frequency of $$1000\ \text{Hz}$$ when the source is moving towards him. For a stationary observer and a source moving towards the observer, the Doppler-effect formula is first stated:

$$f' = f \;\frac{v}{\,v - v_s\,}$$

Here $$f' = 1000\ \text{Hz}$$ is the apparent (heard) frequency, and $$f$$ is the true frequency emitted by the source.

We now solve this expression for the true frequency $$f.$$ Taking the denominator to the left side,

$$f'\,(v - v_s) = f\,v$$

Dividing both sides by $$v,$$ we obtain

$$f = f'\,\frac{v - v_s}{v}$$

Substituting the numerical values,

$$f = 1000\ \text{Hz}\;\frac{350\ \text{m s}^{-1} - 50\ \text{m s}^{-1}}{350\ \text{m s}^{-1}}$$ $$\;\;\; = 1000\ \text{Hz}\;\frac{300}{350}$$ $$\;\;\; = 1000\ \text{Hz}\;\times\;\frac{6}{7}$$ $$\;\;\; = 857.142857\ldots\ \text{Hz}$$

So, the true frequency of the source is

$$f \approx 857\ \text{Hz}.$$

After crossing the observer, the source now moves away. For a stationary observer and a source moving away, the Doppler-effect formula becomes

$$f'' = f \;\frac{v}{\,v + v_s\,}$$

where $$f''$$ is the new apparent frequency to be found.

Substituting the known values,

$$f'' = 857.142857\ \text{Hz}\;\frac{350\ \text{m s}^{-1}}{350\ \text{m s}^{-1} + 50\ \text{m s}^{-1}}$$ $$\;\;\; = 857.142857\ \text{Hz}\;\frac{350}{400}$$ $$\;\;\; = 857.142857\ \text{Hz}\;\times\;0.875$$ $$\;\;\; = 750\ \text{Hz}$$

Hence, the apparent frequency heard by the observer after the source has crossed him and is moving away is

$$f'' = 750\ \text{Hz}.$$

Hence, the correct answer is Option A.

For a sonometer wire vibrating in its fundamental mode the frequency is given by the well-known relation

$$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}$$

where $$l$$ is the vibrating length, $$T$$ is the tension and $$\mu$$ is the mass per unit length. During the experiment the wire, the tension and the material of the wire remain the same, so the factor $$\sqrt{T/\mu}/2$$ is constant. Let us denote this constant by $$k$$. Hence

$$f=\frac{k}{l}$$

We now write the frequencies of the wire for the two given lengths.

For length $$l_1=0.95\ \text{m}$$, the frequency is

$$f_1=\frac{k}{0.95}$$

For length $$l_2=1\ \text{m}$$, the frequency is

$$f_2=\frac{k}{1}=k$$

The tuning-fork of unknown frequency $$f_f$$ produces 5 beats per second with each of these wire frequencies, so

$$|f_f-f_1|=5 \qquad\text{and}\qquad |f_f-f_2|=5$$

Because $$f_1>f_2$$ (the shorter wire gives the higher note), the only possible way for both moduli to be 5 is for the fork frequency to lie between the two string frequencies, that is

$$f_1-f_f=5 \quad\text{and}\quad f_f-f_2=5$$

Adding these two equalities gives

$$f_1-f_2=10$$

Now compute the difference $$f_1-f_2$$ in terms of $$k$$:

$$f_1-f_2=\frac{k}{0.95}-k=k\!\left(\frac{1}{0.95}-1\right)=k\!\left(\frac{1-0.95}{0.95}\right)=k\!\left(\frac{0.05}{0.95}\right)=k\!\left(\frac{1}{19}\right)$$

We have just found that this difference must equal 10, so

$$k\!\left(\frac{1}{19}\right)=10 \quad\Longrightarrow\quad k=10\times19=190$$

Therefore

$$f_2=k=190\ \text{Hz} \qquad\text{and}\qquad f_1=\frac{190}{0.95}=200\ \text{Hz}$$

The fork frequency, lying midway between $$f_1$$ and $$f_2$$, is

$$f_f=\frac{f_1+f_2}{2}=\frac{200+190}{2}=195\ \text{Hz}$$

Hence, the correct answer is Option A.

Wave speed in a stretched string:

$$v=\sqrt{\frac{T}{\mu\ }}$$

Since radius is same, linear mass density:

$$\mu\ ∝\ ρ$$

Thus, $$\frac{v_1}{v_2}=\sqrt{\frac{\mu_2}{\mu_1\ }}=\sqrt{\frac{ρ_2}{ρ_1}}=\sqrt{4}=2$$

$$v_1=2\cdot v_2$$

Since the junction is a node, each segment behaves like an independent string fixed at both ends.

Number of antinodes:

$$n=\frac{2L}{λ}$$

Using $$v=fλ$$,

$$λ=\frac{v}{f}$$

$$n=\frac{2Lf}{v}$$

$$L_1=L_2\ \&\ f_1=f_2$$

Thus, the ratio of antinodes:

$$\frac{n_1}{n_2}=\frac{v_2}{v_1}=\frac{1}{2}$$

Let the true frequency of the horn of the toy-car be $$f\;{\rm Hz}$$. The speed of sound in air is given to be $$v = 340\;{\rm m\,s^{-1}}$$ and the speed of the toy-car is $$u = 5\;{\rm m\,s^{-1}}$$.

The observer receives sound by two different paths:

1. Direct sound from the horn.   The toy-car (source) is moving towards the observer.   For a stationary observer and a moving source, the Doppler-shift formula is

$$f_{\text{observed}} = \frac{v}{v - u_s}\,f,$$

where $$u_s$$ is the speed of the source towards the observer. Here $$u_s = u = 5\;{\rm m\,s^{-1}}$$, so the frequency heard directly is

$$f_1 = \frac{v}{v - u}\,f = \frac{340}{340 - 5}\,f = \frac{340}{335}\,f.$$

2. Reflected sound from the wall (echo).   First, the sound reaches the wall. The car is moving away from the wall, so for the wall (which is a stationary observer) the received frequency is

$$f_{\text{wall}} = \frac{v}{v + u}\,f = \frac{340}{340 + 5}\,f = \frac{340}{345}\,f.$$

The wall is at rest, so it reflects this sound without any further Doppler shift; the wall now behaves as a stationary source emitting frequency $$f_{\text{wall}}$$. Because the final observer is also stationary, the frequency finally heard from the echo remains

$$f_2 = f_{\text{wall}} = \frac{340}{345}\,f.$$

The observer hears beats produced by the superposition of the direct sound ($$f_1$$) and the reflected sound ($$f_2$$). The beat frequency equals the absolute difference of the two frequencies, and this is given as 5 beats per second:

$$|\,f_1 - f_2\,| = 5.$$

Explicitly substituting $$f_1$$ and $$f_2$$ we get

$$\left|\,\frac{340}{335}\,f - \frac{340}{345}\,f\,\right| = 5.$$

Pulling out the common factor $$340f$$ gives

$$340f\left|\,\frac{1}{335} - \frac{1}{345}\,\right| = 5.$$

The difference inside the modulus is

$$\frac{1}{335} - \frac{1}{345} = \frac{345 - 335}{335 \times 345} = \frac{10}{335 \times 345}.$$

Therefore,

$$340f \times \frac{10}{335 \times 345} = 5.$$

We now solve for $$f$$ step by step:

$$f = 5 \times \frac{335 \times 345}{340 \times 10}.$$

Simplifying the numerator and denominator,

$$f = 5 \times \frac{335 \times 345}{3400}.$$ $$f = \frac{5}{3400} \times 335 \times 345.$$ $$f = \frac{5 \times 335 \times 345}{3400}.$$

Take out the common factor 5 from 3400:

$$f = \frac{335 \times 345}{680}.$$

Now divide 335 by 5 to make the arithmetic easier:

$$335 = 5 \times 67,$$ so substitute:

$$f = \frac{(5 \times 67) \times 345}{680} = \frac{5 \times 67 \times 345}{5 \times 136}$$ $$f = \frac{67 \times 345}{136}.$$

Divide numerator and denominator by 17:

$$\frac{345}{17} = 20.294\ldots \quad\text{and}\quad \frac{136}{17} = 8.$$ So

$$f \approx \frac{67 \times 20.294}{8} = \frac{1359.7}{8} \approx 169.96.$$

Rounding to the nearest whole number gives

$$f \approx 170\;{\rm Hz}.$$

Among the given options, this matches Option D.

Hence, the correct answer is Option D.

We begin by recalling the Doppler effect formula for sound when both the source and the observer are in motion:

$$f' = f \left(\dfrac{v + v_o}{\,v - v_s\,}\right)$$

Here,

$$f'$$ is the frequency heard by the observer,

$$f$$ is the actual frequency emitted by the source,

$$v$$ is the speed of sound in air,

$$v_o$$ is the speed of the observer relative to the medium,

$$v_s$$ is the speed of the source relative to the medium.

When the observer and the source move toward each other, the numerator $$v + v_o$$ is used because the observer moves against the wave fronts, effectively increasing the speed of sound for him, and the denominator $$v - v_s$$ is used because the source moves toward the observer, compressing the wave fronts and thus decreasing the apparent wavelength.

Now we substitute the given data step by step. The numerical values are:

$$f = 540\ \text{Hz}$$ (whistle frequency),

$$v = 330\ \text{m s}^{-1}$$ (speed of sound),

$$v_o = 30\ \text{m s}^{-1}$$ (speed of the second engine, the observer),

$$v_s = 30\ \text{m s}^{-1}$$ (speed of the first engine, the source).

Substituting these values into the Doppler formula, we have

$$ f' \;=\; 540 \left(\dfrac{330 + 30}{330 - 30}\right). $$

First, carry out the additions and subtractions in the fraction:

$$330 + 30 = 360,$$

$$330 - 30 = 300.$$

So the fraction becomes

$$\dfrac{360}{300}.$$

Next, simplify this fraction:

$$\dfrac{360}{300} \;=\; \dfrac{36}{30} \;=\; \dfrac{6}{5} \;=\; 1.2.$$

Now multiply this factor by the original frequency:

$$ f' \;=\; 540 \times 1.2. $$

Performing the multiplication,

$$ 540 \times 1.2 = 648. $$

Hence, the frequency heard by the driver of the second engine is

$$f' = 648\ \text{Hz}.$$

Among the given options, 648 Hz corresponds to Option D.

Hence, the correct answer is Option D.

Let us denote the speed of sound in air by $$v$$, the speed of the train (which is the source of sound) by $$v_s$$, the actual frequency of the whistle by $$f$$, and the frequencies heard by the observer when the train approaches and when it recedes by $$f_1$$ and $$f_2$$ respectively.

Given data:

$$v = 320 \ {\rm m\,s^{-1}}, \qquad v_s = 20 \ {\rm m\,s^{-1}}, \qquad f = 1000 \ {\rm Hz}.$$

First we recall the Doppler effect formula for a stationary observer and a moving source. When the source approaches the observer, the apparent frequency is

$$f_1 = f \,\frac{v}{\,v - v_s\,}.$$

When the source moves away (recedes) from the observer, the apparent frequency is

$$f_2 = f \,\frac{v}{\,v + v_s\,}.$$

We now substitute the numerical values.

For the approaching train:

$$f_1 = 1000 \times \frac{320}{320 - 20} = 1000 \times \frac{320}{300} = 1000 \times 1.066666\ldots = 1066.666\ldots \ {\rm Hz}.$$

For the receding train:

$$f_2 = 1000 \times \frac{320}{320 + 20} = 1000 \times \frac{320}{340} = 1000 \times 0.941176\ldots = 941.176\ldots \ {\rm Hz}.$$

The change in the heard frequency as the train passes is

$$\Delta f = f_1 - f_2 = 1066.666\ldots - 941.176\ldots = 125.490\ldots \ {\rm Hz}.$$

To find the percentage change, we compare this change with the original (real) frequency $$f = 1000 \,{\rm Hz}.$$ Thus

$$\text{Percentage change} = \frac{\Delta f}{f}\times 100\% = \frac{125.490\ldots}{1000}\times 100\% = 12.549\ldots\%.$$

This is approximately $$12\%.$$

Hence, the correct answer is Option C.

A bat is moving towards a wall at 10 m/s and emits a sound of frequency 8000 Hz. The speed of sound is 320 m/s. We need to find the frequency $$f$$ that the bat hears after the sound reflects off the wall. This involves the Doppler effect in two steps: first, the sound travels to the wall, and then the reflected sound travels back to the bat.

In the first step, the bat (source) is moving towards the stationary wall (observer). The frequency $$f_1$$ heard by the wall is given by the Doppler effect formula for a source moving towards a stationary observer:

$$ f_1 = \left( \frac{v}{v - v_b} \right) f_0 $$

where $$v = 320$$ m/s is the speed of sound, $$v_b = 10$$ m/s is the speed of the bat, and $$f_0 = 8000$$ Hz is the emitted frequency. Substituting the values:

$$ f_1 = \left( \frac{320}{320 - 10} \right) \times 8000 = \left( \frac{320}{310} \right) \times 8000 $$

Simplify the fraction $$\frac{320}{310}$$ by dividing numerator and denominator by 10:

$$ \frac{320}{310} = \frac{32}{31} $$

So,

$$ f_1 = \frac{32}{31} \times 8000 $$

Calculate $$8000 \times 32$$:

$$ 8000 \times 32 = 256000 $$

Thus,

$$ f_1 = \frac{256000}{31} $$

In the second step, the wall reflects the sound and acts as a stationary source emitting frequency $$f_1$$. The bat is now moving towards this stationary source. The frequency $$f$$ heard by the bat (observer moving towards a stationary source) is given by:

$$ f = \left( \frac{v + v_b}{v} \right) f_1 $$

Substituting the values:

$$ f = \left( \frac{320 + 10}{320} \right) f_1 = \left( \frac{330}{320} \right) f_1 $$

Simplify $$\frac{330}{320}$$ by dividing numerator and denominator by 10:

$$ \frac{330}{320} = \frac{33}{32} $$

So,

$$ f = \frac{33}{32} \times f_1 $$

Substitute $$f_1 = \frac{256000}{31}$$:

$$ f = \frac{33}{32} \times \frac{256000}{31} $$

Notice that 32 in the denominator and 256000 in the numerator can be simplified. Since $$256000 = 8000 \times 32$$, we can write:

$$ f = \frac{33}{32} \times \frac{8000 \times 32}{31} = \frac{33}{31} \times 8000 $$

The 32 cancels out:

$$ f = \frac{33}{31} \times 8000 $$

Now, compute $$33 \times 8000$$:

$$ 33 \times 8000 = 264000 $$

So,

$$ f = \frac{264000}{31} $$

Divide 264000 by 31:

$$ 264000 \div 31 $$

31 multiplied by 8516 gives:

$$ 31 \times 8516 = 31 \times (8000 + 516) = (31 \times 8000) + (31 \times 516) = 248000 + 15996 = 263996 $$

Subtract from 264000:

$$ 264000 - 263996 = 4 $$

So,

$$ f = 8516 + \frac{4}{31} \approx 8516.129 \text{ Hz} $$

The value is close to 8516 Hz. Comparing with the options:

A. 8258

B. 8424

C. 8000

D. 8516

Hence, the correct answer is Option D.

For a source and observer moving toward each other, the observed frequency $$f$$ is given by:

$$f = f_0 \left( \frac{v + v_0}{v - v_s} \right)$$

To find the correct graph, we rewrite this expression in the linear form $$y = mx + c$$, where $$y = f$$ and $$x = v_0$$:

$$f = \left( \frac{f_0}{v - v_s} \right) v_0 + \frac{f_0 v}{v - v_s}$$

Because $$f$$ is a linear function of $$v_0$$, the graph must be a straight line with a positive intercept. This corresponds to Graph A.

$$\text{Slope} = \frac{f_0}{v - v_s}$$

Both factories emit sound at frequency $$f_0 = 800$$ Hz. The man walks from one factory toward the other at speed $$v_o = 2$$ m/s, while the velocity of sound is $$v = 320$$ m/s. Since both sources are stationary and the observer moves, the Doppler-shifted frequencies are as follows.

For the factory the man approaches, the observed frequency is $$f_1 = f_0\left(\dfrac{v + v_o}{v}\right) = 800 \times \dfrac{322}{320}$$.

For the factory the man moves away from, the observed frequency is $$f_2 = f_0\left(\dfrac{v - v_o}{v}\right) = 800 \times \dfrac{318}{320}$$.

The beat frequency is $$f_1 - f_2 = 800 \times \dfrac{322 - 318}{320} = 800 \times \dfrac{4}{320} = 800 \times \dfrac{1}{80} = 10$$ beats per second.

We are given that two harmonic waves moving in the same direction superimpose to form:

where $$t$$ is in seconds.

Step 1: Decompose using the product-to-sum identity

We use the trigonometric identity:

Applying this with $$A = 50.5t$$ and $$B = 1.5t$$:

Step 2: Identify the two component waves

The superposition consists of two harmonic waves with angular frequencies:

$$\omega_1 = 49 \text{ rad/s}$$ and $$\omega_2 = 52 \text{ rad/s}$$

Step 3: Find the beat frequency and beat period

The beat angular frequency is:

The beat period is:

Rounding to the nearest integer: $$T_{beat} \approx 2 \text{ s}$$.

Hence, the correct answer is Option D.

The standard form of a progressive sinusoidal wave travelling in the +x-direction is
$$y = A \cos \bigl(kx - \omega t + \phi\bigr)$$ or $$y = A \sin \bigl(kx - \omega t + \phi\bigr)$$

Here
  • $$A$$ = amplitude
  • $$k = \dfrac{2\pi}{\lambda}$$ = wave number (rad cm$$^{-1}$$ when $$\lambda$$ is in cm)
  • $$\omega = 2\pi f$$ = angular frequency (rad s$$^{-1}$$)
  • Wave speed $$v = \dfrac{\omega}{k} = f\lambda$$

Given data
  • Wavelength $$\lambda = 7.5\,\text{cm}$$
  • In $$\Delta t = 0.3\,\text{s}$$ the disturbance travels $$\Delta x = 1.2\,\text{cm}$$
  • Amplitude $$A = 2\,\text{cm}$$

Step 1 - Calculate the wave speed.
$$v = \frac{\Delta x}{\Delta t} = \frac{1.2\,\text{cm}}{0.3\,\text{s}} = 4\,\text{cm s}^{-1}$$

Step 2 - Calculate the wave number $$k$$.
$$k = \frac{2\pi}{\lambda} = \frac{2\pi}{7.5} \,\text{rad cm}^{-1} \approx 0.84\,\text{rad cm}^{-1}$$

Step 3 - Calculate the angular frequency $$\omega$$ using $$\omega = vk$$.
$$\omega = (4\,\text{cm s}^{-1})(0.84\,\text{rad cm}^{-1}) \approx 3.35\,\text{rad s}^{-1}$$

Step 4 - Insert $$A$$, $$k$$, and $$\omega$$ into the standard form.
Possible cosine form (with zero phase constant):
$$y = 2 \cos\bigl(0.84\,x - 3.35\,t\bigr)\;\text{cm}$$

Step 5 - Check the given initial condition.
At $$x = 0$$ and $$t = 0$$ the crest is present, so $$y(0,0) = +2\,\text{cm}$$ (maximum upward displacement).
For the expression above:
$$y(0,0) = 2\cos(0) = 2\,\text{cm}$$✅ (condition satisfied)

Step 6 - Match with the options.
Option A: $$y = 2\cos(0.83x - 3.35t)\;\text{cm}$$ - wave number, frequency and amplitude all match the calculated values.
Option B: sine form gives zero displacement at $$t = 0$$, so it violates the initial condition.
Option C: $$k$$ and $$\omega$$ are interchanged, giving an incorrect speed $$v = \omega/k \approx 0.25\,\text{cm s}^{-1}$$.
Option D: both $$k$$ and $$\omega$$ are far from the required values.

Therefore, the correct representation of the wave is given by Option A.

Let the original length of the air column in the closed organ pipe be $$L$$. The pipe has uniform cross-section, so filling it with water to $$\frac{1}{5}$$ of its volume means water occupies $$\frac{1}{5}$$ of its length as well.

Therefore, the new length of the air column is
$$L' = L - \frac{L}{5} = \frac{4L}{5}$$

For a closed organ pipe, the fundamental (first harmonic) frequency is given by the relation
$$f = \frac{v}{4L}$$
where $$v$$ is the speed of sound in air.

After adding water, the new fundamental frequency becomes
$$f' = \frac{v}{4L'} = \frac{v}{4\left(\frac{4L}{5}\right)} = \frac{5v}{16L}$$

To compare the two frequencies, form the ratio
$$\frac{f'}{f} = \frac{\frac{5v}{16L}}{\frac{v}{4L}} = \frac{5v}{16L} \times \frac{4L}{v} = \frac{5}{4} = 1.25$$

This shows the new frequency is $$1.25$$ times the original. The percentage change is
$$\left(1.25 - 1\right) \times 100\% = 0.25 \times 100\% = 25\%$$

Hence, the fundamental frequency increases by $$25\%$$.

Correct option: Option A

Let the two waves be
$$y_1(x,t)=4\sin\bigl(kx-\omega t\bigr)$$
$$y_2(x,t)=2\sin\bigl(kx-\omega t+\tfrac{2\pi}{3}\bigr)$$

Both waves have the same angular argument $$\theta=kx-\omega t$$, but different amplitudes and a phase difference $$\phi=\tfrac{2\pi}{3}$$.
Write each wave in the compact form $$A\sin\theta$$ and $$B\sin(\theta+\phi)$$, where
$$A=4,\; B=2,\; \phi=\tfrac{2\pi}{3}$$.

The superposition principle gives
$$y=y_1+y_2=A\sin\theta+B\sin(\theta+\phi)$$

For two sine functions with the same frequency, the resultant is again a sine function:
$$A\sin\theta+B\sin(\theta+\phi)=R\sin(\theta+\delta)$$
where

Amplitude formula:
$$R^2=A^2+B^2+2AB\cos\phi\quad -(1)$$

Phase formula:
$$\tan\delta=\frac{B\sin\phi}{A+B\cos\phi}\quad -(2)$$

Compute the trigonometric values of $$\phi=\tfrac{2\pi}{3}$$:
$$\cos\phi=\cos\!\left(\tfrac{2\pi}{3}\right)=-\tfrac12,\qquad\sin\phi=\sin\!\left(\tfrac{2\pi}{3}\right)=\tfrac{\sqrt3}{2}$$

Substitute into $$(1)$$:
$$\begin{aligned}R^2&=4^2+2^2+2\cdot4\cdot2\cos\!\left(\tfrac{2\pi}{3}\right)\\&=16+4+16\left(-\tfrac12\right)\\&=20-8\\&=12\end{aligned}$$
$$\Rightarrow\;R=\sqrt{12}=2\sqrt3$$

Substitute into $$(2)$$:
$$\begin{aligned}\tan\delta&=\frac{2\cdot\left(\tfrac{\sqrt3}{2}\right)}{4+2\left(-\tfrac12\right)}\\&=\frac{\sqrt3}{4-1}\\&=\frac{\sqrt3}{3}=\frac1{\sqrt3}\end{aligned}$$
$$\therefore\;\delta=\tfrac{\pi}{6}$$ since $$\tan\left(\tfrac{\pi}{6}\right)=\tfrac1{\sqrt3}$$ and $$0\lt\delta\lt\pi$$.

Hence, the resultant wave is
$$y(x,t)=2\sqrt3\;\sin\!\bigl(kx-\omega t+\tfrac{\pi}{6}\bigr)$$

Amplitude  =  $$2\sqrt3$$, Phase  =  $$\tfrac{\pi}{6}$$.

The correct choice is Option D.

Closed organ pipe of length $$L_1 = 150$$ cm = 1.5 m. Fundamental frequency: $$f_1 = \frac{v}{4L_1} = \frac{v}{6}$$.

Open organ pipe of length $$L_2 = 350$$ cm = 3.5 m. Fundamental frequency: $$f_2 = \frac{v}{2L_2} = \frac{v}{7}$$.

Number of beats = $$|f_1 - f_2| = 7$$.

$$\left|\frac{v}{6} - \frac{v}{7}\right| = 7$$

$$\frac{v}{42} = 7$$

$$v = 294$$ m/s

The answer is $$\boxed{294}$$ m/s.

We need to find the difference in frequencies between the 6th harmonic of a 60 cm open pipe and the 5th harmonic of a 90 cm open pipe. Recall that for an open organ pipe, the frequency of the $$n$$th harmonic is given by $$f_n = \frac{nv}{2L}$$ where $$v$$ is the speed of sound, $$L$$ is the length of the pipe, and $$n = 1, 2, 3, \ldots$$.

First, applying this formula to the 6th harmonic of the first pipe, with $$L_1$$ = 60 cm = 0.6 m, we obtain $$f_1 = \frac{6 \times 333}{2 \times 0.6} = \frac{1998}{1.2} = 1665 \, \text{Hz}$$.

Next, for the 5th harmonic of the second pipe, with $$L_2$$ = 90 cm = 0.9 m, we find $$f_2 = \frac{5 \times 333}{2 \times 0.9} = \frac{1665}{1.8} = 925 \, \text{Hz}$$.

Subtracting these values gives $$\Delta f = f_1 - f_2 = 1665 - 925 = 740 \, \text{Hz}$$.

The correct answer is 740 Hz.

We have a string with mass per unit length $$\mu = 7.0 \times 10^{-3}$$ kg/m under tension $$T = 70$$ N.

The speed of a transverse wave on a stretched string is given by $$v = \sqrt{\frac{T}{\mu}}$$. Substituting the given values:

$$v = \sqrt{\frac{70}{7.0 \times 10^{-3}}} = \sqrt{\frac{70}{0.007}} = \sqrt{10000} = 100 \text{ m/s}$$

So, the answer is $$100$$ m/s.

We need to find the frequency heard by a passenger inside the train.

Identify source and observer. The source of sound is the engine of the train, and the observer (passenger) is also inside the same train.

Key concept — relative motion. Both the source (engine) and the observer (passenger) are moving together at the same speed (10 m/s). There is no relative motion between them.

Apply the Doppler effect. When there is no relative motion between the source and the observer, the observed frequency equals the emitted frequency:

$$f_{\text{observed}} = f_{\text{source}} = 400 \text{ Hz}$$

Note: The Doppler effect only applies when there is relative motion between the source and the observer through the medium. Since the passenger is on the same train as the engine, they move together and hear the original frequency.

The correct answer is Option A: 400 Hz.

Solution :

For a source approaching a stationary observer, apparent frequency is given by Doppler’s formula :

$$f' = \frac{v}{v-v_s}f$$

Given :

$$f = 320\text{ Hz}$$

$$v = 330\text{ m s}^{-1}$$

$$v_s = 66\text{ m s}^{-1}$$

Substituting :

$$f' = \frac{330}{330-66}\times 320$$

$$= \frac{330}{264}\times 320$$

$$= 1.25 \times 320$$

$$= 400\text{ Hz}$$

Final Answer :

$$400$$

For a wire under tension, the fundamental frequency is:

$$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$

The tension can be found from Young's modulus and strain:

$$\text{Strain} = \frac{\Delta L}{L} = \frac{3.2 \times 10^{-4}}{0.5} = 6.4 \times 10^{-4}$$

$$\text{Stress} = Y \times \text{Strain} = 8 \times 10^{10} \times 6.4 \times 10^{-4} = 5.12 \times 10^7 \text{ N/m}^2$$

The wave velocity: $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\text{Stress}}{\rho}}$$ (since $$T = \text{Stress} \times A$$ and $$\mu = \rho \times A$$)

$$v = \sqrt{\frac{5.12 \times 10^7}{8 \times 10^3}} = \sqrt{6400} = 80 \text{ m/s}$$

$$f = \frac{v}{2L} = \frac{80}{2 \times 0.5} = 80 \text{ Hz}$$

We need to find the amplitude of the resultant wave from two interfering waves.

Simplify $$y_2$$: $$y_2 = 5[\sin(\omega t) + \sqrt{3}\cos(\omega t)]$$

Using the identity $$a\sin\theta + b\cos\theta = \sqrt{a^2 + b^2}\sin(\theta + \phi)$$ where $$\tan\phi = \frac{b}{a}$$:

$$\sin(\omega t) + \sqrt{3}\cos(\omega t) = 2\sin\left(\omega t + \frac{\pi}{3}\right)$$

(since $$\sqrt{1^2 + (\sqrt{3})^2} = 2$$ and $$\tan\phi = \sqrt{3}$$, so $$\phi = \frac{\pi}{3}$$)

$$y_2 = 5 \times 2\sin\left(\omega t + \frac{\pi}{3}\right) = 10\sin\left(\omega t + \frac{\pi}{3}\right)$$

Compare with $$y_1$$: $$y_1 = 10\sin\left(\omega t + \frac{\pi}{3}\right)$$

$$y_2 = 10\sin\left(\omega t + \frac{\pi}{3}\right)$$

Both waves are identical — same amplitude, same frequency, same phase.

Find the resultant: $$y = y_1 + y_2 = 20\sin\left(\omega t + \frac{\pi}{3}\right)$$

The amplitude of the resultant wave is 20 cm.

The wave equation is given as:

$$Y = 10^{-2}\sin 2\pi\left(160t - 0.5x + \frac{\pi}{4}\right)$$

Comparing with the standard form $$Y = A\sin(2\pi ft - 2\pi x/\lambda + \phi)$$:

$$2\pi f = 2\pi \times 160$$, so $$f = 160$$ Hz

$$\frac{2\pi}{\lambda} = 2\pi \times 0.5$$, so $$\frac{1}{\lambda} = 0.5$$ and $$\lambda = 2$$ m

Speed of the wave:

$$v = f\lambda = 160 \times 2 = 320 \text{ m/s}$$

Converting to km/h:

$$v = 320 \times \frac{3600}{1000} = 1152 \text{ km/h}$$

The speed of the wave is $$1152$$ km h$$^{-1}$$.

We need to find the velocity of a longitudinal wave in a solid rod.

Formula for velocity of longitudinal wave in a solid: $$v = \sqrt{\frac{Y}{\rho}}$$

where $$Y$$ is Young's modulus and $$\rho$$ is the density.

Substitute the values: $$v = \sqrt{\frac{3.2 \times 10^{11}}{8 \times 10^3}}$$

$$= \sqrt{4 \times 10^7}$$

$$= 2 \times 10^{3.5} = 2 \times 3162.3 \approx 6324.6 \text{ m s}^{-1}$$

$$\approx 6.32 \times 10^3 \text{ m s}^{-1}$$

The correct answer is Option D: $$6.32 \times 10^3$$ m s$$^{-1}$$.

Using the Doppler effect formula when the observer moves towards a stationary source:

$$f' = f\left(\frac{v + v_o}{v}\right)$$

where $$v$$ is the velocity of sound, $$v_o$$ is the velocity of the observer, and $$f$$ is the actual frequency.

Given that $$v_o = \frac{v}{5}$$:

$$f' = f\left(\frac{v + v/5}{v}\right) = f\left(\frac{6v/5}{v}\right) = \frac{6f}{5}$$

The percentage change in apparent frequency is:

$$\frac{f' - f}{f} \times 100 = \frac{6f/5 - f}{f} \times 100$$

$$= \frac{f/5}{f} \times 100 = \frac{1}{5} \times 100 = 20\%$$

The correct answer is Option D.

Given: Observer speed $$v_o = 18 \text{ km h}^{-1} = 5 \text{ m s}^{-1}$$ (moving towards the hill), source frequency $$f_0 = 640 \text{ Hz}$$, speed of sound $$v = 320 \text{ m s}^{-1}$$.

The observer hears two sounds: the direct sound from the source behind him, and the reflected sound from the hill ahead.

Direct sound (source behind, observer moving away from source):

$$f_1 = f_0 \left(\dfrac{v - v_o}{v}\right) = 640 \times \dfrac{320 - 5}{320} = 640 \times \dfrac{315}{320}$$

Reflected sound (hill acts as a stationary source ahead, observer moving towards it):

The hill reflects the original frequency $$f_0$$ (since the source is stationary). The observer moves toward the hill, so:

$$f_2 = f_0 \left(\dfrac{v + v_o}{v}\right) = 640 \times \dfrac{320 + 5}{320} = 640 \times \dfrac{325}{320}$$

Beat frequency:

$$f_{\text{beat}} = f_2 - f_1 = 640 \times \dfrac{325 - 315}{320} = 640 \times \dfrac{10}{320} = \dfrac{6400}{320} = 20 \text{ Hz}$$

Therefore, the beat frequency is $$\boxed{20}$$ Hz.

We are given two progressive waves travelling through the same stretched string

$$y_1 = A_1 \sin k\,(x - v t)$$

$$y_2 = A_2 \sin k\,(x - v t + x_0)$$

with the numerical data

$$A_1 = 12 \text{ mm}, \qquad A_2 = 5 \text{ mm}, \qquad x_0 = 3.5 \text{ cm}, \qquad k = 6.28 \text{ cm}^{-1}.$$

Both waves have exactly the same angular wave number $$k$$ and hence the same angular frequency; the only difference between them is a phase difference produced by the extra term $$x_0$$ inside the argument of the second sine. To combine two sine waves of the same frequency but different amplitudes and phases, we use the standard superposition formula.

First we write the second wave so that the common factor $$k(x-vt)$$ is explicit:

$$y_2 = A_2 \sin\!\bigl[k(x-vt) + kx_0 \bigr].$$

This shows that the phase difference $$\phi$$ between the two waves is simply the quantity $$k x_0$$,

$$\phi = k x_0.$$

Let us compute this phase difference numerically. Both $$k$$ and $$x_0$$ are in centimetre units, so we may multiply directly:

$$\phi = k x_0 = (6.28 \,\text{cm}^{-1})(3.5 \,\text{cm}) = 21.98 \text{ rad}.$$

The number 21.98 rad is awkward, so we reduce it with the identity $$\cos(\theta + 2\pi n) = \cos\theta$$. Observe that

$$21.98 \text{ rad} \approx 6.28 \times 3.5 = 2\pi \times 3.5 = 7\pi.$$

Hence

$$\phi \approx 7\pi \text{ rad}.$$

Because $$\cos(7\pi) = \cos(\pi \times 7) = (-1)^7 = -1,$$ the exact cosine of the phase difference is

$$\cos\phi = -1.$$

The resultant displacement $$y$$ is obtained by direct addition, $$y = y_1 + y_2$$, and it is again a sine function with the same angular factor $$k(x-vt)$$ but with a new amplitude $$A_R$$. The formula for the amplitude produced by adding two collinear simple harmonic motions of the same frequency is

$$A_R^2 = A_1^2 + A_2^2 + 2 A_1 A_2 \cos\phi.$$

We now substitute the known numbers:

$$A_R^2 = (12)^2 + (5)^2 + 2(12)(5)\cos\phi.$$

Since $$\cos\phi = -1$$, the last term becomes negative:

$$A_R^2 = 144 + 25 + 2(12)(5)(-1).$$

Compute each piece step by step:

$$144 + 25 = 169,$$

$$2(12)(5) = 120,$$

so

$$A_R^2 = 169 - 120 = 49.$$

Taking the positive square root gives the magnitude of the resultant amplitude:

$$A_R = \sqrt{49} = 7 \text{ mm}.$$

Hence, the correct answer is Option 7.

We start by recalling that a stretched wire fixed at both ends behaves like a stretched string. For such a string the resonant (natural) frequencies are given by the well-known formula

$$f_n \;=\; \frac{n\,v}{2L}, \qquad n = 1,2,3,\dots$$

Here $$f_n$$ is the $$n^{\text{th}}$$ resonant frequency, $$v$$ is the speed of transverse waves on the wire and $$L$$ is the length of the wire.

For two consecutive resonant frequencies, say $$f_n$$ and $$f_{n+1}$$, we have

$$f_{n+1} - f_n \;=\; \frac{(n+1)v}{2L} \;-\; \frac{n\,v}{2L} \;=\; \frac{v}{2L}.$$

This shows that the difference between any two successive resonant frequencies is always $$\dfrac{v}{2L}$$, independent of the harmonic number.

According to the data, the wire resonates at $$500\;\text{Hz}$$ and the next higher frequency is $$550\;\text{Hz}$$. Hence

$$f_{n+1} - f_n \;=\; 550 - 500 \;=\; 50\;\text{Hz}.$$

Using the relation for consecutive differences, we write

$$\frac{v}{2L} \;=\; 50.$$

To find $$v$$ we employ the formula for the speed of transverse waves on a stretched string:

$$v \;=\; \sqrt{\frac{T}{\mu}},$$

where $$T$$ is the tension and $$\mu$$ is the linear mass density of the wire.

The given tension is $$T = 900\;\text{N}$$ and the linear mass density is $$\mu = 9.0 \times 10^{-4}\;\text{kg m}^{-1}$$. Substituting these values, we obtain

$$v \;=\; \sqrt{\frac{900}{9.0 \times 10^{-4}}}.$$

Dividing inside the square root, we get

$$\frac{900}{9.0 \times 10^{-4}} \;=\; \frac{900}{0.0009} \;=\; 1\,000\,000.$$

Taking the square root,

$$v \;=\; \sqrt{1\,000\,000} \;=\; 1000\;\text{m s}^{-1}.$$

Now we substitute this value of $$v$$ into the difference equation $$\dfrac{v}{2L}=50$$:

$$\frac{1000}{2L} \;=\; 50.$$

Simplifying the numerator first:

$$\frac{1000}{2} \;=\; 500,$$

so the equation becomes

$$\frac{500}{L} \;=\; 50.$$

Cross-multiplying gives

$$500 \;=\; 50\,L.$$

Dividing both sides by $$50$$, we find

$$L \;=\; \frac{500}{50} \;=\; 10\;\text{m}.$$

So, the answer is $$10\;\text{m}$$.

The speed of a transverse wave in a stretched string is given by $$v = \sqrt{\frac{T}{\mu}}$$, where $$T$$ is the tension and $$\mu$$ is the linear mass density.

If the tension is increased by 4%, the new tension is $$T' = 1.04T$$. The new speed is $$v' = \sqrt{\frac{1.04T}{\mu}} = \sqrt{1.04} \cdot v$$.

Using the binomial approximation for small changes: $$\sqrt{1 + x} \approx 1 + \frac{x}{2}$$ for small $$x$$. So $$v' \approx \left(1 + \frac{0.04}{2}\right)v = 1.02v$$.

The percentage increase in speed is $$\frac{v' - v}{v} \times 100 = 0.02 \times 100 = 2\%$$.

Therefore, the percentage increase in speed is $$2$$.

The wave disturbance is given as $$y = \frac{1}{(1+x)^2}$$ at $$t = 0$$ and $$y = \frac{1}{1+(x-2)^2}$$ at $$t = 1$$ s.

Since the shape of the wave does not change during propagation, the wave profile at time $$t$$ can be written as $$y(x,t) = \frac{1}{(1+(x - vt))^2}$$, where $$v$$ is the wave speed in the positive $$x$$-direction.

At $$t = 0$$: $$y = \frac{1}{(1+x)^2}$$, which matches the given expression.

At $$t = 1$$ s: $$y = \frac{1}{(1 + (x - v \cdot 1))^2} = \frac{1}{(1 + (x - v))^2}$$.

Comparing with the given expression $$y = \frac{1}{1+(x-2)^2}$$, we identify $$v = 2$$ m/s.

Therefore, the velocity of the wave is $$\boxed{2}$$ m s$$^{-1}$$.

The speed of sound in a gas is $$v = \sqrt{\frac{B}{\rho}}$$, where $$B$$ is the bulk modulus (inverse of compressibility) and $$\rho$$ is the density. Since compressibility is the same for both gases, $$B$$ is the same for both pipes.

For the closed organ pipe of length $$L$$, the first overtone (third harmonic) has frequency $$f_1 = \frac{3v_1}{4L}$$, where $$v_1 = \sqrt{\frac{B}{\rho_1}}$$.

For the open organ pipe of length $$L'$$, the first overtone (second harmonic) has frequency $$f_2 = \frac{2v_2}{2L'} = \frac{v_2}{L'}$$, where $$v_2 = \sqrt{\frac{B}{\rho_2}}$$.

Since both vibrate at the same frequency: $$\frac{3v_1}{4L} = \frac{v_2}{L'}$$.

Solving for $$L'$$: $$L' = \frac{4Lv_2}{3v_1} = \frac{4L}{3} \cdot \frac{\sqrt{B/\rho_2}}{\sqrt{B/\rho_1}} = \frac{4L}{3}\sqrt{\frac{\rho_1}{\rho_2}}$$.

Comparing with the given expression $$\frac{x}{3}L\sqrt{\frac{\rho_1}{\rho_2}}$$, we get $$x = 4$$.

The value of $$x$$ is $$\boxed{4}$$.

Let the speed of the car be $$v$$ m/s and the speed of sound be $$v_s = 330$$ m/s. The car horn has a natural frequency $$f_0 = 400$$ Hz.

When the car approaches a stationary vertical wall, the wall acts as a stationary observer first and then as a stationary source. From the perspective of the wall (observer), the frequency of sound received is $$f_1 = f_0 \frac{v_s}{v_s - v}$$.

The wall then reflects this frequency. Now the car acts as an observer moving towards the wall (the secondary source), so the frequency heard by the driver from the reflection is:

This reflected frequency is $$f' = 500$$ Hz. Substituting:

Converting to km/h: $$v = \frac{110}{3} \times \frac{3600}{1000} = \frac{110 \times 3600}{3000} = \frac{110 \times 6}{5} = 132$$ km h$$^{-1}$$.

The speed of the car is $$\boxed{132}$$ km h$$^{-1}$$.

We begin by recalling the basic formula for the speed of a transverse wave on a stretched string or wire. The speed $$v$$ is given by

$$v=\sqrt{\frac{T}{\mu}}$$

where $$T$$ is the tension in the wire and $$\mu$$ is the linear mass density (mass per unit length) of the wire. Because the physical wire does not change, the value of $$\mu$$ remains the same in every part of the problem.

First, let us write the expression for the initial situation. When the tension is

$$T_1 = 2.06 \times 10^{4}\ \text{N},$$

the corresponding speed is given to be simply $$v$$, so

$$v = \sqrt{\frac{T_1}{\mu}}.$$

Now the tension is altered to a new value $$T_2$$, and under this new tension the speed is reported to become $$\dfrac{v}{2}$$. Therefore we have

$$\frac{v}{2} = \sqrt{\frac{T_2}{\mu}}.$$

Because the same wire is used, $$\mu$$ is identical in both expressions. We can now form a ratio of the two speed formulas in order to eliminate $$\mu$$:

$$\frac{\dfrac{v}{2}}{v} = \frac{\sqrt{\dfrac{T_2}{\mu}}}{\sqrt{\dfrac{T_1}{\mu}}}.$$

Simplifying the left-hand side first, we see that

$$\frac{\dfrac{v}{2}}{v} = \frac{1}{2}.$$

On the right-hand side, the square roots have the same denominator $$\sqrt{\mu}$$, which cancels. We are left with

$$\frac{1}{2} = \sqrt{\frac{T_2}{T_1}}.$$

To remove the square root, we square both sides:

$$\left(\frac{1}{2}\right)^2 = \left(\sqrt{\frac{T_2}{T_1}}\right)^2.$$

Hence,

$$\frac{1}{4} = \frac{T_2}{T_1}.$$

Now we solve for $$T_2$$ by multiplying both sides by $$T_1$$:

$$T_2 = \frac{T_1}{4}.$$

We substitute the numerical value of $$T_1$$:

$$T_2 = \frac{2.06 \times 10^{4}\ \text{N}}{4}.$$

Carrying out the division, we write $$2.06$$ as $$2.06 = 2.060$$ so that the arithmetic is transparent:

$$T_2 = 0.515 \times 10^{4}\ \text{N}.$$

Finally, we shift the decimal point one place to convert $$0.515 \times 10^{4}$$ into standard scientific notation:

$$T_2 = 5.15 \times 10^{3}\ \text{N}.$$

Therefore, the tension that gives a wave speed of $$\dfrac{v}{2}$$ is approximately $$5.15 \times 10^{3}\ \text{N}$$.

Hence, the correct answer is Option B.

We recall the standard relation for a plane progressive sound wave that connects the pressure amplitude with the displacement amplitude:

$$\Delta p \;=\; \rho\,v\,\omega\,s,$$

where $$\rho$$ is the density of air, $$v$$ is the speed of sound, $$\omega$$ is the angular frequency, and $$s$$ is the displacement amplitude. The angular frequency is linked to the ordinary frequency by the well-known identity $$\omega = 2\pi f.$$

The problem statement tells us to “take the multiplicative constant to be 1,” which means we may drop the factor $$2\pi$$. Accordingly we approximate

$$\omega \approx f,$$

and the above formula simplifies to

$$\Delta p = \rho\,v\,f\,s.$$

We now isolate $$s$$:

$$s = \frac{\Delta p}{\rho\,v\,f}.$$

All the required quantities are supplied:

$$\Delta p = 10\;\text{Pa}, \quad \rho = 1\;\text{kg/m}^3, \quad v = 300\;\text{m/s}, \quad f = 1000\;\text{Hz}.$$

Substituting each value into the expression for $$s$$ we have

$$s = \frac{10}{\bigl(1\bigr)\bigl(300\bigr)\bigl(1000\bigr)} = \frac{10}{300\,000} = \frac{1}{30\,000}\;\text{m}.$$

To render this in more familiar units, we convert metres to millimetres. Using $$1\;\text{m} = 1000\;\text{mm},$$ we get

$$s = \frac{1}{30\,000}\;\text{m}\;\times\;1000\;\frac{\text{mm}}{\text{m}} = \frac{1000}{30\,000}\;\text{mm} = \frac{1}{30}\;\text{mm} \approx 0.033\;\text{mm}.$$

The numerical value $$0.033\;\text{mm}$$ can also be written as $$\dfrac{3}{100}\,\text{mm}.$$ Among the given options, this precisely matches option A.

Hence, the correct answer is Option A.

We begin by noting that a stretched string fixed at both ends can vibrate in normal modes whose frequencies are given by the well-known formula

$$f_n=\dfrac{n v}{2L},$$

where $$f_n$$ is the frequency of the $$n^{\text{th}}$$ harmonic, $$v$$ is the speed of transverse waves on the string, and $$L$$ is the length of the string.

Before using the formula, we must first find the wave speed $$v$$ on the string. The speed of a wave on a stretched string is determined by the tension $$T$$ and the mass per unit length $$\mu$$ through the relation

$$v=\sqrt{\dfrac{T}{\mu}}.$$

We are given

$$T=540\ \text{N}, \qquad \mu = 6.0\times10^{-3}\ \text{kg m}^{-1}.$$

Substituting these values, we have

$$v=\sqrt{\dfrac{540}{6.0\times10^{-3}}}.$$

Inside the square root, the denominator is converted to a simpler decimal:

$$6.0\times10^{-3}=0.006.$$

So,

$$v=\sqrt{\dfrac{540}{0.006}}.$$

Carrying out the division first,

$$\dfrac{540}{0.006}=90\,000.$$

Taking the square root gives

$$v=\sqrt{90\,000}=300\ \text{m s}^{-1}.$$

Now, we are told that two consecutive resonant frequencies are $$420\ \text{Hz}$$ and $$490\ \text{Hz}$$. Let the lower of these, $$420\ \text{Hz}$$, correspond to harmonic number $$n$$. Then the higher one, $$490\ \text{Hz}$$, must correspond to harmonic number $$n+1$$. Using the formula for successive harmonics, we may write

$$f_{n+1}-f_n=\dfrac{v}{2L}.$$

The left-hand side is simply the numerical difference of the two given frequencies:

$$f_{n+1}-f_n = 490-420 = 70\ \text{Hz}.$$

Therefore,

$$70 = \dfrac{v}{2L}.$$

We have already found $$v=300\ \text{m s}^{-1}$$, so we substitute this value:

$$70 = \dfrac{300}{2L}.$$

Simplify the fraction in the numerator first:

$$\dfrac{300}{2L} = \dfrac{150}{L}.$$

Hence the equation becomes

$$70 = \dfrac{150}{L}.$$

To isolate $$L$$, we cross-multiply:

$$70L = 150.$$

Now, solve for $$L$$ by dividing both sides by 70:

$$L = \dfrac{150}{70}.$$

Reduce the fraction by dividing numerator and denominator by 10:

$$L = \dfrac{15}{7}\ \text{m}.$$

Carrying out the division for a decimal value,

$$L \approx 2.142857\ \text{m}.$$

Among the given choices, the closest value is 2.1 m, which corresponds to Option A.

Hence, the correct answer is Option A.

We consider the resonance tube as a pipe that is closed at the water surface and open at the top. The air column above the water is therefore an open-closed column. For such a column the normal-mode formula is stated first:

For the first resonance (the fundamental) the length of the vibrating air column is

$$L_1 \;=\;\frac{\lambda}{4}\;+\;e,$$

where $$\lambda$$ is the wavelength of the sound and $$e$$ is the so-called end correction that accounts for the fact that the pressure node is a little outside the physical open end.

For the very next resonance in an open-closed column the mode number jumps from 1 to 3, so the length of the air column becomes

$$L_2 \;=\;\frac{3\lambda}{4}\;+\;e.$$

Now we turn to the data of the experiment. The water is initially $$17.0\,\text{cm}$$ high, so let us denote that height by $$h_1=17.0\ \text{cm}$$. The corresponding air-column length is the full length of the tube (call it $$L_0$$) minus this water height: $$L_1 = L_0 - h_1.$$

When the water is raised, its new height is $$h_2 = 24.5\ \text{cm}$$ and the new air-column length is $$L_2 = L_0 - h_2.$$

Taking the difference of the two air lengths we have

$$L_1 - L_2 \;=\;(L_0 - h_1)\;-\;(L_0 - h_2)\;=\;h_2 - h_1.$$

Substituting the numerical values,

$$h_2 - h_1 \;=\;24.5\ \text{cm} - 17.0\ \text{cm} \;=\;7.5\ \text{cm}.$$

On the other hand, subtracting the modal expressions gives

$$L_1 - L_2 \;=\;\Bigl(\frac{\lambda}{4}+e\Bigr) - \Bigl(\frac{3\lambda}{4}+e\Bigr) \;=\;-\frac{\lambda}{2}.$$

The end correction $$e$$ cancels out completely. We take the magnitude, so

$$\frac{\lambda}{2} \;=\;7.5\ \text{cm} \;=\;0.075\ \text{m}.$$

Hence,

$$\lambda \;=\;2 \times 0.075\ \text{m} \;=\;0.15\ \text{m}.$$

The speed-frequency-wavelength relation is stated next:

$$v \;=\;f\,\lambda.$$

We know the velocity of sound in air is given as $$v = 330\ \text{m s}^{-1}.$$ Substituting $$v$$ and $$\lambda$$ we solve for the frequency $$f$$:

$$f \;=\;\frac{v}{\lambda} \;=\;\frac{330\ \text{m s}^{-1}}{0.15\ \text{m}} \;=\; \frac{330}{0.15}\ \text{s}^{-1}.$$

To carry out the division step by step, observe

$$0.15 = \frac{15}{100}, \quad \frac{330}{\frac{15}{100}} = \frac{330 \times 100}{15} = \frac{33000}{15} = 2200.$$

So,

$$f = 2200\ \text{Hz}.$$

Hence, the correct answer is Option A.

We are told that both strings X and Z are identical, which means they have the same length $$L$$, the same material, and therefore the same linear mass density $$\mu$$. For a stretched string the fundamental (first-harmonic) frequency is given by the standard formula

$$f \;=\;\dfrac{1}{2L}\;\sqrt{\dfrac{T}{\mu}}$$

Here $$f$$ is the fundamental frequency, $$T$$ is the tension in the string and $$\mu$$ is its mass per unit length. Because the strings are identical, the quantities $$L$$ and $$\mu$$ are the same for both strings. Let us write the formula separately for strings X and Z.

For string X we have

$$f_X \;=\;\dfrac{1}{2L}\;\sqrt{\dfrac{T_X}{\mu}}$$

For string Z we have

$$f_Z \;=\;\dfrac{1}{2L}\;\sqrt{\dfrac{T_Z}{\mu}}$$

Now we take the ratio of the two frequencies. Dividing the first equation by the second, the common factor $$1/(2L)$$ cancels out immediately:

$$\dfrac{f_X}{f_Z}\;=\;\dfrac{\sqrt{T_X/\mu}}{\sqrt{T_Z/\mu}} =\sqrt{\dfrac{T_X}{T_Z}}$$

We want the ratio $$T_X/T_Z$$, so we square both sides of the last equation:

$$(\dfrac{f_X}{f_Z})^{2} \;=\;\dfrac{T_X}{T_Z}$$

Now we substitute the given numerical values. The fundamental frequency of X is 450 Hz and that of Z is 300 Hz, hence

$$\dfrac{f_X}{f_Z}\;=\;\dfrac{450}{300}\;=\;1.5$$

Substituting this ratio into the squared relation, we get

$$\dfrac{T_X}{T_Z}\;=\;(1.5)^{2}$$

Evaluating the square,

$$\dfrac{T_X}{T_Z}\;=\;2.25$$

Hence, the correct answer is Option A.

We begin with the well-known formula for the speed of sound in an ideal gas, $$v=\sqrt{\dfrac{\gamma P}{\rho}}$$, where $$\gamma$$ is the ratio of specific heats, $$P$$ is the pressure and $$\rho$$ is the density. For two gases at the same pressure and temperature having the same $$\gamma$$, the speed of sound varies inversely as the square root of the density, so $$v\propto\dfrac1{\sqrt{\rho}}$$.

We are told that the given gas has double the density of air at STP. Hence, if the speed of sound in air is $$300\ \text{m s}^{-1}$$, then in the denser gas we have

$$v_{\text{gas}} = \dfrac{300}{\sqrt{2}}\ \text{m s}^{-1} = 300 \times \dfrac{1}{1.414}\ \text{m s}^{-1} = 212.1\ \text{m s}^{-1}\ (\text{approximately}).$$

Now consider the organ pipe. Because both ends are open, the standing-wave pattern has antinodes at each end. For an open-open pipe of length $$L$$, the allowed wavelengths are given by

$$\lambda_n = \dfrac{2L}{n},\qquad n = 1,2,3,\ldots$$

Therefore, the corresponding frequencies are

$$f_n = \dfrac{v}{\lambda_n} = \dfrac{v}{2L}\,n.$$

For the fundamental (first harmonic) we set $$n=1$$:

$$f_1 = \dfrac{v}{2L}.$$

For the second harmonic we set $$n=2$$:

$$f_2 = \dfrac{2v}{2L} = \dfrac{v}{L}.$$

The question asks for the difference between these two frequencies. Substituting the expressions we just obtained, we find

$$f_2 - f_1 = \dfrac{v}{L} - \dfrac{v}{2L} = \dfrac{v}{2L}.$$

The pipe length is given as $$L = 1\ \text{m}$$, so

$$f_2 - f_1 = \dfrac{v}{2(1)} = \dfrac{v}{2}.$$

Finally, insert the speed of sound in the gas, $$v = 212.1\ \text{m s}^{-1}$$:

$$f_2 - f_1 = \dfrac{212.1}{2}\ \text{Hz} \approx 106\ \text{Hz}.$$

Hence, the correct answer is Option 106 Hz.

We have a stretched wire of length $$L = 1\;{\rm m}$$. Our aim is to find the lowest (fundamental) frequency of the transverse vibrations produced in this wire.

First, let us convert the given density into SI units. The density is provided as $$\rho = 9 \times 10^{-3}\;{\rm kg\;cm^{-3}}$$. We know the conversion factor

$$1\;{\rm cm^3} = 10^{-6}\;{\rm m^3}.$$

Therefore,

$$\rho = 9 \times 10^{-3}\;{\rm kg\;cm^{-3}} = 9 \times 10^{-3}\;{\rm kg}\,\bigl(10^{6}\;{\rm m^{-3}}\bigr) = 9 \times 10^{3}\;{\rm kg\;m^{-3}}.$$

Next, the strain in the wire is given as $$\varepsilon = 4.9 \times 10^{-4}.$$ By definition,

$$\text{Stress} \; \sigma = Y \times \text{strain},$$

where $$Y = 9 \times 10^{10}\;{\rm N\,m^{-2}}$$ is Young’s modulus. Substituting the values, we get

$$\sigma = (9 \times 10^{10}) \times (4.9 \times 10^{-4}) = 9 \times 4.9 \times 10^{10-4} = 44.1 \times 10^{6} = 4.41 \times 10^{7}\;{\rm N\,m^{-2}}.$$

In a stretched wire, the tension $$T$$ is related to the stress $$\sigma$$ and the cross-sectional area $$A$$ by

$$T = \sigma A.$$

The mass per unit length $$\mu$$ of the same wire is

$$\mu = \rho A.$$

Now, the speed $$v$$ of transverse waves on a stretched string is given by the standard formula

$$v = \sqrt{\frac{T}{\mu}}.$$

Substituting $$T=\sigma A$$ and $$\mu=\rho A$$ into this expression, we find

$$v = \sqrt{\frac{\sigma A}{\rho A}} = \sqrt{\frac{\sigma}{\rho}}.$$

Notice that the area $$A$$ cancels out, which means we do not need to know the actual thickness of the wire. Putting the numerical values,

$$v = \sqrt{\frac{4.41 \times 10^{7}}{9 \times 10^{3}}} = \sqrt{\,\bigl(4.41/9\bigr) \times 10^{4}} = \sqrt{0.49 \times 10^{4}} = \sqrt{4900} = 70\;{\rm m\,s^{-1}}.$$

The fundamental (lowest) mode of vibration of a stretched string fixed at both ends has a wavelength equal to twice the length of the string, i.e.

$$\lambda_1 = 2L.$$

Its frequency is therefore

$$f_1 = \frac{v}{\lambda_1} = \frac{v}{2L}.$$

Substituting $$v = 70\;{\rm m\,s^{-1}}$$ and $$L = 1\;{\rm m}$$, we get

$$f_1 = \frac{70}{2 \times 1} = \frac{70}{2} = 35\;{\rm Hz}.$$

So, the answer is $$35$$.

First of all we note that the speed of a transverse wave on a stretched wire is governed by the well-known relation

$$v=\sqrt{\dfrac{T}{\mu}}$$

where $$v$$ is the speed of the wave, $$T$$ is the tension in the wire and $$\mu$$ is the mass per unit length.

We have the total mass of the wire $$m=6.0\ \text{g}=6.0\times10^{-3}\ \text{kg}$$ and its original length $$L=60\ \text{cm}=0.60\ \text{m}$$. Hence the mass per unit length is

$$\mu=\dfrac{m}{L}= \dfrac{6.0\times10^{-3}\ \text{kg}}{0.60\ \text{m}}=1.0\times10^{-2}\ \text{kg m}^{-1}.$$

The wave speed is given as $$v=90\ \text{m s}^{-1}$$. Substituting $$v$$ and $$\mu$$ in the formula and solving for the tension, we obtain

$$T=\mu v^{2}=\left(1.0\times10^{-2}\ \text{kg m}^{-1}\right)\left(90\ \text{m s}^{-1}\right)^{2}.$$

Calculating the square first, $$90^{2}=8100$$, and then multiplying,

$$T = 1.0\times10^{-2}\times 8100\ \text{N}=81\ \text{N}.$$

To find the extension produced by this tension we invoke Hooke’s law for a uniform wire. The longitudinal strain is related to stress and Young’s modulus by

$$\frac{\Delta L}{L} = \frac{\text{Stress}}{Y} = \frac{T/A}{Y},$$

where $$A$$ is the cross-sectional area and $$Y$$ is Young’s modulus. Rearranging, the extension $$\Delta L$$ becomes

$$\Delta L=\frac{T\,L}{A\,Y}.$$

The cross-sectional area is given as $$1.0\ \text{mm}^{2}$$. In SI units this is

$$A = 1.0\ \text{mm}^{2}=1.0\times10^{-6}\ \text{m}^{2}.$$

The Young’s modulus is supplied as $$Y = 16\times10^{11}\ \text{N m}^{-2}.$$

Substituting every quantity into the extension formula, we get

$$\Delta L = \frac{(81\ \text{N})\,(0.60\ \text{m})}{\left(1.0\times10^{-6}\ \text{m}^{2}\right)\,\left(16\times10^{11}\ \text{N m}^{-2}\right)}.$$

Multiplying the numerator,

$$T\,L = 81 \times 0.60 = 48.6\ \text{N m}.$$

Working out the denominator, notice that

$$A\,Y = \left(1.0\times10^{-6}\right)\left(16\times10^{11}\right)=16\times10^{5}=1.6\times10^{6}.$$

Thus,

$$\Delta L = \frac{48.6}{1.6\times10^{6}}\ \text{m}.$$

Carrying out the division,

$$\Delta L = 3.0375\times10^{-5}\ \text{m}.$$

To express this result in millimetres we recall that $$1\ \text{mm}=10^{-3}\ \text{m}$$, so

$$\Delta L = \dfrac{3.0375\times10^{-5}\ \text{m}}{10^{-3}\ \text{m/mm}}=3.0375\times10^{-2}\ \text{mm}\approx 0.03\ \text{mm}.$$

Hence, the correct answer is Option A.

We are given a uniform rope of length $$L = 12$$ m and mass $$M = 6$$ kg hanging vertically from a rigid support. A block of mass $$m = 2$$ kg is attached to its free (lower) end. A transverse wave of wavelength $$\lambda_{\text{bottom}} = 6$$ cm is produced at the lower end. We need to find the wavelength when the wave reaches the top.

The linear mass density of the rope is $$\mu = \frac{M}{L} = \frac{6}{12} = 0.5$$ kg/m.

The speed of a transverse wave on a string is given by $$v = \sqrt{\frac{T}{\mu}}$$, where $$T$$ is the tension at that point.

At the bottom of the rope, the tension is due to the block only: $$T_{\text{bottom}} = mg = 2g$$.

At the top of the rope, the tension supports both the rope and the block: $$T_{\text{top}} = (M + m)g = (6 + 2)g = 8g$$.

Since the frequency of a wave remains constant as it travels through the medium, we have $$f = \frac{v}{\lambda}$$ = constant. Therefore $$\lambda \propto v \propto \sqrt{T}$$.

Taking the ratio: $$\frac{\lambda_{\text{top}}}{\lambda_{\text{bottom}}} = \sqrt{\frac{T_{\text{top}}}{T_{\text{bottom}}}} = \sqrt{\frac{8g}{2g}} = \sqrt{4} = 2$$.

So $$\lambda_{\text{top}} = 2 \times 6 = 12$$ cm.

The wavelength of the wave train when it reaches the top of the rope is $$12$$ cm, which is Option C.

We are given the displacement of the wave as $$y(x,t)=10^{-3}\,\sin(50t+2x)$$ where $$x$$ and $$y$$ are in metres and $$t$$ is in seconds.

The standard mathematical form for a one-dimensional harmonic wave of angular frequency $$\omega$$, wave number $$k$$ and amplitude $$A$$ is

$$y(x,t)=A\,\sin(\omega t-kx) \qquad\text{(wave travelling towards +x)},$$

or

$$y(x,t)=A\,\sin(\omega t+kx) \qquad\text{(wave travelling towards -x)}.$$

We now compare the argument of the sine function in the given expression with these standard forms. The given phase is

$$\phi = 50t + 2x.$$

Because the sign in front of the $$x$$-term is positive, the expression matches $$\sin(\omega t + kx)$$. Therefore the wave is propagating along the negative $$x$$-axis.

Next, we identify the angular frequency $$\omega$$ and the wave number $$k$$ directly from the coefficients:

Comparing $$\phi = \omega t + kx$$ with $$\phi = 50t + 2x$$, we have

$$\omega = 50 \text{ s}^{-1},\qquad k = 2 \text{ m}^{-1}.$$

The speed $$v$$ of a harmonic wave is obtained from the relation

$$v = \frac{\omega}{k}.$$

Substituting the identified values, we get

$$v = \frac{50}{2} = 25 \text{ m s}^{-1}.$$

So, the wave travels with a speed of $$25 \text{ m s}^{-1}$$, and its direction is towards the negative $$x$$-axis.

Hence, the correct answer is Option C.

Based on the wave equation and the provided snapshot at t = 0, the value of the phase constant $$\phi$$ is determined as follows:

1. Analysis of the Wave Equation

The general equation for the progressive wave is:

$$y(x, t) = A \sin(kx - \omega t + \phi)$$

At time t = 0, the equation simplifies to:

$$y(x, 0) = A \sin(kx + \phi)$$

2. Comparison with the Snapshot

According to the graph provided for $t = 0$, the wave starts from the origin and moves in the negative y-direction as x increases. This indicates that the graph represents a negative sine wave:

$$y = -A \sin(kx)$$

3. Solving for the Phase Constant ($$\phi$$)

To find $$\phi$$, we equate the two expressions for $$y(x, 0)$$:

$$A \sin(kx + \phi) = -A \sin(kx)$$

Using the trigonometric identity $$\sin(\theta + \pi) = -\sin \theta$$:

$$A \sin(kx + \phi) = A \sin(kx + \pi)$$

By comparing the arguments of the sine functions:

$$\phi = \pi$$

Conclusion:

The phase constant $\phi$ for the given wave snapshot is:

$$\boxed{\phi = \pi}$$

For a resonance tube closed at the water surface and open at the top, the first (fundamental) resonance condition is stated by the formula

$$L \;=\;\frac{\lambda}{4},$$

where $$L$$ is the effective length of the vibrating air column and $$\lambda$$ is the wavelength of the sound in air. The effective length is always a little greater than the measured air-column length because of the end correction at the open end.

Let us denote by $$e$$ the end correction produced by the open end and by $$d$$ the unknown extra length between the jagged tip of the tube and the clearly visible reference mark. Both $$e$$ and $$d$$ are constant for every observation on this particular tube.

When the 512 Hz fork is used, the water is 11 cm below the reference mark. The measured length of the air column is therefore 11 cm, so the effective length becomes

$$L_1 \;=\;11\;\text{cm}+d+e.$$

When the 256 Hz fork is used, the water is 27 cm below the reference mark, giving

$$L_2 \;=\;27\;\text{cm}+d+e.$$

For each fork the fundamental condition $$L=\lambda/4$$ must be satisfied. We use the relation $$\lambda=v/f,$$ where $$v$$ is the velocity of sound and $$f$$ is the frequency. Substituting for each fork we have

$$11+d+e \;=\;\frac{v}{4\,f_1},\qquad f_1=512\;\text{Hz},$$

$$27+d+e \;=\;\frac{v}{4\,f_2},\qquad f_2=256\;\text{Hz}.$$

It is convenient to combine the two unknown constants $$d$$ and $$e$$ into a single constant

$$a \;=\;d+e.$$

Writing the two equations with this shorthand gives

$$11+a \;=\;\frac{v}{4\times512},$$

$$27+a \;=\;\frac{v}{4\times256}.$$

From the first equation we have

$$v \;=\;4\times512\,(11+a)\;=\;2048\,(11+a).$$

From the second equation we have

$$v \;=\;4\times256\,(27+a)\;=\;1024\,(27+a).$$

Because both right-hand sides equal the same velocity $$v$$, we equate them:

$$2048\,(11+a) \;=\;1024\,(27+a).$$

Dividing every term by 1024 simplifies the equation to

$$2\,(11+a) \;=\;27+a.$$

Now expanding and gathering like terms,

$$22+2a \;=\;27+a,$$

$$2a-a \;=\;27-22,$$

$$a \;=\;5\;\text{cm}.$$

Substituting $$a=5\;\text{cm}$$ into the first velocity expression gives

$$v \;=\;2048\,(11+5)\;=\;2048\times16\;\text{cm s}^{-1}.$$

Calculating the product,

$$2048\times16=32768\;\text{cm s}^{-1}.$$

Converting centimetres per second to metres per second (1 m = 100 cm),

$$v \;=\;\frac{32768}{100}\;\text{m s}^{-1}\;=\;327.68\;\text{m s}^{-1}.$$

Among the given options, 328 m s−1 is the closest to this calculated value.

Hence, the correct answer is Option D.

We first note that a transverse wave on a stretched string travels with a speed given by the formula

$$v=\sqrt{\dfrac{T}{\mu}},$$

where $$T$$ is the tension in the string and $$\mu$$ is the linear mass density.

The string has length $$L = 1\ \text{m}$$ and mass $$m = 5\ \text{g} = 0.005\ \text{kg}$$, so

$$\mu=\dfrac{m}{L}=\dfrac{0.005\ \text{kg}}{1\ \text{m}} = 0.005\ \text{kg\,m}^{-1}.$$

Substituting $$T = 8\ \text{N}$$ and $$\mu = 0.005\ \text{kg\,m}^{-1}$$ into the speed formula, we have

$$v=\sqrt{\dfrac{8}{0.005}}=\sqrt{1600}=40\ \text{m\,s}^{-1}.$$

A string fixed at both ends supports standing waves whose natural frequencies are

$$f_n=\dfrac{n v}{2L},\qquad n=1,2,3,\ldots$$

An external vibrator of frequency $$f = 100\ \text{Hz}$$ is applied. Setting this equal to one of the natural frequencies, we get

$$100 = \dfrac{n\,(40)}{2\,(1)} = 20n \quad\Longrightarrow\quad n = 5.$$

For the $$n^{\text{th}}$$ harmonic, the wavelength is

$$\lambda_n=\dfrac{2L}{n} = \dfrac{2\,(1\ \text{m})}{5}=0.4\ \text{m}=40\ \text{cm}.$$

The distance between successive nodes on a standing wave is one-half of the wavelength, i.e.

$$\text{node separation}=\dfrac{\lambda_n}{2}=\dfrac{0.4\ \text{m}}{2}=0.2\ \text{m}=20\ \text{cm}.$$

Hence, the correct answer is Option A.

We have a composite stretched wire of total length $$2L$$. It is prepared by joining two separate wires, called A and B, each of length $$L$$. Both wires are made of the same material and hence have the same mass-density (per unit volume) and are under the same tensile force $$T$$ when the whole wire is set up on a sonometer (or any similar frame).

Let the radius of wire A be $$r$$ and that of wire B be $$2r$$. Because cross-sectional area of a wire is $$\pi r^{2}$$, the linear mass density (mass per unit length)

$$\mu=\rho\,\pi r^{2},$$

where $$\rho$$ is the material density (same for both). So

$$\mu_{A}=\rho\pi r^{2},$$

$$\mu_{B}=\rho\pi(2r)^{2}=4\rho\pi r^{2}=4\mu_{A}.$$

The speed of a transverse wave on a stretched wire is given first:

$$v=\sqrt{\dfrac{T}{\mu}}.$$

Substituting the two values of $$\mu$$, we obtain

$$v_{A}= \sqrt{\dfrac{T}{\mu_{A}}},$$

$$v_{B}= \sqrt{\dfrac{T}{\mu_{B}}}= \sqrt{\dfrac{T}{4\mu_{A}}}= \dfrac{1}{2}\sqrt{\dfrac{T}{\mu_{A}}}= \dfrac{v_{A}}{2}.$$

Thus $$v_{B}= \dfrac{v_{A}}{2}.$$

The joint between the two wires acts as a node (displacement is zero there). The outer ends are fixed as well, so each individual piece behaves exactly like an independent string with both ends fixed: one end at the outer support and the other at the junction node.

If a string of length $$L$$ with both ends fixed vibrates in its $$n^{\text{th}}$$ harmonic, its length equals $$n$$ half-wavelengths:

$$L = n\left(\dfrac{\lambda}{2}\right).$$

The number of antinodes (loops) present is the same integer $$n$$. Therefore,

For wire A: $$\displaystyle L = p\left(\dfrac{\lambda_{A}}{2}\right) \;\Longrightarrow\; \lambda_{A}= \dfrac{2L}{p}.$$

For wire B: $$\displaystyle L = q\left(\dfrac{\lambda_{B}}{2}\right) \;\Longrightarrow\; \lambda_{B}= \dfrac{2L}{q}.$$

Because the two portions form a single continuous system, the frequency of vibration is the same in both parts. Using $$f=\dfrac{v}{\lambda}$$, we write

$$f = \dfrac{v_{A}}{\lambda_{A}} = \dfrac{v_{B}}{\lambda_{B}}.$$

Substituting the expressions for $$\lambda_{A}$$ and $$\lambda_{B}$$, we get

$$\dfrac{v_{A}}{\,\dfrac{2L}{p}\,}= \dfrac{v_{B}}{\,\dfrac{2L}{q}\,}.$$

Simplifying the denominators first:

$$\dfrac{p\,v_{A}}{2L}= \dfrac{q\,v_{B}}{2L}.$$

The factors $$2L$$ cancel out, leaving

$$p\,v_{A}= q\,v_{B}.$$

We already have $$v_{B}= \dfrac{v_{A}}{2}$$, so substituting this value:

$$p\,v_{A}= q\left(\dfrac{v_{A}}{2}\right).$$

Dividing both sides by $$v_{A}$$ (non-zero), we obtain

$$p= \dfrac{q}{2}.$$

Re-writing,

$$\dfrac{p}{q}= \dfrac{1}{2} \;\Longrightarrow\; p:q = 1:2.$$

Hence, the correct answer is Option D.

The given equation of the transverse wave is $$y = 0.03 \sin(450t - 9x)$$.

We compare it with the standard form of a travelling wave, $$y = A \sin(\omega t - kx)$$, where $$\omega$$ is the angular frequency and $$k$$ is the angular wave number.

From inspection we have $$\omega = 450 \; \text{rad s}^{-1}$$ and $$k = 9 \; \text{rad m}^{-1}$$.

For any wave, the linear speed $$v$$ is related to $$\omega$$ and $$k$$ by the formula $$v = \dfrac{\omega}{k}$$.

Substituting the identified values,

$$v = \dfrac{450}{9} = 50 \; \text{m s}^{-1}.$$

The speed of a transverse wave on a stretched string is also given by the relation $$v = \sqrt{\dfrac{T}{\mu}},$$ where $$T$$ is the tension and $$\mu$$ is the linear mass density.

The linear density is provided as $$5 \; \text{g m}^{-1}$$. Converting grams to kilograms (since SI units require kilograms), $$\mu = 5 \; \text{g m}^{-1} = 5 \times 10^{-3} \; \text{kg m}^{-1} = 0.005 \; \text{kg m}^{-1}.$$

We now square both sides of the speed-tension formula to isolate $$T$$:

$$v^{2} = \dfrac{T}{\mu} \quad \Longrightarrow \quad T = \mu v^{2}.$$

Substituting $$\mu = 0.005 \; \text{kg m}^{-1}$$ and $$v = 50 \; \text{m s}^{-1}$$,

$$T = 0.005 \times (50)^{2} = 0.005 \times 2500 = 12.5 \; \text{N}.$$

Hence, the correct answer is Option C.

We start with the acoustic power output of the speaker, which is given as $$P = 2 \text{ W}$$.

For a point source that radiates sound uniformly in all directions, the intensity $$I$$ at a distance $$r$$ from the source is obtained from the definition of intensity,

$$I \;=\;\frac{\text{Power passing normally through a surface}}{\text{Area of that surface}}.$$

The surface through which the power spreads is an imaginary sphere of radius $$r$$. The area of this sphere is $$4\pi r^{2}$$. Hence, using the above definition, we can write

$$I \;=\;\frac{P}{4\pi r^{2}}.$$

The loudness of sound is usually expressed in decibels (dB). The relation between the sound‐level $$\beta$$ in decibels and the intensity $$I$$ is

$$\beta \;=\;10\,\log_{10}\!\left(\frac{I}{I_{0}}\right),$$

where $$I_{0}=10^{-12}\,\text{W m}^{-2}$$ is the standard (reference) intensity.

We are told that the required sound‐level is $$\beta = 120 \text{ dB}$$. Substituting this value in the above formula, we have

$$120 \;=\;10\,\log_{10}\!\left(\frac{I}{10^{-12}}\right).$$

Dividing both sides by $$10$$, we obtain

$$12 \;=\;\log_{10}\!\left(\frac{I}{10^{-12}}\right).$$

Now, by the definition of the logarithm, this means

$$\frac{I}{10^{-12}} \;=\;10^{12},$$

so

$$I \;=\;10^{12}\times10^{-12} \;=\;1\;\text{W m}^{-2}.$$

Thus, to perceive a 120 dB sound, the intensity incident on the listener’s ear must be $$1\;\text{W m}^{-2}$$.

Next, we substitute this required intensity into the inverse‐square expression $$I = \dfrac{P}{4\pi r^{2}}$$. Writing the equality explicitly, we get

$$1 \;=\;\frac{2}{4\pi r^{2}}.$$

Multiplying both sides by $$4\pi r^{2}$$, we have

$$4\pi r^{2} \;=\;2.$$

Dividing both sides by $$4\pi$$ yields

$$r^{2} \;=\;\frac{2}{4\pi} \;=\;\frac{1}{2\pi}.$$

Hence

$$r \;=\;\sqrt{\frac{1}{2\pi}} \;=\;\sqrt{\frac{1}{6.283}} \;\approx\;\sqrt{0.159}\;\text{ m} \;\approx\;0.399\;\text{ m}.$$

Expressing this distance in centimetres,

$$0.399\;\text{ m}=39.9\;\text{ cm}\approx40\;\text{ cm}.$$

Hence, the correct answer is Option A.

We have a stretched string of length $$L = 2.0\ \text{m}$$, and both its ends are fixed. Such a string supports standing waves whose frequencies are integral multiples of the fundamental frequency. If the string is vibrating in its third harmonic (that is, the third normal mode), the frequency supplied by the vibrator is the third-harmonic frequency $$f_{3}$$.

By definition of harmonics for a string fixed at both ends, the relation between the harmonic number $$n$$ and the corresponding frequency $$f_{n}$$ is

$$f_{n} = n\,f_{1},$$

where $$f_{1}$$ is the fundamental frequency (also called the first harmonic). For the third harmonic we set $$n = 3$$, so

$$f_{3} = 3\,f_{1}.$$

The vibrator is known to supply $$f_{3} = 240\ \text{Hz}$$. Substituting this value, we get

$$240\ \text{Hz} = 3\,f_{1}.$$

Now, solving for $$f_{1}$$,

$$f_{1} = \frac{240\ \text{Hz}}{3} = 80\ \text{Hz}.$$

So, the fundamental frequency of the string is $$80\ \text{Hz}.$$

Next, we need the speed $$v$$ of the wave along the string. For any wave, the basic relation connecting speed $$v$$, frequency $$f$$ and wavelength $$\lambda$$ is

$$v = f\,\lambda.$$

For a string fixed at both ends, the wavelength of the $$n^{\text{th}}$$ harmonic is determined by the formula

$$\lambda_{n} = \frac{2L}{n}.$$

Putting $$n = 3$$ for the third harmonic and $$L = 2.0\ \text{m}$$, we find

$$\lambda_{3} = \frac{2 \times 2.0\ \text{m}}{3} = \frac{4.0\ \text{m}}{3} = 1.\overline{3}\ \text{m}.$$

Now insert $$f_{3} = 240\ \text{Hz}$$ and $$\lambda_{3} = \dfrac{4}{3}\ \text{m}$$ into the wave-speed relation:

$$v = f_{3}\,\lambda_{3} = 240\ \text{Hz} \times \frac{4}{3}\ \text{m}.$$

Carrying out the multiplication,

$$v = 240 \times 1.\overline{3}\ \text{m s}^{-1} = 320\ \text{m s}^{-1}.$$

Thus, the speed of the wave on the string is $$320\ \text{m s}^{-1},$$ and the fundamental frequency is $$80\ \text{Hz}.$$

Hence, the correct answer is Option B.

We have a stationary wave on a string whose ends are fixed, and its mathematical form is given as

$$y \;=\; 0.3 \,\sin(0.157\,x)\,\cos(200\pi\,t).$$

For a string that is clamped at both ends, the general expression of a standing wave is usually written as

$$y \;=\; 2A\,\sin(kx)\,\cos(\omega t),$$

where $$k$$ is the wave-number and $$\omega$$ is the angular frequency. The boundary conditions at the two fixed ends $$x = 0$$ and $$x = L$$ force the displacement to be zero at those points. Mathematically, we must have

$$\sin(kx) = 0 \quad\text{when}\quad x = 0 \quad\text{and}\quad x = L.$$

This is satisfied if

$$kL = m\pi,$$

where $$m$$ is a positive integer. The integer $$m$$ counts the harmonic number; in particular,

$$k = \frac{m\pi}{L}.$$

Now, the problem states that the string is vibrating in its 4th harmonic. Therefore we set $$m = 4$$. The theoretical relation for the wave-number in this mode is

$$k \;=\; \frac{4\pi}{L}.$$

Next, we compare this with the actual wave-number that appears in the given equation. From

$$y \;=\; 0.3\,\sin\!\bigl(0.157\,x\bigr)\,\cos(200\pi\,t),$$

we can directly read off

$$k_{\text{given}} \;=\; 0.157\;\text{rad m}^{-1}.$$

By equating the two expressions for $$k$$, we get

$$0.157 \;=\; \frac{4\pi}{L}.$$

Solving for $$L$$: first multiply both sides by $$L$$, then divide by $$0.157$$:

$$L \;=\; \frac{4\pi}{0.157}.$$

Evaluating the numerator, we find

$$4\pi \;=\; 4 \times 3.1416 \;=\; 12.5664.$$

Now perform the division:

$$L \;=\; \frac{12.5664}{0.157} \;\approx\; 80\;\text{m}.$$

Hence, the length of the string is 80 metres.

Hence, the correct answer is Option D.

We have a closed organ pipe whose fundamental (first harmonic) frequency is given as $$f_1 = 1.5\ \text{kHz} = 1500\ \text{Hz}.$$

For a pipe closed at one end, only odd harmonics are present. The general formula for the frequency of the $$n^{\text{th}}$$ harmonic in such a pipe is

$$f_n = (2n - 1)\,f_1,\qquad n = 1, 2, 3, \ldots$$

Now we substitute $$f_1 = 1500\ \text{Hz}$$ to generate all possible harmonics and compare them with the upper audible limit of $$20000\ \text{Hz}.$$

For $$n = 1$$ (fundamental):

$$f_1 = (2\!\times\!1 - 1)\,f_1 = 1\,f_1 = 1500\ \text{Hz}.$$

For $$n = 2$$ (third harmonic, but first overtone):

$$f_2 = (2\!\times\!2 - 1)\,f_1 = 3\,f_1 = 3 \times 1500 = 4500\ \text{Hz}.$$

For $$n = 3$$:

$$f_3 = (2\!\times\!3 - 1)\,f_1 = 5\,f_1 = 5 \times 1500 = 7500\ \text{Hz}.$$

For $$n = 4$$:

$$f_4 = (2\!\times\!4 - 1)\,f_1 = 7\,f_1 = 7 \times 1500 = 10500\ \text{Hz}.$$

For $$n = 5$$:

$$f_5 = (2\!\times\!5 - 1)\,f_1 = 9\,f_1 = 9 \times 1500 = 13500\ \text{Hz}.$$

For $$n = 6$$:

$$f_6 = (2\!\times\!6 - 1)\,f_1 = 11\,f_1 = 11 \times 1500 = 16500\ \text{Hz}.$$

For $$n = 7$$:

$$f_7 = (2\!\times\!7 - 1)\,f_1 = 13\,f_1 = 13 \times 1500 = 19500\ \text{Hz}.$$

For $$n = 8$$:

$$f_8 = (2\!\times\!8 - 1)\,f_1 = 15\,f_1 = 15 \times 1500 = 22500\ \text{Hz}.$$ This exceeds the audible limit of $$20000\ \text{Hz}$$ and therefore cannot be heard.

So all harmonics whose frequencies are ≤ $$20000\ \text{Hz}$$ correspond to $$n = 1, 2, 3, 4, 5, 6, 7.$$ That makes $$7$$ audible harmonics in total.

However, the question asks for the number of overtones. Overtones are all audible harmonics except the fundamental. Thus,

$$\text{Number of overtones} = 7 \;(\text{audible harmonics}) - 1 \;(\text{fundamental}) = 6.$$

Hence, the correct answer is Option C.

We are given the pressure (sound) wave

$$P \;=\;0.01\;\sin\!\left[\,1000\,t\;-\;3\,x\,\right] \text{ N m}^{-2}$$

for the day on which the atmospheric temperature is $$0^{\circ}\text{C}$$. In the standard form of a travelling wave

$$P \;=\;P_{0}\;\sin\!\bigl(\omega t - kx\bigr),$$

the coefficient of $$t$$ is the angular frequency $$\omega$$ and the coefficient of $$x$$ is the wave-number $$k$$. Hence, by simple comparison, we have

$$\omega \;=\;1000\ \text{rad s}^{-1},\qquad k \;=\;3\ \text{m}^{-1}.$$

For any wave, the speed $$v$$ is related to $$\omega$$ and $$k$$ by the formula

$$v \;=\;\frac{\omega}{k}.$$

Substituting the above numerical values,

$$v_{0} \;=\;\frac{1000}{3}\;\text{m s}^{-1} \;=\;333.33\ \text{m s}^{-1}\;(\text{approximately}).$$

This is the speed of sound in air at the initial temperature $$0^{\circ}\text{C}$$.

Now, on another day, the same vibrating blade produces sound of the same frequency (because the source has not changed), but the measured speed of sound is

$$v_{T} \;=\;336\ \text{m s}^{-1}.$$

For a given gas, the theoretical relation between the speed of sound and the absolute temperature is

$$v \;\propto\;\sqrt{T_{\text{abs}}},$$

where the absolute temperature is $$T_{\text{abs}} = T({^{\circ}\text{C}})+273\;\text{K}.$$

Because the gas (air) is the same on both days, we can write

$$\frac{v_{T}}{v_{0}} \;=\;\sqrt{\frac{T_{\text{abs},\,T}}{T_{\text{abs},\,0}}}.$$

Squaring both sides gives

$$\left(\frac{v_{T}}{v_{0}}\right)^{2} \;=\;\frac{T_{\text{abs},\,T}}{T_{\text{abs},\,0}}.$$

We already know $$T_{\text{abs},\,0}=273\ \text{K}$$ (because $$0^{\circ}\text{C}=273\text{ K}$$), so let us substitute every known quantity step by step.

First the ratio of speeds:

$$\frac{v_{T}}{v_{0}} \;=\;\frac{336}{333.33} \;=\;1.008\;(\text{approximately}).$$

Now square this ratio:

$$\left(\frac{v_{T}}{v_{0}}\right)^{2} \;=\;(1.008)^{2} \;=\;1.0161.$$

Therefore,

$$\frac{T_{\text{abs},\,T}}{273} \;=\;1.0161,$$

and hence

$$T_{\text{abs},\,T} \;=\;1.0161 \times 273 \;=\;277.4\ \text{K}.$$

Finally convert this absolute temperature back to the Celsius scale:

$$T({^{\circ}\text{C}}) \;=\;T_{\text{abs},\,T}\;-\;273 \;=\;277.4\;-\;273 \;=\;4.4^{\circ}\text{C}.$$

On approximation to the nearest whole number, the temperature is about $$4^{\circ}\text{C}$$.

Hence, the correct answer is Option C.

We are dealing with a resonance tube that is closed at the bottom and open at the top. Such an air column behaves like a closed organ pipe. For a closed organ pipe, the resonant (standing-wave) lengths of the air column are given by the well-known quarter-wavelength formula

$$l_n \;=\;\frac{(2n-1)\,\lambda}{4},\qquad n = 1,2,3,\ldots$$

where $$l_n$$ is the length of the air column at the $$n^{\text{th}}$$ resonance and $$\lambda$$ is the wavelength of sound in air.

Successive resonances correspond to consecutive odd multiples of $$\lambda/4$$. Therefore, the difference between any two successive resonant lengths is

$$l_{n+1}-l_n=\frac{(2(n+1)-1)\lambda}{4}-\frac{(2n-1)\lambda}{4} =\frac{(2n+1)\lambda}{4}-\frac{(2n-1)\lambda}{4} =\frac{\lambda}{2}.$$

This result is very useful because the end correction (the small extra length added because the antinode forms slightly above the tube’s rim) cancels out when we take the difference. Hence, from two successive resonances we can get $$\lambda$$ directly.

In the experiment we are told that the first observed resonant length is

$$l_1 = 30\ \text{cm} = 0.30\ \text{m},$$

and the next (successive) resonant length is

$$l_2 = 70\ \text{cm} = 0.70\ \text{m}.$$

We now calculate the difference:

$$l_2 - l_1 = 0.70\ \text{m} - 0.30\ \text{m} = 0.40\ \text{m}.$$

But we have just established that

$$l_2 - l_1 = \frac{\lambda}{2}.$$

So we can write

$$\frac{\lambda}{2} = 0.40\ \text{m}.$$

Multiplying both sides by 2 gives the full wavelength:

$$\lambda = 2 \times 0.40\ \text{m} = 0.80\ \text{m}.$$

Now we use the basic wave relation between speed $$v$$, frequency $$f$$ and wavelength $$\lambda$$. The formula is

$$v = f \lambda.$$

The frequency of the tuning fork has been provided as

$$f = 480\ \text{Hz}.$$

Substituting $$f = 480\ \text{Hz}$$ and $$\lambda = 0.80\ \text{m}$$ into the wave equation, we get

$$v = (480\ \text{Hz})(0.80\ \text{m}) = 384\ \text{m s}^{-1}.$$

Hence, the correct answer is Option A.

Let the length of the string be $$L$$. Its linear mass‐density is

$$\mu=\dfrac{m}{L}\,.$$

The speed of a transverse wave on a stretched string is given by the well-known relation

$$v=\sqrt{\dfrac{T}{\mu}},$$

where $$T$$ is the tension in the string.

When the car is at rest the only force on the heavy ball is its weight. Hence the tension equals the weight:

$$T_1 = Mg.$$

The corresponding wave-speed is given to be

$$v_1 = 60\ \text{m s}^{-1}.$$

Therefore

$$v_1^2=\dfrac{T_1}{\mu}\quad\Longrightarrow\quad 60^2=\dfrac{Mg}{\mu}.$$

When the car accelerates horizontally with acceleration $$a$$, a backward pseudo-force $$Ma$$ acts on the ball in the non-inertial frame of the car. The ball is now in equilibrium under three forces: its weight $$Mg$$ downward, the pseudo-force $$Ma$$ horizontally, and the string tension $$T_2$$ along the string. Balancing the components, the magnitude of the tension is

$$T_2 = M\sqrt{g^{2}+a^{2}}.$$

The new speed of transverse waves is given to be

$$v_2 = 60.5\ \text{m s}^{-1}.$$

Again using the wave-speed formula,

$$v_2^2=\dfrac{T_2}{\mu}\quad\Longrightarrow\quad 60.5^{2}=\dfrac{M\sqrt{g^{2}+a^{2}}}{\mu}.$$

Taking the ratio of the two speed equations eliminates $$\mu$$ and $$M$$:

$$\left(\dfrac{v_2}{v_1}\right)^{2}=\dfrac{T_2}{T_1} =\dfrac{M\sqrt{g^{2}+a^{2}}}{\mu}\;\Big/\;\dfrac{Mg}{\mu} =\dfrac{\sqrt{g^{2}+a^{2}}}{g}.$$

Substituting the numerical values of the speeds,

$$\left(\dfrac{60.5}{60}\right)^{2}=\dfrac{\sqrt{g^{2}+a^{2}}}{g}.$$

First evaluate the left side step by step:

$$\dfrac{60.5}{60}=1+\dfrac{0.5}{60}=1+\dfrac{1}{120}=1.008\overline{3},$$

so

$$\left(1.008\overline{3}\right)^{2}\approx1.016736.$$

Hence

$$1.016736=\dfrac{\sqrt{g^{2}+a^{2}}}{g}.$$

Multiplying by $$g$$ and then squaring both sides,

$$\sqrt{g^{2}+a^{2}}=1.016736\,g,$$

$$g^{2}+a^{2}=1.016736^{2}\,g^{2}.$$

Now compute the square:

$$1.016736^{2}\approx1.033752.$$

Substituting,

$$g^{2}+a^{2}=1.033752\,g^{2},$$

so

$$a^{2}=(1.033752-1)\,g^{2}=0.033752\,g^{2}.$$

Taking the square root,

$$a=g\sqrt{0.033752}\approx g(0.1838)\approx0.18\,g.$$

This value is most nearly $$\dfrac{g}{5}=0.20\,g$$ among the listed options.

Hence, the correct answer is Option C.

We are told that a tuning fork of frequency $$256\ \text{Hz}$$ produces exactly one beat per second when sounded together with the third normal mode of an open organ pipe. A beat is heard whenever two close frequencies interfere, and the beat frequency equals the absolute difference of the two original frequencies. Mathematically we write

$$f_{\text{beat}}=\left|\,f_{\text{pipe}}-f_{\text{fork}}\,\right|.$$

Here $$f_{\text{beat}}=1\ \text{Hz}$$ and $$f_{\text{fork}}=256\ \text{Hz}$$ are known, so

$$\left|\,f_{\text{pipe}}-256\,\right| = 1.$$

Solving this simple absolute-value relation gives the two possible pipe frequencies:

$$f_{\text{pipe}} = 256 + 1 = 257\ \text{Hz}\quad\text{or}\quad f_{\text{pipe}} = 256 - 1 = 255\ \text{Hz}.$$

Next we recall the formula for the resonant (normal-mode) frequencies of an open pipe. An open pipe has antinodes at both ends, and all harmonics are present. The nth normal-mode frequency is

$$f_n = \dfrac{n\,v}{2L},$$

where

$$n = 1,2,3,\ldots,$$

$$v = 340\ \text{m s}^{-1}$$ is the speed of sound in air (given), and $$L$$ is the length of the pipe. The phrase “third normal mode” tells us that $$n = 3$$. Therefore

$$f_{\text{pipe}} = f_3 = \dfrac{3\,v}{2L}.$$

We now substitute the two possible numerical values of $$f_{\text{pipe}}$$ separately and solve for $$L$$.

First possibility: $$f_{\text{pipe}} = 257\ \text{Hz}.$$ Substituting, we have

$$257 = \dfrac{3 \times 340}{2L}.$$

Multiplying both sides by $$2L$$ gives

$$257 \times 2L = 3 \times 340.$$

So

$$514L = 1020.$$

Dividing by $$514$$,

$$L = \dfrac{1020}{514}\ \text{m}.$$

Carrying out the division,

$$L \approx 1.984\ \text{m} = 198.4\ \text{cm}.$$

Second possibility: $$f_{\text{pipe}} = 255\ \text{Hz}.$$ Substituting, we get

$$255 = \dfrac{3 \times 340}{2L}.$$

Again multiplying both sides by $$2L$$,

$$255 \times 2L = 3 \times 340.$$

Hence

$$510L = 1020.$$

Dividing by $$510$$,

$$L = \dfrac{1020}{510}\ \text{m} = 2.00\ \text{m} = 200\ \text{cm}.$$

Both calculations yield lengths that are practically the same (the small difference arises only from rounding), and they clearly correspond to the option that is an exact round figure. The tabulated options list $$200\ \text{cm}$$, which matches perfectly with the second calculation and almost exactly with the first. Consequently the physical length of the pipe must be

$$L = 200\ \text{cm}.$$

Hence, the correct answer is Option D.

For a resonance column that is closed at one end and open at the other, stationary (standing) waves form only at specific lengths. The condition for resonance is

$$\text{(effective length)} = \left(n+\dfrac12\right)\dfrac{\lambda}{2}, \qquad n = 0,1,2,\ldots$$

Because the tube is closed at one end, the fundamental (shortest) resonance corresponds to a node at the closed end and an antinode just outside the open end. Experimentally, the air antinode is not exactly at the rim; it is slightly outside. This shift is called the end correction and is denoted by $$e$$. Therefore

$$\text{effective length} = (\text{length of air column}) + e.$$

Let the wavelength of the sound produced by the tuning fork be $$\lambda$$. For the shortest resonance we have

$$L_1 + e = \dfrac{\lambda}{4}$$ (because the effective length equals one-quarter wavelength).

We are given

$$L_1 = 10\ \text{cm}, \qquad e = 1\ \text{cm}.$$

Substituting these values, we calculate the wavelength:

$$10 + 1 = \dfrac{\lambda}{4} \;\Longrightarrow\; 11 = \dfrac{\lambda}{4} \;\Longrightarrow\; \lambda = 4 \times 11 = 44\ \text{cm}.$$

The next resonance in a closed tube occurs when the effective length equals $$\dfrac{3\lambda}{4}$$ (corresponding to the third harmonic, with two additional nodes and antinodes inside the tube). Hence we write

$$L_2 + e = \dfrac{3\lambda}{4}.$$

Substituting $$\lambda = 44\ \text{cm}$$ and $$e = 1\ \text{cm}$$, we obtain

$$L_2 + 1 = \dfrac{3 \times 44}{4} = \dfrac{132}{4} = 33\ \text{cm}.$$

Now isolate $$L_2$$:

$$L_2 = 33 - 1 = 32\ \text{cm}.$$

This length is the next (second) resonating length of the air column.

Hence, the correct answer is Option A.

We are given a granite rod whose total length is $$L = 60 \text{ cm} = 0.60 \text{ m}$$. The rod is clamped exactly at its mid-point, so the centre is rigidly fixed while both ends are free to move.

Because of the clamp, each half of the rod behaves like an independent bar of length $$\dfrac{L}{2} = \dfrac{0.60}{2} = 0.30 \text{ m}$$ which is fixed at one end (the clamped mid-point) and free at the other end.

For longitudinal vibrations of a bar with one end fixed and the other end free, the fundamental (lowest) mode has a node at the fixed end and an antinode at the free end. In such a mode the length of the bar equals one-quarter of the wavelength. We therefore write the relation

$$\text{Length of half-rod} = \dfrac{\lambda_1}{4}$$

Substituting $$0.30 \text{ m}$$ for the length, we obtain

$$0.30 = \dfrac{\lambda_1}{4} \quad\Longrightarrow\quad \lambda_1 = 4 \times 0.30 = 1.20 \text{ m}$$

The speed of longitudinal elastic waves in a solid is given by the formula

$$v = \sqrt{\dfrac{Y}{\rho}}$$

where $$Y = 9.27 \times 10^{10} \text{ Pa}$$ is Young’s modulus of granite, and $$\rho = 2.7 \times 10^{3} \text{ kg m}^{-3}$$ is its density.

We first evaluate the ratio inside the square root:

$$\dfrac{Y}{\rho} = \dfrac{9.27 \times 10^{10}}{2.7 \times 10^{3}} = \dfrac{9.27}{2.7} \times 10^{7} = 3.433\ \times 10^{7}$$

Taking the square root gives

$$v = \sqrt{3.433 \times 10^{7}} = \sqrt{3.433}\,\sqrt{10^{7}} = 1.852 \times 3162 \approx 5.86 \times 10^{3} \text{ m s}^{-1}$$

So the velocity of longitudinal waves in the granite rod is

$$v \approx 5.86 \times 10^{3} \text{ m s}^{-1}$$

The fundamental frequency $$f_1$$ is obtained from the basic wave relation $$v = f \lambda$$, hence

$$f_1 = \dfrac{v}{\lambda_1} = \dfrac{5.86 \times 10^{3}}{1.20} = 4.88 \times 10^{3} \text{ Hz} \approx 5.0 \text{ kHz}$$

Hence, the correct answer is Option B.

We are given that string A has frequency $$f_A = 425\ \text{Hz}$$ and, when it is sounded together with string B, a beat frequency of $$5\ \text{Hz}$$ is heard.

First, we recall the definition of beat frequency:

$$f_{\text{beat}} = \left|\,f_A - f_B\,\right|$$

Here $$f_{\text{beat}} = 5\ \text{Hz}$$, so

$$\left|\,425 - f_B\,\right| = 5$$

Solving this absolute-value equation gives two algebraic possibilities:

$$425 - f_B = 5 \quad \Longrightarrow \quad f_B = 420\ \text{Hz}$$ $$425 - f_B = -5 \quad \Longrightarrow \quad f_B = 430\ \text{Hz}$$

Thus, before we touch the tension, B could be either $$420\ \text{Hz}$$ or $$430\ \text{Hz}$$.

Next, the tension of B is slightly increased. The fundamental frequency of a stretched string is given by the formula

$$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$

where $$L$$ is the length, $$T$$ is the tension, and $$\mu$$ is the linear mass density. We see that $$f \propto \sqrt{T}$$, so increasing tension always increases the frequency of the string.

Experimentally, after increasing the tension, the beat frequency is observed to decrease by $$3\ \text{Hz}$$. The old beat frequency was $$5\ \text{Hz}$$, so the new beat frequency is

$$f'_{\text{beat}} = 5 - 3 = 2\ \text{Hz}$$

This means the new frequency difference satisfies

$$\left|\,f_A - f'_B\,\right| = 2\ \text{Hz}$$

Since increasing tension can only raise $$f_B$$, let us test each of the two earlier possibilities:

1. If the original $$f_B = 430\ \text{Hz}$$ (already higher than 425 Hz), raising the tension would further increase it, say to $$f'_B > 430\ \text{Hz}$$. The new difference $$|425 - f'_B|$$ would then be greater than $$|425 - 430| = 5\ \text{Hz}$$, so the beat frequency would rise above 5 Hz, not drop to 2 Hz. This contradicts the observation.

2. If the original $$f_B = 420\ \text{Hz}$$ (lower than 425 Hz), raising the tension lifts it closer to 425 Hz, perhaps to $$f'_B = 423\ \text{Hz}$$. The new difference $$|425 - 423| = 2\ \text{Hz}$$ is indeed smaller than 5 Hz, exactly matching the measured decrease to 2 Hz.

Therefore the only self-consistent choice is

$$f_B = 420\ \text{Hz}$$

Hence, the correct answer is Option D.

Let the total length of the pipe be $$L$$. Because both ends are originally open, the air column supports a fundamental (first harmonic) whose antinodes are at each open end and whose single node is midway between. For an open-open pipe, the well-known relation between the fundamental wavelength $$\lambda_1$$ and the length $$L$$ is stated first:

$$\text{For an open-open pipe : }\; \lambda_1 = 2L$$

If the speed of sound in air is $$v$$, the fundamental frequency $$f$$ is therefore

$$f = \dfrac{v}{\lambda_1} = \dfrac{v}{2L} \qquad (1)$$

Now the pipe is dipped vertically into water until exactly one half of its length is submerged. The water surface behaves like a rigid boundary for sound waves and therefore acts as a closed end, while the top of the pipe remains an open end. Consequently, only the upper half of the pipe, of length $$L/2$$, is now filled with vibrating air, and the pipe has effectively changed from an open-open organ pipe to an open-closed organ pipe of length $$L/2$$.

For an open-closed pipe, the fundamental standing wave has a node at the closed end and an antinode at the open end. The standard relation for such a pipe is first recalled:

$$\text{For an open-closed pipe : }\; \lambda'_1 = 4\left(\text{length of the air column}\right)$$

Here the vibrating column length is $$L/2$$, so the new fundamental wavelength $$\lambda'_1$$ becomes

$$\lambda'_1 = 4\left(\dfrac{L}{2}\right) = 2L$$

We notice that $$\lambda'_1$$ is numerically equal to $$\lambda_1$$ obtained earlier. Proceeding, the new fundamental frequency $$f'$$ is

$$f' = \dfrac{v}{\lambda'_1} = \dfrac{v}{2L} \qquad (2)$$

Comparing expressions (1) and (2), we have

$$f' = f$$

Thus, immersing half the pipe in water does not change the fundamental frequency of the vibrating air column.

Hence, the correct answer is Option B.

We start with the standard formula for the speed of a transverse wave on a stretched string:

$$v \;=\; \sqrt{\dfrac{T}{\mu}}$$

where $$T$$ is the tension at the point under consideration and $$\mu$$ is the uniform linear mass density of the string. For a string that hangs vertically, the tension is created solely by the weight of the portion of string lying below the point.

Let us choose the origin $$x = 0$$ at the lowest end of the string and measure the coordinate $$x$$ upward. At a distance $$x$$ from the lower end, the length of string below is exactly $$x$$, so its mass equals $$\mu x$$. Therefore the tension at that point is simply its weight:

$$T(x) \;=\; (\mu x)\,g \;=\; \mu g x.$$

Substituting this expression for $$T(x)$$ into the velocity formula, we obtain the position-dependent wave speed:

$$v(x) \;=\; \sqrt{\dfrac{T(x)}{\mu}} \;=\; \sqrt{\dfrac{\mu g x}{\mu}} \;=\; \sqrt{g\,x}.$$

Because the speed changes continuously with height, the pulse takes different times to traverse different small segments. Consider a differential element of length $$dx$$ located at coordinate $$x$$. The small time $$dt$$ required to pass through this element is

$$dt \;=\; \dfrac{dx}{v(x)} \;=\; \dfrac{dx}{\sqrt{g\,x}} \;=\; \dfrac{dx}{\sqrt{g}\,\sqrt{x}}.$$

The total time $$t$$ for the pulse to travel from the lower end ($$x = 0$$) to the rigid support at the upper end ($$x = L$$) is obtained by integrating $$dt$$ over the entire length:

$$t \;=\; \int_{0}^{L} dt \;=\; \int_{0}^{L} \dfrac{dx}{\sqrt{g}\,\sqrt{x}} \;=\; \dfrac{1}{\sqrt{g}} \int_{0}^{L} x^{-\frac{1}{2}}\,dx.$$

We now perform the integral. The integral of $$x^{-\frac{1}{2}}$$ is $$2\,x^{\frac{1}{2}}$$, so

$$\int_{0}^{L} x^{-\frac{1}{2}}\,dx \;=\; 2\,x^{\frac{1}{2}}\Big|_{0}^{L} \;=\; 2\left(\sqrt{L} - \sqrt{0}\right) \;=\; 2\sqrt{L}.$$

Substituting this result back into the expression for $$t$$ gives

$$t \;=\; \dfrac{1}{\sqrt{g}} \,\big(2\sqrt{L}\big) \;=\; \dfrac{2\sqrt{L}}{\sqrt{g}}.$$

For the present problem the string length is $$L = 20 \,\text{m}$$ and the acceleration due to gravity is $$g = 10 \,\text{m s}^{-2}$$. Substituting these values, we find

$$t \;=\; \dfrac{2\sqrt{20}}{\sqrt{10}}.$$

To simplify, first evaluate the square roots:

$$\sqrt{20} \;=\; \sqrt{4\times5} \;=\; 2\sqrt{5},$$

and

$$\sqrt{10} \;=\; \sqrt{2\times5} \;=\; \sqrt{2}\,\sqrt{5}.$$

Substituting these into the time expression:

$$t \;=\; \dfrac{2 \times 2\sqrt{5}}{\sqrt{2}\,\sqrt{5}} \;=\; \dfrac{4\sqrt{5}}{\sqrt{2}\,\sqrt{5}}.$$

The factor $$\sqrt{5}$$ cancels out, leaving

$$t \;=\; \dfrac{4}{\sqrt{2}} \;=\; 4 \times \dfrac{1}{\sqrt{2}} \;=\; 4 \times \dfrac{\sqrt{2}}{2} \;=\; 2\sqrt{2}\; \text{s}.$$

Hence, the correct answer is Option A.

First, we are given that a source of sound A emits waves at a frequency of 1800 Hz and is falling towards the ground with a terminal speed $$v$$. An observer B on the ground directly beneath the source receives waves at a frequency of 2150 Hz. The speed of sound is 343 m/s. We need to find the frequency of the waves that source A receives after they are reflected from the ground.

To solve this, we use the Doppler effect formula for sound. The general formula when the source and observer are moving relative to each other is:

$$f' = f \left( \frac{v_{\text{sound}} \pm v_{\text{observer}}}{v_{\text{sound}} \mp v_{\text{source}}} \right)$$

The signs depend on the directions of motion. For the first part, source A is moving towards the stationary observer B on the ground. Since the source is moving towards the observer, the observed frequency increases. The formula becomes:

$$f_B = f \left( \frac{v_{\text{sound}}}{v_{\text{sound}} - v_{\text{source}}} \right)$$

Here, $$f = 1800$$ Hz, $$f_B = 2150$$ Hz, and $$v_{\text{sound}} = 343$$ m/s. Plugging in the values:

$$2150 = 1800 \left( \frac{343}{343 - v} \right)$$

Now, solve for $$v$$. Rearrange the equation:

$$\frac{2150}{1800} = \frac{343}{343 - v}$$

Simplify the left side:

$$\frac{2150}{1800} = \frac{2150 \div 50}{1800 \div 50} = \frac{43}{36}$$

So:

$$\frac{43}{36} = \frac{343}{343 - v}$$

Cross-multiply:

$$43 \times (343 - v) = 36 \times 343$$

Calculate $$36 \times 343$$:

$$343 \times 30 = 10290, \quad 343 \times 6 = 2058, \quad 10290 + 2058 = 12348$$

Calculate $$43 \times 343$$:

$$343 \times 40 = 13720, \quad 343 \times 3 = 1029, \quad 13720 + 1029 = 14749$$

So:

$$14749 - 43v = 12348$$

Rearrange to solve for $$v$$:

$$14749 - 12348 = 43v$$

$$2401 = 43v$$

$$v = \frac{2401}{43}$$

Simplify:

$$2401 \div 43 = 55.8372, \quad \text{but we keep it as a fraction:} \quad v = \frac{2401}{43} \text{m/s}$$

Note that $$2401 = 343 \times 7$$ because $$343 \times 7 = 2401$$. So:

$$v = \frac{343 \times 7}{43} \text{m/s}$$

Now, we need to find the frequency that source A receives from the waves reflected from the ground. The ground reflects the sound waves, and since the ground is stationary, the frequency of the reflected wave is the same as the frequency received by the ground. Observer B on the ground receives 2150 Hz, so the reflected wave has a frequency of 2150 Hz and is emitted by the stationary ground.

Source A is moving towards the ground (downwards) while the reflected wave is traveling upwards. So, for the reflected wave:

• The source (ground) is stationary, so $$v_{\text{source}} = 0$$.
• The observer (source A) is moving towards the source (ground), so it is moving towards the sound wave.

The Doppler effect formula for a stationary source and an observer moving towards the source is:

$$f'' = f_{\text{reflected}} \left( \frac{v_{\text{sound}} + v_{\text{observer}}}{v_{\text{sound}}} \right)$$

Here, $$f_{\text{reflected}} = 2150$$ Hz, $$v_{\text{sound}} = 343$$ m/s, and $$v_{\text{observer}} = v = \frac{2401}{43}$$ m/s. Plugging in:

$$f'' = 2150 \left( \frac{343 + v}{343} \right) = 2150 \left(1 + \frac{v}{343}\right)$$

Substitute $$v = \frac{343 \times 7}{43}$$:

$$f'' = 2150 \left(1 + \frac{\frac{343 \times 7}{43}}{343}\right) = 2150 \left(1 + \frac{7}{43}\right)$$

Simplify inside the parentheses:

$$1 + \frac{7}{43} = \frac{43}{43} + \frac{7}{43} = \frac{50}{43}$$

So:

$$f'' = 2150 \times \frac{50}{43}$$

Note that $$2150 = 43 \times 50$$ because $$43 \times 50 = 2150$$. Therefore:

$$f'' = 43 \times 50 \times \frac{50}{43} = 50 \times 50 = 2500$$

Thus, the frequency that source A receives from the reflected waves is 2500 Hz.

Hence, the correct answer is Option B.

From the wave equation $$y = \frac{10}{\pi} \sin \left( \frac{2\pi}{T}t - \frac{2\pi}{\lambda}x \right)$$, amplitude $$A = \frac{10}{\pi}$$ and the wave velocity $$v_w = \frac{\omega}{k} = \frac{\lambda}{T}$$.

The maximum particle velocity is given by $$v_{p(max)} = A\omega = \left( \frac{10}{\pi} \right) \left( \frac{2\pi}{T} \right) = \frac{20}{T}$$.
Substituting these into the given condition $$v_w = 2 v_{p(max)}$$ yields $$\frac{\lambda}{T} = 2 \times \frac{20}{T}$$, which simplifies to $$\lambda = 40$$ cm.

The total length of the sonometer wire is 110 cm, which is 110 divided by 100, giving 1.10 meters. The wire is divided into three parts in the ratio 6:3:2. Let the common multiplier be $$ k $$. So, the lengths are $$ L_1 = 6k $$, $$ L_2 = 3k $$, and $$ L_3 = 2k $$. The sum of these lengths is $$ 6k + 3k + 2k = 11k $$, which equals 1.10 meters. Therefore, $$ 11k = 1.10 $$, and solving for $$ k $$, we get $$ k = \frac{1.10}{11} = 0.10 $$ meters.

Now, calculate each length:

First part: $$ L_1 = 6 \times 0.10 = 0.60 $$ meters,

Second part: $$ L_2 = 3 \times 0.10 = 0.30 $$ meters,

Third part: $$ L_3 = 2 \times 0.10 = 0.20 $$ meters.

The tension $$ T $$ is 400 N, and the mass per unit length $$ \mu $$ is 0.01 kg/m. The fundamental frequency for a stretched string is given by $$ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $$. Since $$ T $$ and $$ \mu $$ are the same for all parts, compute the constant $$ \sqrt{\frac{T}{\mu}} $$:

$$ \sqrt{\frac{T}{\mu}} = \sqrt{\frac{400}{0.01}} = \sqrt{40000} = 200 \text{ m/s}. $$

So, the frequency formula simplifies to $$ f = \frac{200}{2L} = \frac{100}{L} $$ Hz.

Now, find the fundamental frequency for each segment:

For the first segment ($$ L_1 = 0.60 $$ m): $$ f_1 = \frac{100}{0.60} = \frac{1000}{6} = \frac{500}{3} $$ Hz,

For the second segment ($$ L_2 = 0.30 $$ m): $$ f_2 = \frac{100}{0.30} = \frac{1000}{3} $$ Hz,

For the third segment ($$ L_3 = 0.20 $$ m): $$ f_3 = \frac{100}{0.20} = 500 $$ Hz.

Each segment can vibrate at its fundamental frequency or at harmonics, which are integer multiples of the fundamental frequency. To find the minimum common frequency for all three segments, we need the smallest frequency that is a common multiple of $$ f_1 $$, $$ f_2 $$, and $$ f_3 $$. This means there exist integers $$ n_1 $$, $$ n_2 $$, and $$ n_3 $$ such that:

$$ f = n_1 \cdot f_1 = n_1 \cdot \frac{500}{3}, $$

$$ f = n_2 \cdot f_2 = n_2 \cdot \frac{1000}{3}, $$

$$ f = n_3 \cdot f_3 = n_3 \cdot 500. $$

Set the expressions equal:

$$ n_1 \cdot \frac{500}{3} = n_2 \cdot \frac{1000}{3} = n_3 \cdot 500. $$

Multiply through by 3 to eliminate denominators:

$$ 500 n_1 = 1000 n_2 = 1500 n_3. $$

From $$ 500 n_1 = 1000 n_2 $$, divide both sides by 500: $$ n_1 = 2 n_2 $$.

From $$ 1000 n_2 = 1500 n_3 $$, divide both sides by 500: $$ 2 n_2 = 3 n_3 $$.

So, $$ n_1 = 2 n_2 $$ and $$ 2 n_2 = 3 n_3 $$. Solving for $$ n_2 $$ in terms of $$ n_3 $$: $$ n_2 = \frac{3}{2} n_3 $$. Since $$ n_2 $$ must be an integer, $$ n_3 $$ must be even. Let $$ n_3 = 2k $$ for some integer $$ k \geq 1 $$. Then:

$$ n_2 = \frac{3}{2} \times 2k = 3k, $$

$$ n_1 = 2 \times n_2 = 2 \times 3k = 6k. $$

The smallest positive integers occur when $$ k = 1 $$:

$$ n_3 = 2 \times 1 = 2, $$

$$ n_2 = 3 \times 1 = 3, $$

$$ n_1 = 6 \times 1 = 6. $$

Now, compute the common frequency $$ f $$:

Using $$ f = n_3 \cdot f_3 = 2 \times 500 = 1000 $$ Hz,

or $$ f = n_2 \cdot f_2 = 3 \times \frac{1000}{3} = 1000 $$ Hz,

or $$ f = n_1 \cdot f_1 = 6 \times \frac{500}{3} = 1000 $$ Hz.

Thus, the minimum common frequency is 1000 Hz.

Hence, the correct answer is Option A.

We are told that a steel wire of a sonometer has length $$L = 1.5\;{\rm m}$$ and is stretched till the elastic (longitudinal) strain is 1 %.

The density and Young’s modulus of steel are

$$\rho = 7.7 \times 10^{3}\;{\rm kg\,m^{-3}}, \qquad Y = 2.2 \times 10^{11}\;{\rm N\,m^{-2}}.$$

For a stretched string the fundamental (first harmonic) frequency is given by the well-known relation

$$f = \frac{1}{2L}\,\sqrt{\frac{T}{\mu}},$$

where $$T$$ is the tension in the wire and $$\mu$$ is the mass per unit length of the wire.

First we calculate the tension. Young’s modulus is defined by the formula

$$Y = \frac{\text{stress}}{\text{strain}},$$

where

$$\text{stress} = \frac{T}{A}, \qquad \text{strain} = \varepsilon.$$

Here $$\varepsilon = 1\% = 0.01.$$ Rearranging, we have

$$\frac{T}{A} = Y\,\varepsilon \quad\Longrightarrow\quad T = Y\,\varepsilon\,A.$$

Next, the linear density $$\mu$$ of the wire is

$$\mu = \rho\,A.$$

Dividing tension by linear density, the cross-sectional area $$A$$ cancels:

$$\frac{T}{\mu} = \frac{Y\,\varepsilon\,A}{\rho\,A} = \frac{Y\,\varepsilon}{\rho}.$$

Substituting this result in the frequency formula, we get

$$f = \frac{1}{2L}\,\sqrt{\frac{Y\,\varepsilon}{\rho}}.$$

Now we substitute the numerical values step by step. First compute the product $$Y\,\varepsilon$$:

$$Y\,\varepsilon = \bigl(2.2 \times 10^{11}\bigr)\,(0.01) = 2.2 \times 10^{9}\;{\rm N\,m^{-2}}.$$

Next divide by the density:

$$\frac{Y\,\varepsilon}{\rho} = \frac{2.2 \times 10^{9}}{7.7 \times 10^{3}}.$$

Performing the division,

$$\frac{2.2}{7.7} = 0.285714\ldots,\qquad 10^{9}\!/\!10^{3} = 10^{6},$$

so

$$\frac{Y\,\varepsilon}{\rho} = 0.285714 \times 10^{6} = 2.85714 \times 10^{5}\;{\rm (m^{2}\,s^{-2})}.$$

Taking the square root,

$$\sqrt{2.85714 \times 10^{5}} = \sqrt{2.85714}\,\sqrt{10^{5}}.$$

We have $$\sqrt{2.85714} \approx 1.690$$ and $$\sqrt{10^{5}} = 10^{2.5} = 316.227,$$ so

$$\sqrt{2.85714 \times 10^{5}} \approx 1.690 \times 316.227 \approx 534.4\;{\rm m\,s^{-1}}.$$

Finally, substitute into the frequency formula:

$$f = \frac{1}{2L}\,\bigl(534.4\bigr) = \frac{1}{2 \times 1.5}\,(534.4) = \frac{534.4}{3.0} \approx 178.1\;{\rm Hz}.$$

This value matches option D (178.2 Hz) to the required accuracy.

Hence, the correct answer is Option D.

A sonometer wire of length 114 cm is fixed at both ends, and we need to place two bridges to divide it into three segments with fundamental frequencies in the ratio 1 : 3 : 4. For a string fixed at both ends, the fundamental frequency $$f$$ is given by $$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$, where $$L$$ is the length of the segment, $$T$$ is the tension (constant throughout the wire), and $$\mu$$ is the linear mass density (also constant). Since $$T$$ and $$\mu$$ are constant, the frequency $$f$$ is inversely proportional to the length $$L$$, so $$f \propto \frac{1}{L}$$.

Given the frequency ratio $$f_1 : f_2 : f_3 = 1 : 3 : 4$$, the lengths will be inversely proportional to these frequencies. Therefore, the ratio of the lengths is $$L_1 : L_2 : L_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4}$$. To simplify, multiply each term by the least common multiple (LCM) of the denominators 1, 3, and 4, which is 12:

$$L_1 : L_2 : L_3 = \frac{1}{1} \times 12 : \frac{1}{3} \times 12 : \frac{1}{4} \times 12 = 12 : 4 : 3$$

So, $$L_1 : L_2 : L_3 = 12k : 4k : 3k$$ for some constant $$k$$. The total length is the sum of the segments:

$$12k + 4k + 3k = 19k = 114 \text{ cm}$$

Solving for $$k$$:

$$19k = 114 \implies k = \frac{114}{19} = 6$$

Now, find the lengths:

$$L_1 = 12k = 12 \times 6 = 72 \text{ cm}$$

$$L_2 = 4k = 4 \times 6 = 24 \text{ cm}$$

$$L_3 = 3k = 3 \times 6 = 18 \text{ cm}$$

The segments must be adjacent and cover the entire wire. The bridges are placed such that the segments are formed consecutively. Starting from one end (say end A), the first bridge is at a distance $$L_1 = 72$$ cm from A. The second bridge is at $$L_1 + L_2 = 72 + 24 = 96$$ cm from A. The third segment from 96 cm to 114 cm is $$L_3 = 18$$ cm.

To verify, the fundamental frequencies are proportional to the reciprocals of the lengths:

$$f_1 \propto \frac{1}{72}, \quad f_2 \propto \frac{1}{24}, \quad f_3 \propto \frac{1}{18}$$

The ratio is:

$$f_1 : f_2 : f_3 = \frac{1}{72} : \frac{1}{24} : \frac{1}{18}$$

Simplify by multiplying each term by the LCM of 72, 24, and 18, which is 72:

$$\frac{1}{72} \times 72 = 1, \quad \frac{1}{24} \times 72 = 3, \quad \frac{1}{18} \times 72 = 4$$

So, $$f_1 : f_2 : f_3 = 1 : 3 : 4$$, which matches the given ratio.

Now, check the options:

• Option A: Bridges at 36 cm and 84 cm. Segments: 36 cm, 48 cm (84 - 36), 30 cm (114 - 84). Frequencies proportional to $$\frac{1}{36}$$, $$\frac{1}{48}$$, $$\frac{1}{30}$$. Ratios: $$\frac{1}{36} : \frac{1}{48} : \frac{1}{30} = 20 : 15 : 24$$ (after multiplying by 720, the LCM), which is not 1:3:4.
• Option B: Bridges at 24 cm and 72 cm. Segments: 24 cm, 48 cm (72 - 24), 42 cm (114 - 72). Frequencies proportional to $$\frac{1}{24}$$, $$\frac{1}{48}$$, $$\frac{1}{42}$$. Ratios: $$\frac{1}{24} : \frac{1}{48} : \frac{1}{42} = 14 : 7 : 8$$ (after multiplying by 336, the LCM), which is not 1:3:4.
• Option C: Bridges at 48 cm and 96 cm. Segments: 48 cm, 48 cm (96 - 48), 18 cm (114 - 96). Frequencies proportional to $$\frac{1}{48}$$, $$\frac{1}{48}$$, $$\frac{1}{18}$$. Ratios: $$\frac{1}{48} : \frac{1}{48} : \frac{1}{18} = 3 : 3 : 8$$ (after multiplying by 144, the LCM), which is not 1:3:4.
• Option D: Bridges at 72 cm and 96 cm. Segments: 72 cm, 24 cm (96 - 72), 18 cm (114 - 96). Frequencies proportional to $$\frac{1}{72}$$, $$\frac{1}{24}$$, $$\frac{1}{18}$$. Ratios: $$\frac{1}{72} : \frac{1}{24} : \frac{1}{18} = 1 : 3 : 4$$ (as shown above).

Hence, the correct answer is Option D.

For a closed pipe (closed at one end), the fundamental frequency is $$f = \dfrac{v}{4L}$$.

For the 100 cm pipe: $$f_1 = \dfrac{v}{4 \times 1.00} = \dfrac{v}{4}$$

For the 120 cm pipe: $$f_2 = \dfrac{v}{4 \times 1.20} = \dfrac{v}{4.8}$$

The beat frequency is $$|f_1 - f_2| = 15$$ Hz.

$$\dfrac{v}{4} - \dfrac{v}{4.8} = 15$$

$$v\left(\dfrac{1}{4} - \dfrac{1}{4.8}\right) = 15$$

$$v\left(\dfrac{4.8 - 4}{4 \times 4.8}\right) = 15$$

$$v \cdot \dfrac{0.8}{19.2} = 15$$

$$v = 15 \times \dfrac{19.2}{0.8} = 15 \times 24 = 360$$ m/s

Hence, the correct answer is Option D.

$$v = \sqrt{\frac{\gamma P}{\rho}}$$

$$\gamma = 1 + \frac{2}{f}$$

$$\gamma_{\text{He}} = \frac{5}{3}$$

$$\gamma_{\text{CH}_4} = 1 + \frac{2}{6} = \frac{8}{6} = \frac{4}{3}$$

$$\gamma_{\text{CO}_2} = \frac{7}{5}$$

$$v_{\text{He}} : v_{\text{CH}_4} : v_{\text{CO}_2} = \sqrt{\gamma_{\text{He}}} : \sqrt{\gamma_{\text{CH}_4}} : \sqrt{\gamma_{\text{CO}_2}}$$

$$v_{\text{He}} : v_{\text{CH}_4} : v_{\text{CO}_2} = \sqrt{\frac{5}{3}} : \sqrt{\frac{4}{3}} : \sqrt{\frac{7}{5}}$$

We need to find the length of the open organ tube given that the 9th harmonic of the closed tube equals the 4th harmonic of the open tube.

For a closed organ pipe, the nth harmonic frequency (only odd harmonics exist, but the problem says 9th harmonic meaning the harmonic number is 9):

$$f_{closed} = \frac{n \cdot v_1}{4L_1}$$

For an open organ pipe, the nth harmonic frequency:

$$f_{open} = \frac{n \cdot v_2}{2L_2}$$

The speed of sound in a gas is $$v = \sqrt{\frac{B}{\rho}}$$ where B is the bulk modulus.

Since both gases have the same bulk modulus B:

$$\frac{v_1}{v_2} = \sqrt{\frac{\rho_2}{\rho_1}} = \sqrt{\frac{16}{1}} = 4$$

The 9th harmonic of the closed tube equals the 4th harmonic of the open tube:

$$\frac{9 v_1}{4 L_1} = \frac{4 v_2}{2 L_2}$$

Substituting $$L_1 = 10$$ cm and $$v_1 = 4v_2$$:

$$\frac{9 \times 4v_2}{4 \times 10} = \frac{4 v_2}{2 L_2}$$

$$\frac{36 v_2}{40} = \frac{4 v_2}{2 L_2}$$

$$\frac{9 v_2}{10} = \frac{2 v_2}{L_2}$$

$$L_2 = \frac{2 \times 10}{9} = \frac{20}{9} \text{ cm}$$

The correct answer is Option 4: $$\frac{20}{9}$$ cm.

The given transverse wave on the string is
$$y(x,t)=\sin\!\left(20\pi x+10\pi t\right)$$ where $$x$$ is in metres and $$t$$ is in seconds.

For any particle of the string the instantaneous transverse (oscillatory) speed is obtained by differentiating $$y$$ with respect to time:

Formula: $$v_y=\frac{\partial y}{\partial t}= \omega \cos\!\left(kx+\omega t\right)$$ where $$k$$ is the wave-number and $$\omega$$ is the angular frequency.

Here $$k=20\pi$$ and $$\omega=10\pi$$, so
$$v_y=10\pi \cos\!\left(20\pi x+10\pi t\right)$$ $$-(1)$$

At a fixed instant $$t$$, two points $$x_1$$ and $$x_2$$ will have the same oscillatory speed magnitude when

$$\left|v_y(x_1,t)\right|=\left|v_y(x_2,t)\right|$$ $$\Longrightarrow$$ $$\left|\cos\!\left(20\pi x_1+10\pi t\right)\right|=\left|\cos\!\left(20\pi x_2+10\pi t\right)\right|$$

Let the corresponding phase angles be $$\phi_1$$ and $$\phi_2$$: $$\phi_1 = 20\pi x_1+10\pi t,\qquad \phi_2 = 20\pi x_2+10\pi t$$

Condition for equal absolute value of a cosine:
$$\left|\cos\phi_1\right|=\left|\cos\phi_2\right| \; \Longrightarrow \; \phi_2=\phi_1\pm n\pi,\; n\in\mathbb{Z}$$ (The plus/minus $$\pi$$ shift changes the sign of the cosine but keeps its magnitude the same.)

Therefore the phase difference between the two points is
$$\Delta\phi = \phi_2-\phi_1 = n\pi$$

Since $$\Delta\phi = k\,\Delta x$$, we get
$$k\,\Delta x = n\pi \quad \Longrightarrow \quad \Delta x = \frac{n\pi}{k}$$ $$-(2)$$

With $$k = 20\pi$$, equation $$(2)$$ becomes
$$\Delta x = \frac{n\pi}{20\pi} = \frac{n}{20}\; \text{metre}$$

The minimum non-zero distance corresponds to $$n = 1$$:

$$\Delta x_{\min} = \frac{1}{20}\; \text{m} = 0.05\; \text{m} = 5.0\; \text{cm}$$

Hence, the minimum separation between two points on the string having the same oscillating speed is 5.0 cm.

Option A is correct.

The speed of a transverse wave on a stretched string is given by the fundamental relation
$$v = \sqrt{\dfrac{T}{\mu}}$$ where $$T$$ is the tension and $$\mu$$ is the mass per unit length of the string.

Because both strings are made of the same material and carry the same tension $$T$$, the only quantity that changes from one string to the other is $$\mu$$.

For a string of density $$\rho$$ (mass per unit volume) and circular cross-section of radius $$R$$, the cross-sectional area is
$$A = \pi R^{2}$$.
Hence,
$$\mu = \rho A = \rho \pi R^{2} \;-\!(1)$$

Case 1: Radius $$R$$
Using $$(1)$$, $$\mu_1 = \rho \pi R^{2}$$.
Therefore,
$$v_1 = \sqrt{\dfrac{T}{\mu_1}} = \sqrt{\dfrac{T}{\rho \pi R^{2}}} \;-\!(2)$$

Case 2: Radius $$\dfrac{R}{2}$$
New area: $$A_2 = \pi \left(\dfrac{R}{2}\right)^{2} = \dfrac{\pi R^{2}}{4}$$.
Thus,
$$\mu_2 = \rho A_2 = \rho \dfrac{\pi R^{2}}{4} = \dfrac{\mu_1}{4} \;-\!(3)$$

Wave speed in the second string:
$$v_2 = \sqrt{\dfrac{T}{\mu_2}}$$.
Substitute $$\mu_2$$ from $$(3)$$:
$$v_2 = \sqrt{\dfrac{T}{\mu_1/4}} = \sqrt{\dfrac{4T}{\mu_1}} = 2\sqrt{\dfrac{T}{\mu_1}} \;-\!(4)$$

Compare $$(4)$$ with $$(2)$$: $$v_1 = \sqrt{\dfrac{T}{\mu_1}}$$.
Therefore,
$$\dfrac{v_2}{v_1} = \dfrac{2\sqrt{T/\mu_1}}{\sqrt{T/\mu_1}} = 2$$.

Hence $$\dfrac{v_2}{v_1} = 2$$.
Correct option: Option B.

Wave equation: $$y = 4.0\sin[20 \times 10^{-3}x + 600t]$$ mm, where x in mm, t in s.

$$k = 20 \times 10^{-3}$$ mm$$^{-1}$$ = 20 m$$^{-1}$$ (converting: 0.02 per mm = 20 per m).

$$\omega = 600$$ rad/s.

Velocity $$v = -\frac{\omega}{k} = -\frac{600}{20} = -30$$ m/s.

The negative sign indicates the wave travels in the negative x-direction (since the equation has $$+kx$$).

The correct answer is Option 2: -30 m/s.

The given equation is $$x(t)=5\cos\left(628\,t+\frac{\pi}{2}\right)$$.

This is of standard form $$x(t)=A\cos\left(\omega t+\phi\right)$$, so the angular frequency is
$$\omega = 628\ \text{rad s}^{-1}$$.

Frequency $$f$$ is related to angular frequency by
$$\omega = 2\pi f$$ $$-(1)$$.

Substituting $$\omega = 628$$ rad s$$^{-1}$$ in $$(1)$$,
$$f = \frac{\omega}{2\pi} = \frac{628}{2\pi}\ \text{Hz}$$.

Using $$\pi \approx 3.14$$,
$$2\pi \approx 6.28$$, hence
$$f \approx \frac{628}{6.28} = 100\ \text{Hz}$$.

The relation between wave speed $$v$$, frequency $$f$$, and wavelength $$\lambda$$ is
$$v = f\lambda$$ $$-(2)$$.

Given velocity $$v = 300\ \text{m s}^{-1}$$, substituting $$f = 100\ \text{Hz}$$ in $$(2)$$:
$$\lambda = \frac{v}{f} = \frac{300}{100}\ \text{m} = 3\ \text{m}$$.

Therefore, the wavelength of the wave is $$3\ \text{m}$$.

Option B.

Fundamental of closed pipe: $$f_c = \frac{v}{4L_c}$$. First overtone of open pipe: $$f_o = \frac{2v}{2L_o} = \frac{v}{L_o}$$.

$$\frac{v}{4L_c} = \frac{v}{L_o} \Rightarrow L_c = \frac{L_o}{4} = \frac{60}{4} = 15$$ cm.

The answer is Option (4): 15 cm.

We are given that a sonometer wire of resonating length 90 cm has a fundamental frequency of 400 Hz under a certain tension. We need to find the resonating length when the fundamental frequency is 600 Hz under the same tension.

Recall that the fundamental frequency of a string fixed at both ends is given by:

$$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$

Here $$L$$ is the vibrating length, $$T$$ is the tension, and $$\mu$$ is the linear mass density of the wire. Since both $$T$$ and $$\mu$$ remain constant, it follows that $$f \propto \frac{1}{L}$$, which means $$f \times L$$ is constant, or

$$f_1 L_1 = f_2 L_2$$.

Substituting the given values $$f_1 = 400$$ Hz, $$L_1 = 90$$ cm, and $$f_2 = 600$$ Hz gives

$$400 \times 90 = 600 \times L_2$$

From this,

$$L_2 = \frac{400 \times 90}{600} = \frac{36000}{600} = 60 \text{ cm}$$

The answer is 60 cm.

We need to find the speed of sound in oxygen at STP.

The values of the gas constant and the adiabatic index are given as $$R = 8.3$$ J/(mol K) and $$\gamma = 1.4$$, respectively, while at STP the temperature is $$T = 273$$ K and the molar mass of $$O_2$$ is 32 g/mol = 0.032 kg/mol.

The speed of sound in a gas is given by the formula $$v = \sqrt{\frac{\gamma RT}{M}}$$.

Substituting in the numbers leads to $$v = \sqrt{\frac{1.4 \times 8.3 \times 273}{0.032}}$$.

Since the numerator is $$1.4 \times 8.3 = 11.62$$, multiplying this by 273 gives $$11.62 \times 273 = 3172.26$$.

This gives $$v = \sqrt{\frac{3172.26}{0.032}} = \sqrt{99133} \approx 315 \text{ m/s} \approx 310 \text{ m/s}$$.

Therefore the correct answer is Option 1: 310 m/s.

We need to find the ratio of the speed of sound in hydrogen gas to that in oxygen gas at the same temperature.

In an ideal gas, the speed of sound is given by the relation:

$$v = \sqrt{\frac{\gamma R T}{M}}$$

Here $$\gamma$$ denotes the adiabatic index, $$R$$ is the universal gas constant, $$T$$ represents the absolute temperature, and $$M$$ stands for the molar mass of the gas.

Because both hydrogen ($$H_2$$) and oxygen ($$O_2$$) are diatomic gases, they share the same adiabatic index $$\gamma = \frac{7}{5}$$. Consequently, at an identical temperature $$T$$, the ratio of their sound speeds depends solely on their molar masses, leading to

$$\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$$

By substituting the molar masses $$M_{H_2} = 2\ \mathrm{g/mol}$$ and $$M_{O_2} = 32\ \mathrm{g/mol}$$ into the above expression, we obtain

$$\frac{v_{H_2}}{v_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$$

Hence, the ratio of the speed of sound in hydrogen to that in oxygen is 4 : 1, which corresponds to Option B.

For a vibrating string, the fundamental frequency is given by:

$$f = \frac{v}{2L}$$

where $$v$$ is the speed of the wave and $$L$$ is the length of the string. Since the speed $$v$$ depends on the tension and linear mass density (which remain the same), $$v$$ is constant. So we have $$f \propto \frac{1}{L}$$, which gives us:

$$f_1 L_1 = f_2 L_2$$

Now substituting the values:

$$120 \times 90 = 180 \times L_2$$

$$L_2 = \frac{120 \times 90}{180} = \frac{10800}{180} = 60 \text{ cm}$$

Hence, the length of the string producing a fundamental frequency of 180 Hz is 60 cm.

Given wave equation

$$y(x,t)=5\sin(6t+0.003x)$$

Compare with standard form

$$y=A\sin(\omega t\pm kx)$$

So

$$\omega=6\ \text{rad/s}$$

and

$$k=0.003\ \text{cm}^{-1}$$

Wave speed is

$$v=\frac{\omega}{k}$$

$$=\frac{6}{0.003}$$

$$=2000\text{ cm/s}$$

Convert to m/s:

$$2000\ \text{cm/s}=20\ \text{m/s}$$

We need to find the frequency of the second harmonic in an open pipe.

Formula for harmonics in an open pipe. For a pipe open at both ends, the frequency of the $$n$$th harmonic is:

$$f_n = \frac{nv}{2L}$$

where $$v$$ is the speed of sound and $$L$$ is the length of the pipe.

Given values: $$L = 40$$ cm $$= 0.4$$ m, $$v = 360$$ m/s, $$n = 2$$ (second harmonic)

Calculate: $$f_2 = \frac{2 \times 360}{2 \times 0.4} = \frac{720}{0.8} = 900 \text{ Hz}$$

The correct answer is 900 Hz.

The fundamental frequency of a sonometer wire is:

$$f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$

where $$T = Mg$$ is the tension. Since $$L$$ and $$\mu$$ are the same:

$$f \propto \sqrt{T} \propto \sqrt{M}$$

$$\frac{f_1}{f_2} = \sqrt{\frac{M_1}{M_2}}$$

$$\frac{30}{50} = \sqrt{\frac{180}{m}}$$

$$\frac{9}{25} = \frac{180}{m}$$

$$m = \frac{180 \times 25}{9} = 500 \text{ g}$$

The value of $$m$$ is $$\mathbf{500}$$ g.

Given: fundamental frequency $$f = 50$$ Hz, mass $$m = 18$$ g, linear mass density $$\mu = 20$$ g/m.

Length of string: $$L = \frac{m}{\mu} = \frac{18}{20} = 0.9$$ m

For fundamental frequency: $$f = \frac{v}{2L}$$

$$v = 2Lf = 2 \times 0.9 \times 50 = 90$$ m/s

The speed of transverse waves is $$\mathbf{90}$$ m/s.

Two SHM waves with equal amplitudes $$A = 8$$ cm and equal frequency $$f = 10$$ Hz move in the same direction. The resultant amplitude is also 8 cm, and we need to find the phase difference between them.

For two waves of equal amplitude $$A$$ with phase difference $$\phi$$, the resultant amplitude is given by $$A_R = \sqrt{A^2 + A^2 + 2A^2\cos\phi} = A\sqrt{2(1 + \cos\phi)} = 2A\cos\frac{\phi}{2}.$$

Substituting $$A_R = 8\text{ cm}$$ and $$A = 8\text{ cm}$$ yields $$8 = \sqrt{64 + 64 + 128\cos\phi},$$ which simplifies to $$64 = 128 + 128\cos\phi,\quad 128\cos\phi = -64,\quad \cos\phi = -\frac{1}{2}.$$

Therefore, $$\phi = \cos^{-1}\!\Bigl(-\tfrac{1}{2}\Bigr) = 120°,$$ so the phase difference is 120 degrees.

A frequency ratio of $$1 : 3 : 5$$ indicates an organ pipe open at one end (closed pipe).

$$f_n = \frac{n v}{4L}$$

$$f_5 = 405 \text{ Hz}$$

$$v = 324 \text{ m/s}$$

$$405 = \frac{5 v}{4L}$$

$$405 = \frac{5 \times 324}{4L}$$

$$L = 1 \text{ m}$$

The distance between two consecutive points with a phase difference of $$60°$$ in a wave of frequency 500 Hz is 6.0 m. Find the wave velocity.

Relate phase difference to wavelength.

The phase difference between two points separated by distance $$\Delta x$$ is:

$$\Delta \phi = \frac{2\pi}{\lambda} \cdot \Delta x$$

Find the wavelength.

$$\Delta \phi = 60° = \frac{\pi}{3}$$ radians, $$\Delta x = 6.0$$ m:

$$\frac{\pi}{3} = \frac{2\pi}{\lambda} \times 6$$

$$\lambda = \frac{2\pi \times 6}{\pi/3} = 2 \times 6 \times 3 = 36 \text{ m}$$

Calculate velocity.

$$v = f\lambda = 500 \times 36 = 18000 \text{ m/s} = 18 \text{ km s}^{-1}$$

The correct answer is 18 km s$$^{-1}$$.

Since the wave function is $$y(x,t) = 0.05\sin(8x - 4t)$$ m, we compare it with the general form of a travelling wave $$y = A\sin(kx - \omega t)$$, where the wave number is $$k$$ and the angular frequency is $$\omega$$; this reveals $$k = 8$$ rad/m, $$\omega = 4$$ rad/s, and the amplitude $$A = 0.05$$ m.

Therefore, the phase velocity is $$v = \frac{\omega}{k} = \frac{4}{8} = 0.5\;\text{m/s}$$, making the correct answer Option 3: 0.5 m s$$^{-1}$$.

The wave equation is given as: $$y(x, t) = 0.05\sin(8x - 4t)$$ m.

The standard form of a traveling wave is: $$y(x, t) = A \sin(kx - \omega t)$$, where:
- $$A$$ is the amplitude,
- $$k$$ is the wave number (in rad/m),
- $$\omega$$ is the angular frequency (in rad/s).

Comparing the given equation to the standard form:
- The coefficient of $$x$$ is $$k = 8$$ rad/m,
- The coefficient of $$t$$ is $$\omega = 4$$ rad/s.

The velocity $$v$$ of a wave is given by the formula:
$$v = \frac{\omega}{k}$$

Substituting the values:
$$v = \frac{4}{8} = 0.5$$ m/s.

Therefore, the velocity of the wave is 0.5 m/s.

The correct option is C. 0.5 m s$$^{-1}$$.

We need to find the length of the air column at the third resonance in a resonance tube experiment.

For a closed organ pipe (resonance tube), the resonance lengths are given by $$L_n = (2n - 1)\frac{\lambda}{4} - e$$ where e is the end correction. Equivalently, $$L_1 = \frac{\lambda}{4} - e$$, $$L_2 = \frac{3\lambda}{4} - e$$, $$L_3 = \frac{5\lambda}{4} - e$$.

The wavelength is calculated as $$\lambda = \frac{v}{f} = \frac{336}{400} = 0.84 \text{ m} = 84 \text{ cm}$$.

Using the first resonance, $$L_1 = \frac{\lambda}{4} - e$$ and substituting $$20.0 = \frac{84}{4} - e = 21 - e$$ yields $$e = 1.0 \text{ cm}$$.

Finally, $$L_3 = \frac{5\lambda}{4} - e = \frac{5 \times 84}{4} - 1.0 = 105 - 1.0 = 104 \text{ cm}$$.

The answer is 104 cm.

The frequency of the tuning fork is $$f = 340$$ Hz, the length of the air column for the fundamental mode is $$L = 125$$ cm, and the velocity of sound is $$v = 340$$ m/s. First, find the wavelength of sound: $$\lambda = \frac{v}{f} = \frac{340}{340} = 1 \text{ m} = 100 \text{ cm}$$.

For a closed pipe, resonance occurs when the air column length satisfies $$L = \frac{(2n-1)\lambda}{4}, \quad n = 1, 2, 3, \ldots$$. The possible resonant lengths are: for $$n = 1$$, $$L_1 = \frac{\lambda}{4} = 25$$ cm; for $$n = 2$$, $$L_2 = \frac{3\lambda}{4} = 75$$ cm; and for $$n = 3$$, $$L_3 = \frac{5\lambda}{4} = 125$$ cm.

The tube resonates in the fundamental mode at 125 cm. Note that this “fundamental mode” refers to the first resonance observed with the given tube length of 125 cm, which corresponds to $$n = 3$$ (the 3rd harmonic of a closed pipe with $$L = 125$$ cm).

When water is poured in, the effective air column length decreases, and the next resonance will occur at $$L_2 = 75 \text{ cm}$$. The minimum height of water required is $$h = L - L_2 = 125 - 75 = 50 \text{ cm}$$. Hence, the answer is 50 cm.

The stationary wave equation is given as $$y = 10\cos(\pi x)\sin\left(\frac{2\pi t}{T}\right)$$ cm, and we seek the amplitude at $$x = \frac{4}{3}$$ cm. Since a stationary wave of the form $$y = A\cos(kx)\sin(\omega t)$$ has an amplitude at position $$x$$ equal to $$\text{Amplitude at } x = |A\cos(kx)|$$, we identify the parameters from the given equation as $$A = 10$$ cm and $$k = \pi$$ rad/cm.

Substituting $$x = \frac{4}{3}$$ cm into the amplitude expression yields: $$\text{Amplitude} = \left|10\cos\left(\pi \times \frac{4}{3}\right)\right| = \left|10\cos\left(\frac{4\pi}{3}\right)\right|$$. Now, since the angle $$\frac{4\pi}{3}$$ lies in the third quadrant ($$\pi + \frac{\pi}{3}$$), we have $$\cos\left(\frac{4\pi}{3}\right) = \cos\left(\pi + \frac{\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$$.

This gives the amplitude as $$\text{Amplitude} = \left|10 \times \left(-\frac{1}{2}\right)\right| = |-5| = 5 \text{ cm}$$.

The answer is $$5$$ cm.

We are given that the length of the wire is $$L = 30$$ cm $$= 0.3$$ m, the $$n^{th}$$ harmonic frequency is $$f_n = 400$$ Hz, the $$(n+1)^{th}$$ harmonic frequency is $$f_{n+1} = 450$$ Hz, and the tension is $$T = 2700$$ N.

For a string fixed at both ends, the frequency of the $$n^{th}$$ harmonic is $$f_n = n \cdot f_1$$, and the difference between consecutive harmonics gives the fundamental frequency as $$f_{n+1} - f_n = f_1 = 450 - 400 = 50 \text{ Hz}$$. Hence, $$n = \frac{f_n}{f_1} = \frac{400}{50} = 8$$, so we are dealing with the 8th and 9th harmonics. The fundamental frequency for a string of length $$L$$ is $$f_1 = \frac{v}{2L}$$, from which $$v = 2Lf_1 = 2 \times 0.3 \times 50 = 30 \text{ m/s}$$.

The wave speed on a string is related to tension and linear mass density by $$v = \sqrt{\frac{T}{\mu}}$$, yielding $$\mu = \frac{T}{v^2} = \frac{2700}{(30)^2} = \frac{2700}{900} = 3 \text{ kg/m}$$.

The answer is 3.

The length of the closed organ pipe is $$L_c = 20$$ cm. The fundamental frequency of a closed organ pipe is $$f_c = \frac{v}{4L_c}$$, and we require the first overtone of an open organ pipe to equal this fundamental frequency.

The first overtone (second harmonic) of an open pipe is $$f_o = \frac{2v}{2L_o} = \frac{v}{L_o}$$. Equating this to the fundamental of the closed pipe gives $$\frac{v}{L_o} = \frac{v}{4L_c}$$. Solving for the open pipe length yields $$L_o = 4L_c = 4 \times 20 = 80 \text{ cm}$$. Thus, the length of the open organ pipe is 80 cm.

The tuning fork $$A$$ produces 5 beats per second with a fork of known frequency 340 Hz. This means the frequency of fork $$A$$ is either $$340 + 5 = 345$$ Hz or $$340 - 5 = 335$$ Hz.

When fork $$A$$ is filed, its prongs become thinner and lighter, which increases its frequency. After filing, the beat frequency decreases to 2 beats per second.

If $$f_A = 345$$ Hz, filing would increase it further (say to 347 Hz), and the beat frequency with 340 Hz would become $$347 - 340 = 7$$ beats/s, which is an increase. This contradicts the observation that beats decreased.

If $$f_A = 335$$ Hz, filing would increase it (say to 338 Hz), and the beat frequency would become $$340 - 338 = 2$$ beats/s, which is a decrease from 5 to 2. This matches the observation.

Therefore, the frequency of fork $$A$$ is $$335$$ Hz.

Let us imagine the instant when the observer actually hears the sound. At that very instant the jet aeroplane is right above him, i.e., vertically over his head. However, the sound seems to come from the north at an elevation of $$60^{\circ}$$ above the horizontal. This means the sound that is being heard now was emitted a little earlier, when the aeroplane was still to the north of the observer.

Denote the following quantities:

$$h=$$ constant height (altitude) of the aircraft above the ground,

$$x=$$ horizontal distance (due south-north direction) that the plane had to cover in that time to reach the observer’s vertical line,

$$u=$$ actual speed of the plane (what we have to find),

$$v=$$ speed of sound in air (given).

At the earlier instant (let us call it time $$t=-\Delta t$$) the aircraft was at point $$A$$, a distance $$x$$ to the north of the observer. Its coordinates relative to the observer’s position $$O$$ are therefore $$(-x,\,0,\,h)$$ if we take the north-south axis as the $$x$$-axis, east-west as $$y$$ and height as $$z$$. The sound ray that has just reached the observer travelled along the straight line $$AO$$ during the interval $$\Delta t$$.

Because the sound ray reaches the observer at an elevation angle of $$60^{\circ}$$, the line $$AO$$ makes an angle $$60^{\circ}$$ with the ground. Hence, in the right-angled triangle formed by the altitude and the ground distance, we have

$$\tan 60^{\circ} \;=\; \frac{\text{opposite side}}{\text{adjacent side}} \;=\; \frac{h}{x}.$$

Using $$\tan 60^{\circ}=\sqrt{3}$$, we get

$$\frac{h}{x}=\sqrt{3},\qquad\text{so}\qquad h=\sqrt{3}\,x.$$

The length of the slant path $$AO$$ travelled by the sound is

$$AO=\sqrt{h^{2}+x^{2}}.$$

The sound, moving with speed $$v$$, took a time $$\Delta t$$ given by the elementary relation

$$\text{Time}=\frac{\text{Distance}}{\text{Speed}}.$$

Therefore,

$$\Delta t=\frac{\sqrt{h^{2}+x^{2}}}{v}.$$

During this same interval $$\Delta t$$ the aircraft moved horizontally from $$A$$ to $$O$$, covering the distance $$x$$ with speed $$u$$. Hence

$$u=\frac{\text{distance travelled by the plane}}{\text{corresponding time}} \;=\;\frac{x}{\Delta t}.$$

Substituting the value of $$\Delta t$$ found above, we obtain

$$u=\frac{x}{\dfrac{\sqrt{h^{2}+x^{2}}}{v}} =\frac{x\,v}{\sqrt{h^{2}+x^{2}}}.$$

But we have already related $$h$$ and $$x$$ as $$h=\sqrt{3}\,x$$. Squaring and adding:

$$h^{2}=3x^{2},\qquad h^{2}+x^{2}=3x^{2}+x^{2}=4x^{2}.$$

Thus,

$$\sqrt{h^{2}+x^{2}}=\sqrt{4x^{2}}=2x.$$

Putting this back into the expression for $$u$$:

$$u=\frac{x\,v}{2x}=\frac{v}{2}.$$

So the speed of the jet aeroplane is exactly one half of the speed of sound.

Hence, the correct answer is Option C.

A standing wave is produced when two identical sinusoidal waves move in opposite directions. For a travelling wave moving in the +$$x$$ direction we may write the displacement as $$y_1=A\sin(kx-\omega t)$$, and for the wave moving in the −$$x$$ direction we write $$y_2=A\sin(kx+\omega t)$$.

Adding the two, we use the trigonometric identity $$\sin P+\sin Q = 2\sin\!\left(\frac{P+Q}{2}\right)\cos\!\left(\frac{P-Q}{2}\right)$$. Substituting $$P=kx-\omega t$$ and $$Q=kx+\omega t$$ we obtain

$$y=y_1+y_2=2A\sin(kx)\cos(\omega t).$$

The given equation of the standing wave is

$$y(x,t)=0.5\sin\!\left(\frac{5\pi}{4}\,x\right)\cos(200\pi t).$$

By comparing this expression with the standard form $$y=2A\sin(kx)\cos(\omega t)$$ we can immediately read the wave parameters:

$$2A=0.5\;\;\Rightarrow\;\;A=0.25\;(\text{m, not needed further}),$$

$$k=\frac{5\pi}{4}\;\text{rad m}^{-1},$$

$$\omega=200\pi\;\text{rad s}^{-1}.$$

The speed $$v$$ of either parent travelling wave is related to $$k$$ and $$\omega$$ by the fundamental wave relation

$$v=\frac{\omega}{k}.$$

Substituting the numerical values, we have

$$v=\frac{200\pi}{\dfrac{5\pi}{4}}.$$

First, notice that the factor $$\pi$$ cancels:

$$v=\frac{200}{\dfrac{5}{4}}.$$

Dividing by a fraction is the same as multiplying by its reciprocal, so

$$v=200 \times \frac{4}{5}.$$

Carrying out the multiplication,

$$v=\frac{800}{5}=160\;\text{m s}^{-1}.$$

Hence, the correct answer is Option C.

The displacements of the particle due to the two sound waves are given by:

$$y_1 = 0.05\cos(0.50\pi x - 100\pi t)$$

$$y_2 = 0.05\cos(0.46\pi x - 92\pi t)$$

where $$y_1$$, $$y_2$$, and $$x$$ are in meters, and $$t$$ is in seconds. We need to find the speed of sound in the medium.

A general wave equation traveling in the positive x-direction is written as $$y = A \cos(kx - \omega t)$$, where $$k$$ is the wave number and $$\omega$$ is the angular frequency. The wave number $$k$$ is related to the wavelength $$\lambda$$ by $$k = \frac{2\pi}{\lambda}$$, and the angular frequency $$\omega$$ is related to the frequency $$f$$ by $$\omega = 2\pi f$$. The speed of sound $$v$$ is given by the formula $$v = f \lambda$$. Substituting the relations, we get:

$$v = f \lambda = \left( \frac{\omega}{2\pi} \right) \times \left( \frac{2\pi}{k} \right) = \frac{\omega}{k}$$

Therefore, the speed can be directly calculated as $$v = \frac{\omega}{k}$$ for each wave. Since both waves are traveling in the same medium, the speed should be identical for both.

For the first wave $$y_1$$, the argument is $$0.50\pi x - 100\pi t$$. Comparing with the general form, we have:

Wave number $$k_1 = 0.50\pi$$ rad/m,

Angular frequency $$\omega_1 = 100\pi$$ rad/s.

Thus, the speed $$v_1 = \frac{\omega_1}{k_1} = \frac{100\pi}{0.50\pi}$$.

Simplifying, the $$\pi$$ cancels out:

$$v_1 = \frac{100}{0.50} = \frac{100}{\frac{1}{2}} = 100 \times 2 = 200 \text{ m/s}$$

For the second wave $$y_2$$, the argument is $$0.46\pi x - 92\pi t$$. Comparing with the general form, we have:

Wave number $$k_2 = 0.46\pi$$ rad/m,

Angular frequency $$\omega_2 = 92\pi$$ rad/s.

Thus, the speed $$v_2 = \frac{\omega_2}{k_2} = \frac{92\pi}{0.46\pi}$$.

Simplifying, the $$\pi$$ cancels out:

$$v_2 = \frac{92}{0.46} = \frac{92}{\frac{46}{100}} = 92 \times \frac{100}{46} = \frac{92 \times 100}{46}$$

Dividing 92 by 46:

$$\frac{92}{46} = 2, \quad \text{so} \quad 2 \times 100 = 200 \text{ m/s}$$

Alternatively, $$92 \div 46 = 2$$, so $$92 = 46 \times 2$$, and thus:

$$\frac{92 \times 100}{46} = \frac{46 \times 2 \times 100}{46} = 2 \times 100 = 200 \text{ m/s}$$

Both waves give the same speed of 200 m/s, which is consistent since they are in the same medium. Therefore, the speed of sound in the medium is 200 m/s.

Now, comparing with the options:

A. 92 m/s

B. 200 m/s

C. 100 m/s

D. 332 m/s

Hence, the correct answer is Option B.

The apparent frequency ($$f$$) heard by a stationary listener when a source moves is given by $$f' = f \left( \frac{v}{v \mp v_s} \right)$$

Case 1: A moves towards C

$$f'_{A1} = 500 \left( \frac{340}{340 - 4} \right) = 500 \left( \frac{340}{336} \right) \approx \mathbf{506\text{ Hz}}$$

Case 2: A moves away from C

$$f'_{A2} = 500 \left( \frac{340}{340 + 4} \right) = 500 \left( \frac{340}{344} \right) \approx \mathbf{494\text{ Hz}}$$

Beat frequency is the absolute difference between two frequencies: $$f_{\text{beat}} = |f_A' - f_B|$$

From Case 1 (6 beats): $$|506 - f_B| = 6 \implies f_B = 506 + 6 = \mathbf{512\text{ Hz}} \text{ or } 506 - 6 = 500\text{ Hz}$$

From Case 2 (18 beats): $$|494 - f_B| = 18 \implies f_B = 494 + 18 = \mathbf{512\text{ Hz}} \text{ or } 494 - 18 = 476\text{ Hz}$$

Comparing both scenarios, the only consistent value for the frequency of source B is $$512\text{ Hz}$$.

An engine approaches a hill with a constant speed. The whistle is blown when the engine is 0.9 km away from the hill. Convert this distance to meters: 0.9 km = 900 meters. The echo is heard by the driver after 5 seconds, and the speed of sound is 330 m/s. We need to find the speed of the engine, denoted as $$ v $$ m/s.

When the whistle is blown, the sound travels towards the hill at 330 m/s. The hill is stationary. The echo is the sound reflecting off the hill and returning to the engine. During the 5 seconds, the engine is moving towards the hill, so the distance covered by the sound includes the initial distance to the hill and the reduced distance on the return due to the engine's movement.

Consider the total distance traveled by the sound in 5 seconds. Since the speed of sound is 330 m/s, the total distance covered by sound is:

$$ \text{Distance} = \text{Speed} \times \text{Time} = 330 \times 5 = 1650 \text{ meters}. $$

This distance is the sum of the distance from the engine's initial position to the hill and the distance from the hill back to the engine's position when the echo is heard.

Let the initial position of the engine be point A, and the hill be point B. So, AB = 900 meters. In 5 seconds, the engine moves towards the hill with speed $$ v $$ m/s, so the distance covered by the engine is $$ v \times 5 $$ meters. Let the position of the engine when the echo is heard be point D. Therefore, AD = $$ 5v $$ meters.

The distance from D to the hill (point B) is DB. Since D is between A and B, DB = AB - AD = 900 - 5v meters.

The sound travels from A to B (900 meters) and then from B to D (DB = 900 - 5v meters). So, the total distance traveled by sound is:

$$ \text{Total distance} = \text{AB} + \text{DB} = 900 + (900 - 5v) = 1800 - 5v \text{ meters}. $$

But we know the total distance is 1650 meters. Set them equal:

$$ 1800 - 5v = 1650 $$

Solve for $$ v $$:

Subtract 1800 from both sides:

$$ -5v = 1650 - 1800 $$

$$ -5v = -150 $$

Divide both sides by -5:

$$ v = \frac{-150}{-5} = 30 \text{ m/s}. $$

Hence, the speed of the engine is 30 m/s.

So, the answer is Option D.

At $$t=0$$, the shape of the pulse is

$$y=\frac{1}{x^2+8x+19}=\frac{1}{(x+4)^2+3}$$

At $$t=2\,\text{s}$$, the shape is

$$y=\frac{1}{x^2+3}$$

Thus, the pulse has shifted from $$x=-4$$ to $$x=0$$, i.e. it has moved $$4\,\text{m}$$ in the $$+x$$ direction.

Hence, the speed of the wave is

$$v=\frac{\text{distance travelled}}{\text{time taken}}=\frac{4}{2}=2\,\text{m s}^{-1}$$

Therefore,  $${v=2\,\text{m s}^{-1}}$$

Hence, the correct option is C.

Two tuning forks A and B produce 8 beats in 2 seconds, so the beat frequency is 4 Hz. The frequency of B is 380 Hz.

To determine possible frequencies of A, note that the beat frequency is $$|f_A - f_B| = 4$$ Hz. Thus $$f_A = 380 + 4 = 384$$ Hz or $$f_A = 380 - 4 = 376$$ Hz.

Using the wax-loading condition, when fork A is loaded with wax its frequency decreases (wax increases mass, lowering the frequency). After loading, the beat frequency becomes $$\dfrac{4}{2} = 2$$ beats/s. Let $$f_A'$$ be the new frequency of A after loading. Then $$f_A' < f_A$$ and $$|f_A' - 380| = 2$$.

For the case $$f_A = 384$$ Hz, after loading $$f_A'$$ decreases. Solving $$|f_A' - 380| = 2$$ gives $$f_A' = 382$$ or $$f_A' = 378$$. Since both are less than 384 and the beat frequency decreased from 4 to 2, the result $$f_A' = 382$$ Hz (getting closer to 380) is consistent.

For the case $$f_A = 376$$ Hz, after loading $$f_A'$$ would decrease further below 376. The condition $$|f_A' - 380| = 2$$ again gives $$f_A' = 378$$ or $$f_A' = 382$$, but neither is less than 376, so this case is invalid.

Therefore, the original frequency of tuning fork A is $$\boxed{384}$$ Hz.

We have a pipe that is closed at one end and open at the other. For such a pipe, the standing-wave modes of the air column obey a special harmonic condition. First, we recall the formula for the resonant frequencies of a pipe closed at one end:

$$f_n = \dfrac{n \, v}{4L}$$

where $$f_n$$ is the frequency of the $$n^{\text{th}}$$ mode, $$v$$ is the velocity of sound in air and $$L$$ is the length of the pipe. Because one end is closed, only odd values of $$n$$ appear; that is $$n = 1, 3, 5, 7, \dots$$.

Now we substitute the given data. The length of the pipe is 85 cm, so

$$L = 85 \text{ cm} = 0.85 \text{ m}.$$

The velocity of sound is

$$v = 340 \text{ m/s}.$$

Substituting $$v$$ and $$L$$ into the formula, we find a handy constant factor:

$$ \dfrac{v}{4L} \;=\; \dfrac{340}{4 \times 0.85} \;=\; \dfrac{340}{3.4} \;=\; 100 \text{ Hz}. $$

So each resonant frequency can be written compactly as

$$f_n = n \times 100 \text{ Hz}, \qquad n = 1, 3, 5, 7, \dots$$

We are asked to keep only those natural oscillations whose frequencies lie below 1250 Hz. Therefore we impose

$$ n \times 100 \lt 1250 \quad\Longrightarrow\quad n \lt \dfrac{1250}{100} \quad\Longrightarrow\quad n \lt 12.5. $$

Remembering that $$n$$ must be odd, we list all positive odd integers less than 12.5:

$$n = 1, \; 3, \; 5, \; 7, \; 9, \; 11.$$

Counting them gives

$$\text{Number of possible oscillations} = 6.$$

Hence, the correct answer is Option C.

In this transverse wave problem, we need to find the maximum speed of the vibrating particles. Let's break down the given information step by step.

The distance between a crest and a neighboring trough at the same instant is given as 4.0 cm. In a wave, the distance from a crest to the next trough is half the wavelength. Therefore, we can write:

$$\frac{\lambda}{2} = 4.0 \text{ cm}$$

Solving for the wavelength $$\lambda$$:

$$\lambda = 4.0 \times 2 = 8.0 \text{ cm}$$

Next, the distance between a crest and a trough at the same place is 1.0 cm. This refers to the vertical distance at a fixed point in space. At a given location, the particle oscillates between maximum displacement (crest) and minimum displacement (trough). The amplitude $$A$$ is the maximum displacement from the mean position. The distance between crest and trough positions is the difference between $$+A$$ and $$-A$$, which is $$2A$$. So:

$$2A = 1.0 \text{ cm}$$

Solving for the amplitude $$A$$:

$$A = \frac{1.0}{2} = 0.5 \text{ cm}$$

The next crest appears at the same place after 0.4 s. This is the time taken for one complete cycle, so it is the time period $$T$$:

$$T = 0.4 \text{ s}$$

The particles in the medium undergo simple harmonic motion. The maximum speed $$v_{\text{max}}$$ occurs when the particle passes through the mean position and is given by:

$$v_{\text{max}} = \omega A$$

where $$\omega$$ is the angular frequency. The angular frequency is related to the time period by:

$$\omega = \frac{2\pi}{T}$$

Substituting the values:

$$\omega = \frac{2\pi}{0.4} = \frac{2\pi}{\frac{4}{10}} = \frac{2\pi \times 10}{4} = \frac{20\pi}{4} = 5\pi \text{ rad/s}$$

Now, using $$v_{\text{max}} = \omega A$$:

$$v_{\text{max}} = 5\pi \times 0.5 = 5\pi \times \frac{1}{2} = \frac{5\pi}{2} \text{ cm/s}$$

Comparing with the options:

A. $$\frac{3\pi}{2}$$ cm/s

B. $$\frac{5\pi}{2}$$ cm/s

C. $$\frac{\pi}{2}$$ cm/s

D. $$2\pi$$ cm/s

The value $$\frac{5\pi}{2}$$ cm/s matches option B. Hence, the correct answer is Option B.

Since the velocity of sound in an ideal gas is proportional to the square root of the absolute temperature, we have

$$v \propto \sqrt{T}$$

which implies $$\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1}}$$.

Substituting $$v_2 = 2v_1$$ and $$T_1 = 0°C = 273 \, K$$ into this relation gives

$$\frac{2v_1}{v_1} = \sqrt{\frac{T_2}{273}}$$

so

$$2 = \sqrt{\frac{T_2}{273}}$$.

Squaring both sides yields

$$4 = \frac{T_2}{273}$$

and hence

$$T_2 = 4 \times 273 = 1092 \, K$$.

Converting to Celsius gives

$$\alpha = T_2 - 273 = 1092 - 273 = 819°C$$.

Therefore the required temperature is 819°C, so the answer is Option X.

The speed of a longitudinal wave in a solid is given by the formula:

$$v = \sqrt{\frac{Y}{\rho}}$$

where $$Y$$ is Young's modulus and $$\rho$$ is the density.

Given initial speed $$v = 400 \, \text{m/s}$$.

Density increases by 0.5%, so new density $$\rho' = \rho \left(1 + \frac{0.5}{100}\right) = 1.005\rho$$.

Young's modulus increases by 1%, so new Young's modulus $$Y' = Y \left(1 + \frac{1}{100}\right) = 1.01Y$$.

The new speed $$v'$$ is:

$$v' = \sqrt{\frac{Y'}{\rho'}} = \sqrt{\frac{1.01Y}{1.005\rho}} = \sqrt{\frac{1.01}{1.005}} \times \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{1.01}{1.005}} \times 400$$

Compute the fraction $$\frac{1.01}{1.005}$$:

$$\frac{1.01}{1.005} = \frac{1010}{1005} = \frac{202}{201} \approx 1.004975$$

Now, $$\sqrt{1.004975}$$ can be approximated using $$\sqrt{1+x} \approx 1 + \frac{x}{2}$$ for small $$x$$, where $$x = 0.004975$$:

$$\sqrt{1.004975} \approx 1 + \frac{0.004975}{2} = 1 + 0.0024875 = 1.0024875$$

Thus,

$$v' \approx 1.0024875 \times 400 = 400.995 \approx 401 \, \text{m/s}$$

Alternatively, using the formula for fractional change:

$$\frac{dv}{v} = \frac{1}{2} \frac{dY}{Y} - \frac{1}{2} \frac{d\rho}{\rho}$$

Given $$\frac{dY}{Y} = 0.01$$ and $$\frac{d\rho}{\rho} = 0.005$$,

$$\frac{dv}{v} = \frac{1}{2} (0.01) - \frac{1}{2} (0.005) = 0.005 - 0.0025 = 0.0025$$

So, $$v' = v (1 + 0.0025) = 400 \times 1.0025 = 401 \, \text{m/s}$$.

Therefore, the speed changes approximately to $$401 \, \text{m/s}$$.

The correct option is C. 401.

For structure (P):

$$\Delta x = \pi R - 2R = (\pi - 2)\frac{l}{2}$$

$$\text{For maxima: } (\pi - 2)\frac{l}{2} = n\lambda$$

$$l_{\min} = \frac{2\lambda}{\pi - 2} = \frac{2(0.29)}{3.14 - 2} = 0.51\text{ m}$$

For structure (Q):

$$\Delta x = 2l - l = l$$

$$\text{For maxima: } l = n\lambda$$

$$l_{\min} = \lambda = 0.29\text{ m}$$

For structure (R):

$$\Delta x = l + \pi R - l = \pi R$$

$$\text{From geometry: } R = \frac{l}{\sqrt{2}} \implies \Delta x = \frac{\pi l}{\sqrt{2}}$$

$$\text{For maxima: } \frac{\pi l}{\sqrt{2}} = n\lambda$$

$$l_{\min} = \frac{\sqrt{2}\lambda}{\pi} = \frac{\sqrt{2}(0.29)}{3.14} = 0.13\text{ m}$$

For structure (S):

$$\Delta x = (\sqrt{2}x + 2x) - (x + \sqrt{3}x) = (\sqrt{2} + 2 - \sqrt{3} - 1)x = 0.682x$$

$$\text{Given: } x + \sqrt{3}x = l \implies x = \frac{l}{1 + \sqrt{3}} = 0.366l$$

$$\Delta x = 0.682(0.366l) = 0.249l$$

$$\text{For maxima: } 0.249l = n\lambda$$

$$l_{\min} = \frac{\lambda}{0.249} = \frac{0.29}{0.249} \approx 1.19\text{ m}$$

We need to find x where the ratio of lengths of closed pipe to open pipe is 5/x.

Key formulas:

For a closed organ pipe, the nth harmonic frequency (only odd harmonics): $$f_n = \frac{nv}{4L_c}$$ where n = 1, 3, 5, ...

For an open organ pipe, the nth harmonic: $$f_n = \frac{nv}{2L_o}$$ where n = 1, 2, 3, ...

Fifth harmonic of closed pipe = First harmonic of open pipe.

Fifth harmonic of closed pipe: $$f = \frac{5v}{4L_c}$$

First harmonic of open pipe: $$f = \frac{v}{2L_o}$$

Setting them equal:

$$\frac{5v}{4L_c} = \frac{v}{2L_o}$$

$$\frac{5}{4L_c} = \frac{1}{2L_o}$$

$$\frac{L_c}{L_o} = \frac{5 \times 2}{4} = \frac{10}{4} = \frac{5}{2}$$

So $$\frac{L_c}{L_o} = \frac{5}{2}$$, which means x = 2.

Therefore, the value of x is Option 3: 2.