Chapter 14: Waves

A wave is transverse if the particles of the medium oscillate perpendicular to the direction in which the wave travels, and longitudinal if they oscillate along the direction of travel.

(1) Kink in a spring displaced sideways: One end of the spring is moved sideways, so the kink (the disturbance) is perpendicular to the length of the spring, while it travels along the spring. The particles move at right angles to the direction of propagation, so this is a transverse wave.

(2) Waves in a liquid produced by a piston: The piston moves back and forth along the axis of the cylinder, producing compressions and rarefactions of the liquid in the same direction in which the wave travels. Hence this is a longitudinal wave.

(3) Waves produced by a motorboat: On the surface of water the particles trace nearly circular paths — they move both up-and-down and back-and-forth. Such surface waves are a combination of both transverse and longitudinal motion.

(4) Ultrasonic waves in air from a quartz crystal: Sound waves (including ultrasonic waves) in a gas propagate as compressions and rarefactions, with the air particles oscillating along the direction of propagation. Hence this is a longitudinal wave.

The given wave is $$y(x,t) = 0.005 \sin(80.0\,x - 3.0\,t)$$. Compare it with the standard form of a travelling harmonic wave

$$y(x,t) = a \sin(kx - \omega t)$$

This identifies the amplitude $$a = 0.005 \, \mathrm{m}$$, the angular wave number $$k = 80.0 \, \mathrm{rad\,m^{-1}}$$ and the angular frequency $$\omega = 3.0 \, \mathrm{rad\,s^{-1}}$$.

(a) Amplitude. The amplitude is the coefficient of the sine function:

$$a = 0.005 \, \mathrm{m} = 5 \, \mathrm{mm}$$

(b) Wavelength. The angular wave number is related to the wavelength by $$k = \dfrac{2\pi}{\lambda}$$, so

$$\lambda = \dfrac{2\pi}{k} = \dfrac{2\pi}{80.0} = 7.85 \times 10^{-2} \, \mathrm{m} = 7.85 \, \mathrm{cm}$$

(c) Period and frequency. The angular frequency is related to the period by $$\omega = \dfrac{2\pi}{T}$$, so

$$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{3.0} = 2.09 \, \mathrm{s}$$

The frequency is the reciprocal of the period:

$$\nu = \dfrac{1}{T} = \dfrac{1}{2.09} = 0.48 \, \mathrm{Hz}$$

Displacement at $$x = 30.0 \, \mathrm{cm} = 0.30 \, \mathrm{m}$$ and $$t = 20 \, \mathrm{s}$$. Substitute these values into the wave equation:

$$y = 0.005 \sin\big(80.0 \times 0.30 - 3.0 \times 20\big)$$

$$y = 0.005 \sin(24 - 60) = 0.005 \sin(-36 \, \mathrm{rad})$$

Since the sine function repeats every $$2\pi$$, we may add a whole number of cycles: $$-36 + 6(2\pi) = 1.70 \, \mathrm{rad}$$, so $$\sin(-36) = \sin(1.70) \approx 0.99$$. Therefore

$$y = 0.005 \times 0.99 \approx 5 \times 10^{-3} \, \mathrm{m} = 5 \, \mathrm{mm}$$

The speed of a transverse wave on a stretched wire is

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

where $$T$$ is the tension and $$\mu$$ is the linear mass density (mass per unit length).

Linear mass density.

$$\mu = \dfrac{m}{L} = \dfrac{5.0 \times 10^{-3} \, \mathrm{kg}}{0.72 \, \mathrm{m}} = 6.9 \times 10^{-3} \, \mathrm{kg\,m^{-1}}$$

Speed of the transverse wave.

$$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{60 \, \mathrm{N}}{6.9 \times 10^{-3} \, \mathrm{kg\,m^{-1}}}}$$

$$v = \sqrt{8.64 \times 10^{3}} \approx 93 \, \mathrm{m\,s^{-1}}$$

Newton's formula. Newton assumed that sound propagates through a gas under isothermal conditions, which gives the speed

$$v = \sqrt{\dfrac{P}{\rho}}$$

At standard temperature and pressure (STP), $$P = 1.01 \times 10^{5} \, \mathrm{Pa}$$ and $$T = 273 \, \mathrm{K}$$. One mole of any gas occupies $$22.4 \times 10^{-3} \, \mathrm{m^3}$$ at STP, and the mass of one mole of air is $$29.0 \times 10^{-3} \, \mathrm{kg}$$. Hence the density of air is

$$\rho = \dfrac{\text{mass of one mole}}{\text{volume of one mole}} = \dfrac{29.0 \times 10^{-3}}{22.4 \times 10^{-3}} = 1.29 \, \mathrm{kg\,m^{-3}}$$

Newton's formula then gives

$$v = \sqrt{\dfrac{1.01 \times 10^{5}}{1.29}} = \sqrt{7.83 \times 10^{4}} \approx 280 \, \mathrm{m\,s^{-1}}$$

This is about 15% lower than the experimentally measured value of about $$331 \, \mathrm{m\,s^{-1}}$$.

Laplace correction. Laplace pointed out that the compressions and rarefactions in a sound wave occur so rapidly that the process is adiabatic, not isothermal. The correct speed is therefore

$$v = \sqrt{\dfrac{\gamma P}{\rho}}$$

where $$\gamma = C_P/C_V$$. For air, which is mostly diatomic, $$\gamma = 1.4$$. Hence

$$v = \sqrt{\dfrac{1.4 \times 1.01 \times 10^{5}}{1.29}} = \sqrt{1.4} \times 280 \approx 331 \, \mathrm{m\,s^{-1}}$$

which agrees closely with the measured speed of sound in air.

Pipe open at both ends. For a pipe open at both ends, the resonant (harmonic) frequencies are

$$\nu_n = \dfrac{n\,v}{2L}, \qquad n = 1, 2, 3, \dots$$

The fundamental frequency ($$n = 1$$) is

$$\nu_1 = \dfrac{v}{2L} = \dfrac{330}{2 \times 0.30} = 550 \, \mathrm{Hz}$$

For resonance with the source of frequency $$1.1 \, \mathrm{kHz} = 1100 \, \mathrm{Hz}$$,

$$n = \dfrac{\nu_n}{\nu_1} = \dfrac{1100}{550} = 2$$

So the source resonates with the second harmonic ($$n = 2$$) of the open pipe.

Pipe closed at one end. For a pipe closed at one end, only the odd harmonics are present:

$$\nu_n = \dfrac{(2n-1)\,v}{4L}, \qquad n = 1, 2, 3, \dots$$

The fundamental frequency is

$$\nu_1 = \dfrac{v}{4L} = \dfrac{330}{4 \times 0.30} = 275 \, \mathrm{Hz}$$

To resonate at $$1100 \, \mathrm{Hz}$$ we would need $$\dfrac{1100}{275} = 4$$, i.e. the 4th harmonic. But a closed pipe supports only the odd harmonics ($$1, 3, 5, \dots$$). Since $$4$$ is even, the closed pipe cannot resonate with the same $$1.1 \, \mathrm{kHz}$$ source.

When two notes of nearly equal frequency are sounded together, the beat frequency equals the difference of the two frequencies:

$$\nu_{\text{beat}} = |\nu_A - \nu_B|$$

Initially the beat frequency is $$5 \, \mathrm{Hz}$$ and $$\nu_A = 427 \, \mathrm{Hz}$$, so the original frequency of B is either

$$\nu_B = 427 + 5 = 432 \, \mathrm{Hz} \quad \text{or} \quad \nu_B = 427 - 5 = 422 \, \mathrm{Hz}$$

To decide which value is correct, use the extra information. The tension of string B is increased. Since the frequency of a stretched string increases with tension ($$\nu \propto \sqrt{T}$$), the frequency of B increases.

• If $$\nu_B = 432 \, \mathrm{Hz}$$: increasing it moves it further from $$427 \, \mathrm{Hz}$$, so the beat frequency would increase. This contradicts the observation.
• If $$\nu_B = 422 \, \mathrm{Hz}$$: increasing it moves it closer to $$427 \, \mathrm{Hz}$$, so the beat frequency decreases (here from $$5 \, \mathrm{Hz}$$ to $$3 \, \mathrm{Hz}$$). This matches the observation.

Hence the original frequency of string B is $$\nu_B = 422 \, \mathrm{Hz}$$.

The transverse jerk travels along the string as a transverse wave with speed

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

where $$T$$ is the tension and $$\mu = m/L$$ is the linear mass density of the string.

Linear mass density.

$$\mu = \dfrac{m}{L} = \dfrac{2.50 \, \mathrm{kg}}{20.0 \, \mathrm{m}} = 0.125 \, \mathrm{kg\,m^{-1}}$$

Speed of the disturbance.

$$v = \sqrt{\dfrac{T}{\mu}} = \sqrt{\dfrac{200}{0.125}} = \sqrt{1600} = 40 \, \mathrm{m\,s^{-1}}$$

Time to travel the length of the string. The disturbance travels the full length $$L = 20.0 \, \mathrm{m}$$ at this constant speed:

$$t = \dfrac{L}{v} = \dfrac{20.0}{40} = 0.50 \, \mathrm{s}$$

The total time between dropping the stone and hearing the splash is the sum of two parts: the time $$t_1$$ for the stone to fall to the water, and the time $$t_2$$ for the sound of the splash to travel back up the tower.

Time of fall of the stone. The stone is dropped from rest, so its initial velocity is zero. Using $$h = \dfrac{1}{2}g t_1^{2}$$ with $$h = 300 \, \mathrm{m}$$:

$$t_1 = \sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2 \times 300}{9.8}} = \sqrt{61.2} = 7.82 \, \mathrm{s}$$

Time for the sound to travel up. The splash sound travels the height $$h$$ at the speed of sound:

$$t_2 = \dfrac{h}{v} = \dfrac{300}{340} = 0.88 \, \mathrm{s}$$

Total time.

$$t = t_1 + t_2 = 7.82 + 0.88 = 8.7 \, \mathrm{s}$$

The speed of a transverse wave on the wire is $$v = \sqrt{\dfrac{T}{\mu}}$$, where $$\mu = m/L$$ is the linear mass density. We require this speed to equal the speed of sound in dry air, $$v = 343 \, \mathrm{m\,s^{-1}}$$.

Linear mass density.

$$\mu = \dfrac{m}{L} = \dfrac{2.10 \, \mathrm{kg}}{12.0 \, \mathrm{m}} = 0.175 \, \mathrm{kg\,m^{-1}}$$

Tension. Squaring $$v = \sqrt{T/\mu}$$ gives $$T = v^{2}\mu$$:

$$T = (343)^{2} \times 0.175$$

$$T = 1.176 \times 10^{5} \times 0.175 \approx 2.06 \times 10^{4} \, \mathrm{N}$$

The frequency emitted by the bat is

$$\nu = 1000 \, \mathrm{kHz} = 1000 \times 10^{3} \, \mathrm{Hz} = 1.0 \times 10^{6} \, \mathrm{Hz}$$

(a) Reflected sound. The reflected sound travels back in air — the same medium — so both its speed and its frequency are unchanged. Its wavelength is

$$\lambda_{\text{air}} = \dfrac{v_{\text{air}}}{\nu} = \dfrac{340}{1.0 \times 10^{6}} = 3.4 \times 10^{-4} \, \mathrm{m}$$

(b) Transmitted sound. When the sound passes into water, its frequency stays the same — the frequency is fixed by the source and does not change on entering a new medium. Only the speed changes, to $$v_{\text{water}} = 1486 \, \mathrm{m\,s^{-1}}$$. Hence

$$\lambda_{\text{water}} = \dfrac{v_{\text{water}}}{\nu} = \dfrac{1486}{1.0 \times 10^{6}} = 1.49 \times 10^{-3} \, \mathrm{m}$$

The wavelength is related to the speed and the frequency by

$$\lambda = \dfrac{v}{\nu}$$

Here the speed of sound in the tissue is

$$v = 1.7 \, \mathrm{km\,s^{-1}} = 1.7 \times 10^{3} \, \mathrm{m\,s^{-1}}$$

and the operating frequency of the scanner is

$$\nu = 4.2 \, \mathrm{MHz} = 4.2 \times 10^{6} \, \mathrm{Hz}$$

Therefore

$$\lambda = \dfrac{v}{\nu} = \dfrac{1.7 \times 10^{3}}{4.2 \times 10^{6}} \approx 4.1 \times 10^{-4} \, \mathrm{m}$$

The wave is $$y(x,t) = 3.0 \sin(36\,t + 0.018\,x + \pi/4)$$, with $$x$$ and $$y$$ in cm and $$t$$ in s. To draw a $$y$$–$$t$$ graph we fix $$x$$ and treat $$y$$ as a function of $$t$$.

At $$x = 0$$: $$\quad y = 3.0 \sin(36\,t + \pi/4)$$

At $$x = 2 \, \mathrm{cm}$$: $$\quad y = 3.0 \sin(36\,t + 0.036 + \pi/4)$$

At $$x = 4 \, \mathrm{cm}$$: $$\quad y = 3.0 \sin(36\,t + 0.072 + \pi/4)$$

Each of these is a sine function of time, so each $$y$$–$$t$$ graph is a sinusoidal curve with amplitude $$3.0 \, \mathrm{cm}$$ and period

$$T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{36} \approx 0.175 \, \mathrm{s}$$

To draw them: plot $$y$$ on the vertical axis (from $$-3.0$$ to $$+3.0 \, \mathrm{cm}$$) and $$t$$ on the horizontal axis, marking off intervals of one period $$\approx 0.175 \, \mathrm{s}$$. The curve for $$x = 0$$ starts at $$t = 0$$ with $$y = 3.0 \sin(\pi/4) = +2.12 \, \mathrm{cm}$$.

The only difference between the three cases is the small extra constant added inside the sine: $$0$$, $$0.036 \, \mathrm{rad}$$ and $$0.072 \, \mathrm{rad}$$. These are tiny (a few thousandths of a radian), so the three sinusoidal graphs are practically indistinguishable — each is shifted from the next by only an extremely small phase.

Hence, as we move from one point on the string to another, the oscillation keeps the same amplitude ($$3.0 \, \mathrm{cm}$$) and the same frequency ($$\nu \approx 5.7 \, \mathrm{Hz}$$); only the phase of the oscillation changes from point to point.

This question refers to the standing wave of Exercise 14.11:

$$y(x,t) = 0.06 \sin\!\left(\dfrac{2\pi}{3}x\right)\cos(120\pi t)$$

(i)(a) Frequency. The time-dependence of every point of the string is the same common factor $$\cos(120\pi t)$$. So all points (except the nodes, which never move) oscillate with the same frequency,

$$\nu = \dfrac{120\pi}{2\pi} = 60 \, \mathrm{Hz}$$

(i)(b) Phase. Again the time factor common to all points is $$\cos(120\pi t)$$ — there is no point-to-point phase lag in time. All points pass through their mean positions, and reach their extreme positions, at the same instants. So all points oscillate in phase (within any one loop between two nodes). Across a node the displacement factor $$\sin(2\pi x/3)$$ changes sign, which simply means the next loop moves in the opposite direction — this is a sign reversal, not a continuously varying phase lag.

(i)(c) Amplitude. The amplitude of the point at position $$x$$ is the magnitude of the $$x$$-dependent factor, $$\big|0.06 \sin(2\pi x/3)\big|$$, which clearly depends on $$x$$. So the points do not all have the same amplitude: it is zero at the nodes and a maximum of $$0.06 \, \mathrm{m}$$ at the antinodes.

(ii) Amplitude at $$x = 0.375 \, \mathrm{m}$$. The amplitude is the magnitude of the displacement factor at that point:

$$A = 0.06 \sin\!\left(\dfrac{2\pi}{3} \times 0.375\right) = 0.06 \sin\!\left(\dfrac{\pi}{4}\right)$$

$$A = 0.06 \times \dfrac{1}{\sqrt{2}} = 0.06 \times 0.707 \approx 0.042 \, \mathrm{m}$$

The wire is fixed at both ends and vibrates in its fundamental mode. In the fundamental mode the wire forms a single loop, so its length is half a wavelength:

$$L = \dfrac{\lambda}{2} \qquad\Rightarrow\qquad \lambda = 2L$$

Length of the wire. From the total mass and the linear mass density,

$$L = \dfrac{m}{\mu} = \dfrac{3.5 \times 10^{-2} \, \mathrm{kg}}{4.0 \times 10^{-2} \, \mathrm{kg\,m^{-1}}} = 0.875 \, \mathrm{m}$$

So the wavelength of the fundamental mode is

$$\lambda = 2L = 2 \times 0.875 = 1.75 \, \mathrm{m}$$

(a) Speed of the transverse wave. Using $$v = \nu\lambda$$ with the fundamental frequency $$\nu = 45 \, \mathrm{Hz}$$:

$$v = 45 \times 1.75 = 78.75 \, \mathrm{m\,s^{-1}} \approx 79 \, \mathrm{m\,s^{-1}}$$

(b) Tension in the string. From $$v = \sqrt{T/\mu}$$ we get $$T = v^{2}\mu$$:

$$T = (78.75)^{2} \times 4.0 \times 10^{-2}$$

$$T = 6.20 \times 10^{3} \times 4.0 \times 10^{-2} \approx 248 \, \mathrm{N}$$

The tube is open at one end and closed by the movable piston at the other end — it therefore behaves as a closed pipe (closed at one end). Resonance occurs when there is a displacement antinode at the open end and a node at the piston, which happens for air-column lengths

$$L = \dfrac{\lambda}{4},\quad \dfrac{3\lambda}{4},\quad \dfrac{5\lambda}{4},\quad \dots$$

Hence two consecutive resonance lengths differ by exactly half a wavelength:

$$\Delta L = \dfrac{\lambda}{2}$$

The two observed resonance lengths are $$L_1 = 25.5 \, \mathrm{cm}$$ and $$L_2 = 79.3 \, \mathrm{cm}$$, so

$$\dfrac{\lambda}{2} = L_2 - L_1 = 79.3 - 25.5 = 53.8 \, \mathrm{cm}$$

$$\lambda = 2 \times 53.8 = 107.6 \, \mathrm{cm} = 1.076 \, \mathrm{m}$$

The frequency is fixed by the tuning fork, $$\nu = 340 \, \mathrm{Hz}$$. The speed of sound in air is therefore

$$v = \nu\lambda = 340 \times 1.076 \approx 366 \, \mathrm{m\,s^{-1}}$$

The steel rod is clamped at its middle. The clamp holds the centre fixed, so there is a displacement node at the middle of the rod, while the two free ends are displacement antinodes.

In the fundamental mode of longitudinal vibration, the rod has exactly one node (at the centre) with an antinode at each end. The distance from one end (antinode) to the other end (antinode) — the full length of the rod — equals half a wavelength:

$$L = \dfrac{\lambda}{2} \qquad\Rightarrow\qquad \lambda = 2L = 2 \times 100 \, \mathrm{cm} = 200 \, \mathrm{cm} = 2.0 \, \mathrm{m}$$

The fundamental frequency is $$\nu = 2.53 \, \mathrm{kHz} = 2.53 \times 10^{3} \, \mathrm{Hz}$$. The speed of sound (longitudinal waves) in steel is

$$v = \nu\lambda = 2.53 \times 10^{3} \times 2.0$$

$$v = 5.06 \times 10^{3} \, \mathrm{m\,s^{-1}} = 5.06 \, \mathrm{km\,s^{-1}}$$

Pipe closed at one end. A pipe closed at one end resonates only at the odd-harmonic frequencies

$$\nu_n = \dfrac{(2n-1)\,v}{4L}, \qquad n = 1, 2, 3, \dots$$

The fundamental frequency ($$n = 1$$) is

$$\nu_1 = \dfrac{v}{4L} = \dfrac{340}{4 \times 0.20} = \dfrac{340}{0.80} = 425 \, \mathrm{Hz}$$

The source frequency, $$430 \, \mathrm{Hz}$$, is essentially equal to this fundamental frequency (the small difference lies within experimental tolerance). Hence the $$430 \, \mathrm{Hz}$$ source excites the fundamental mode (the first harmonic), $$n = 1$$, of the closed pipe.

Pipe open at both ends. A pipe open at both ends contains all harmonics:

$$\nu_n' = \dfrac{n\,v}{2L} = \dfrac{n \times 340}{2 \times 0.20} = n \times 850 \, \mathrm{Hz}, \qquad n = 1, 2, 3, \dots$$

For resonance we would need $$n = \dfrac{430}{850} \approx 0.5$$, which is not a whole number. The lowest possible resonance of the open pipe is its fundamental at $$850 \, \mathrm{Hz}$$. Therefore the $$430 \, \mathrm{Hz}$$ source will not be in resonance with the pipe when both ends are open.

When two notes of nearly equal frequency are sounded together, the beat frequency equals the difference of the two frequencies:

$$\nu_{\text{beat}} = |\nu_A - \nu_B|$$

Initially the beat frequency is $$6 \, \mathrm{Hz}$$ and $$\nu_A = 324 \, \mathrm{Hz}$$, so the frequency of B is either

$$\nu_B = 324 + 6 = 330 \, \mathrm{Hz} \quad \text{or} \quad \nu_B = 324 - 6 = 318 \, \mathrm{Hz}$$

To decide which value is correct, use the extra information. The tension in string A is reduced. Since the frequency of a stretched string varies as $$\nu \propto \sqrt{T}$$, reducing the tension makes the frequency of A decrease below $$324 \, \mathrm{Hz}$$.

• If $$\nu_B = 330 \, \mathrm{Hz}$$: lowering $$\nu_A$$ moves it further from $$330 \, \mathrm{Hz}$$, so the beat frequency would increase. This contradicts the observation.
• If $$\nu_B = 318 \, \mathrm{Hz}$$: lowering $$\nu_A$$ moves it closer to $$318 \, \mathrm{Hz}$$, so the beat frequency decreases (here from $$6 \, \mathrm{Hz}$$ to $$3 \, \mathrm{Hz}$$). This matches the observation.

Hence the frequency of string B is $$\nu_B = 318 \, \mathrm{Hz}$$.