Chapter 13: Oscillations

Both $$\sin \omega t$$ and $$\cos \omega t$$ are periodic, each with period $$2\pi/\omega$$. Their sum can be combined into a single sinusoid:

$$\sin \omega t + \cos \omega t = \sqrt{2}\left(\dfrac{1}{\sqrt{2}}\sin \omega t + \dfrac{1}{\sqrt{2}}\cos \omega t\right) = \sqrt{2}\,\sin\!\left(\omega t + \dfrac{\pi}{4}\right)$$

Adding $$2\pi$$ to the argument leaves the value of the sine unchanged, so

$$\sqrt{2}\,\sin\!\left(\omega t + \dfrac{\pi}{4}\right) = \sqrt{2}\,\sin\!\left[\omega\left(t + \dfrac{2\pi}{\omega}\right) + \dfrac{\pi}{4}\right]$$

The function repeats itself after a time $$2\pi/\omega$$. Hence it represents periodic motion with period $$T = \dfrac{2\pi}{\omega}$$.

Each term is periodic on its own. The period of a function is the least time after which it repeats its value:

• $$\sin \omega t$$ has period $$T_0 = \dfrac{2\pi}{\omega}$$,
• $$\cos 2\omega t$$ has period $$\dfrac{2\pi}{2\omega} = \dfrac{T_0}{2}$$,
• $$\sin 4\omega t$$ has period $$\dfrac{2\pi}{4\omega} = \dfrac{T_0}{4}$$.

The period of the first term, $$T_0$$, is an integral multiple of the periods of the other two terms ($$T_0 = 2 \times \tfrac{T_0}{2} = 4 \times \tfrac{T_0}{4}$$). Therefore, after a time $$T_0$$ all three terms simultaneously return to their starting values, and so does their sum.

Hence the function represents periodic motion with period $$T = \dfrac{2\pi}{\omega}$$.

The function $$e^{-\omega t}$$ is not periodic.

As the time $$t$$ increases, $$e^{-\omega t}$$ decreases steadily (monotonically) and tends to zero as $$t \to \infty$$. It never returns to a value it had earlier.

A function that does not repeat its values at regular intervals cannot represent periodic motion.

The function $$\log(\omega t)$$ is not periodic.

As the time $$t$$ increases, $$\log(\omega t)$$ increases monotonically and diverges to $$\infty$$ as $$t \to \infty$$. It never repeats its values, so it cannot represent periodic motion.

(It also grows without bound, so it could not represent any real physical displacement, which must stay finite.)

Write $$\cos \omega t = \sin\!\left(\dfrac{\pi}{2} - \omega t\right)$$, so that

$$\sin \omega t - \cos \omega t = \sin \omega t - \sin\!\left(\dfrac{\pi}{2} - \omega t\right)$$

Using the identity $$\sin C - \sin D = 2\cos\dfrac{C+D}{2}\,\sin\dfrac{C-D}{2}$$ with $$C = \omega t$$ and $$D = \dfrac{\pi}{2} - \omega t$$:

$$= 2\cos\!\left(\dfrac{\pi}{4}\right)\sin\!\left(\omega t - \dfrac{\pi}{4}\right) = \sqrt{2}\,\sin\!\left(\omega t - \dfrac{\pi}{4}\right)$$

This is exactly the standard SHM form $$A\sin(\omega t + \phi)$$, with amplitude $$A = \sqrt{2}$$ and phase constant $$\phi = -\dfrac{\pi}{4}$$.

Hence it represents simple harmonic motion with period $$T = \dfrac{2\pi}{\omega}$$.

Using the trigonometric identity $$\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$$:

$$\sin^2 \omega t = \dfrac{1}{2} - \dfrac{1}{2}\cos 2\omega t$$

The term $$\cos 2\omega t$$ is periodic with period $$\dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}$$. Hence the whole function is periodic with period $$T = \dfrac{\pi}{\omega}$$.

However it is not simple harmonic motion about the origin. A SHM must be a pure sine or cosine oscillating about $$x = 0$$. Here the constant term $$\tfrac{1}{2}$$ shifts the centre of oscillation: the motion is harmonic only about the point $$\tfrac{1}{2}$$, not about zero. As written, $$\sin^2\omega t$$ therefore represents periodic motion but not SHM in the standard sense.

Comparing the given equation $$x = 5\cos[2\pi t + \pi/4]$$ with the standard SHM form $$x = A\cos(\omega t + \phi)$$ (all quantities in SI units):

amplitude $$A = 5 \, \mathrm{m}$$, angular frequency $$\omega = 2\pi \, \mathrm{s^{-1}}$$, hence period $$T = \dfrac{2\pi}{\omega} = 1 \, \mathrm{s}$$.

The displacement at $$t = 1.5 \, \mathrm{s}$$ is

$$x = 5\cos\!\left[2\pi(1.5) + \dfrac{\pi}{4}\right] = 5\cos\!\left[3\pi + \dfrac{\pi}{4}\right]$$

Since $$\cos(3\pi + \theta) = \cos(\pi + \theta) = -\cos\theta$$,

$$x = 5\left(-\cos\dfrac{\pi}{4}\right) = 5 \times (-0.707) = -3.535 \, \mathrm{m}$$

The negative sign means the body is $$3.535 \, \mathrm{m}$$ from the mean position, on the negative side.

The velocity in SHM is the time derivative of the displacement:

$$v = \dfrac{dx}{dt} = -A\omega\sin(\omega t + \phi)$$

At $$t = 1.5 \, \mathrm{s}$$, with $$A = 5 \, \mathrm{m}$$ and $$\omega = 2\pi \, \mathrm{s^{-1}}$$:

$$v = -(5)(2\pi)\sin\!\left[2\pi(1.5) + \dfrac{\pi}{4}\right] = -10\pi\,\sin\!\left[3\pi + \dfrac{\pi}{4}\right]$$

Since $$\sin(3\pi + \theta) = \sin(\pi + \theta) = -\sin\theta$$,

$$v = -10\pi\left(-\sin\dfrac{\pi}{4}\right) = 10\pi \times 0.707$$

$$v \approx 31.4 \times 0.707 \approx 22 \, \mathrm{m\,s^{-1}}$$

So the speed of the body is about $$22 \, \mathrm{m\,s^{-1}}$$.

In SHM the acceleration is related to the displacement by $$a = -\omega^2 x$$.

Using $$\omega = 2\pi \, \mathrm{s^{-1}}$$ and the displacement found in part (a), $$x = -3.535 \, \mathrm{m}$$:

$$a = -\omega^2 x = -(2\pi)^2 \times (-3.535)$$

$$a = -4\pi^2 \times (-3.535) \approx -39.48 \times (-3.535)$$

$$a \approx 140 \, \mathrm{m\,s^{-2}}$$

The acceleration is positive — directed towards the mean position, since the displacement is negative.

The swimmer makes one return trip and then stops. The motion occurs only once — it is not repeated over and over at regular intervals of time.

For motion to be periodic it must repeat itself indefinitely at equal time intervals. Hence this is not periodic motion.

When a freely suspended bar magnet is displaced from its N–S equilibrium direction and released, the earth's magnetic field exerts a restoring torque on it.

The magnet therefore swings to and fro about the N–S direction, repeating its motion at regular intervals of time.

Hence this represents periodic motion (in fact, oscillatory motion).

A hydrogen molecule rotating about its centre of mass comes back to the same orientation after each complete rotation.

This happens at regular intervals of time (one rotational period). Hence this represents periodic motion.

An arrow released from a bow travels forward in a single flight. The motion happens only once and is not repeated.

Since it does not repeat itself at regular intervals, this is not periodic motion.

The earth spins about its axis, completing one rotation in about 24 hours. Since this repeats at regular intervals, the motion is periodic.

However, there is no to-and-fro motion about a mean position, and there is no restoring force proportional to a displacement. The conditions for SHM are not met.

Hence the rotation of the earth is periodic but not simple harmonic motion.

When the mercury column in a U-tube is displaced from its equilibrium (equal-level) position, the weight of the unbalanced column provides a restoring force.

This restoring force is proportional to the displacement of the column and is directed towards the equilibrium position. This is exactly the condition for SHM.

Hence the oscillating mercury column executes (nearly) simple harmonic motion.

For a small displacement near the lowest point of a smooth curved bowl, the arc behaves like a circular arc, so the ball bearing behaves just like the bob of a simple pendulum.

For such small displacements the restoring force is proportional to the displacement and directed towards the lowest (mean) position.

Hence the ball bearing executes (nearly) simple harmonic motion.

A polyatomic molecule has several natural (normal) modes of vibration, each with its own characteristic frequency.

Its general vibration is a superposition of these individual simple harmonic vibrations of different frequencies. Such a superposition repeats itself, so it is periodic, but it is not a single sinusoid.

Hence the general vibration of a polyatomic molecule is periodic but not simple harmonic motion. (Each individual normal mode, however, is itself SHM.)

Combine the two terms into a single sinusoid. Writing $$\cos\omega t = \sin\!\left(\dfrac{\pi}{2} - \omega t\right)$$ and using $$\sin C - \sin D = 2\cos\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$$:

$$\sin\omega t - \cos\omega t = \sin\omega t - \sin\!\left(\dfrac{\pi}{2} - \omega t\right) = \sqrt{2}\,\sin\!\left(\omega t - \dfrac{\pi}{4}\right)$$

This is of the standard SHM form $$A\sin(\omega t + \phi)$$. Hence it represents simple harmonic motion with period

$$T = \dfrac{2\pi}{\omega}$$

Use the identity $$\sin^3\theta = \dfrac{3\sin\theta - \sin 3\theta}{4}$$:

$$\sin^3\omega t = \dfrac{3\sin\omega t - \sin 3\omega t}{4}$$

This is the sum of two SHMs: $$\sin\omega t$$ (period $$2\pi/\omega$$) and $$\sin 3\omega t$$ (period $$2\pi/3\omega$$). The larger period $$2\pi/\omega$$ is an integral multiple ($$3\times$$) of the smaller one, so the sum repeats after a time $$2\pi/\omega$$.

Hence the motion is periodic with period $$T = \dfrac{2\pi}{\omega}$$, but since it is a superposition of two harmonics of different frequencies it is not simple harmonic.

Since cosine is an even function, $$\cos(-\theta) = \cos\theta$$, so

$$3\cos\!\left(\dfrac{\pi}{4} - 2\omega t\right) = 3\cos\!\left(2\omega t - \dfrac{\pi}{4}\right)$$

This is of the standard SHM form $$A\cos(\Omega t + \phi)$$ with amplitude $$A = 3$$, angular frequency $$\Omega = 2\omega$$ and phase $$\phi = -\pi/4$$.

Hence it represents simple harmonic motion with period

$$T = \dfrac{2\pi}{\Omega} = \dfrac{2\pi}{2\omega} = \dfrac{\pi}{\omega}$$

Each term is a SHM on its own:

• $$\cos\omega t$$ has period $$\dfrac{2\pi}{\omega}$$,
• $$\cos 3\omega t$$ has period $$\dfrac{2\pi}{3\omega}$$,
• $$\cos 5\omega t$$ has period $$\dfrac{2\pi}{5\omega}$$.

The largest period $$\dfrac{2\pi}{\omega}$$ is an integral multiple ($$3\times$$ and $$5\times$$) of the other two periods. Therefore after a time $$\dfrac{2\pi}{\omega}$$ all three terms return together and the sum repeats.

Hence the motion is periodic with period $$T = \dfrac{2\pi}{\omega}$$, but being a superposition of three harmonics of different frequencies it is not simple harmonic.

The function $$\exp(-\omega^2 t^2) = e^{-\omega^2 t^2}$$ is a Gaussian. It has its maximum value $$1$$ at $$t = 0$$ and decreases steadily towards zero as $$|t|$$ increases.

It never returns to a value once it has passed it, so it never repeats. Hence it represents non-periodic motion.

The function $$1 + \omega t + \omega^2 t^2$$ is a polynomial in $$t$$. As $$t$$ increases it grows without bound and never repeats its value.

Hence it represents non-periodic motion. (Being unbounded, it also cannot describe any real oscillatory displacement, which must stay finite.)

The mid-point O of AB is the mean position; OA = OB = 5 cm. The direction A → B is taken as positive. In SHM the acceleration and force always point towards the mean position O, and the force has the same sign as the acceleration ($$F = ma$$).

End A is an extreme point of the motion.

Velocity: at an extreme position the particle is momentarily at rest, so the velocity is zero.

Acceleration and force: they point towards O. From A, the mean position O lies in the direction of B, which is the positive direction. Hence the acceleration is positive and the force is positive.

End B is the other extreme point of the motion.

Velocity: at an extreme position the particle is momentarily at rest, so the velocity is zero.

Acceleration and force: they point towards the mean position O. From B, the mean position O lies in the direction of A, which is the negative direction. Hence the acceleration is negative and the force is negative.

The mid-point of AB is the mean position O.

Velocity: the speed is maximum at the mean position. Here the particle is moving towards A, i.e. in the negative direction, so the velocity is negative.

Acceleration and force: at the mean position the displacement from O is zero, so the acceleration is zero and hence the force is also zero.

A point $$2 \, \mathrm{cm}$$ from B is $$10 - 2 = 8 \, \mathrm{cm}$$ from A. Since O is $$5 \, \mathrm{cm}$$ from A, this point lies on the B-side of the mean position (its displacement from O is $$+3 \, \mathrm{cm}$$, towards B).

Velocity: the particle is moving towards A, i.e. in the negative direction, so the velocity is negative.

Acceleration and force: they point towards O. From a point on the B-side, O lies in the direction of A, which is the negative direction. Hence the acceleration is negative and the force is negative.

A point $$3 \, \mathrm{cm}$$ from A lies on the A-side of the mean position O (which is $$5 \, \mathrm{cm}$$ from A). Its displacement from O is $$2 \, \mathrm{cm}$$, towards A.

Velocity: the particle is moving towards B, i.e. in the positive direction, so the velocity is positive.

Acceleration and force: they point towards O. From a point on the A-side, O lies in the direction of B, which is the positive direction. Hence the acceleration is positive and the force is positive.

A point $$4 \, \mathrm{cm}$$ from B is $$10 - 4 = 6 \, \mathrm{cm}$$ from A. Since O is $$5 \, \mathrm{cm}$$ from A, this point lies on the B-side of the mean position (its displacement from O is $$+1 \, \mathrm{cm}$$, towards B).

Velocity: the particle is moving towards A, i.e. in the negative direction, so the velocity is negative.

Acceleration and force: they point towards O. From a point on the B-side, O lies in the direction of A, which is the negative direction. Hence the acceleration is negative and the force is negative.

For simple harmonic motion the acceleration must satisfy $$a = -\omega^2 x$$ — it must be proportional to the first power of the displacement and directed opposite to it (towards the mean position).

Here $$a = 0.7x$$: the acceleration is in the same direction as the displacement (the coefficient $$+0.7$$ is positive). It is not a restoring acceleration.

Hence this does not represent SHM.

For SHM the acceleration must be proportional to the first power of the displacement: $$a = -\omega^2 x$$.

Here $$a = -200x^2$$ is proportional to $$x^2$$, not to $$x$$. The relation is not linear in $$x$$. (Also, since $$x^2$$ is always positive, the acceleration is negative for both positive and negative $$x$$, so it is not always a restoring acceleration.)

Hence this does not represent SHM.

For SHM the acceleration must satisfy $$a = -\omega^2 x$$.

Here $$a = -10x$$ is proportional to the first power of $$x$$ and is directed opposite to the displacement (towards the mean position). It is exactly of the SHM form, with

$$\omega^2 = 10 \implies \omega = \sqrt{10} \, \mathrm{rad\,s^{-1}}$$

Hence this represents simple harmonic motion.

For SHM the acceleration must be proportional to the first power of the displacement and directed opposite to it: $$a = -\omega^2 x$$.

Here $$a = 100x^3$$ is proportional to $$x^3$$, not to $$x$$, and the coefficient $$+100$$ is positive — so the acceleration is in the same direction as the displacement. It is neither linear in $$x$$ nor restoring.

Hence this does not represent SHM.

From Exercise 13.9 the amplitude is $$A = 2 \, \mathrm{cm}$$ and the angular frequency is $$\omega = 20 \, \mathrm{rad\,s^{-1}}$$.

At $$t = 0$$ the mass is at the mean position, so the displacement is zero at $$t = 0$$: $$x(0) = 0$$.

A function that gives zero at $$t = 0$$ is the sine function. Hence

$$x(t) = A\sin\omega t = 2\sin 20t$$

(here $$x$$ is in cm and $$t$$ in s).

The amplitude is $$A = 2 \, \mathrm{cm}$$ and $$\omega = 20 \, \mathrm{rad\,s^{-1}}$$.

At $$t = 0$$ the mass is at the maximum stretched position. Stretching the spring moves the mass in the positive ($$+x$$) direction (left to right), so $$x = +A$$ at $$t = 0$$.

A function that gives its maximum positive value at $$t = 0$$ is the cosine function. Hence

$$x(t) = A\cos\omega t = 2\cos 20t$$

(here $$x$$ is in cm and $$t$$ in s).

The amplitude is $$A = 2 \, \mathrm{cm}$$ and $$\omega = 20 \, \mathrm{rad\,s^{-1}}$$.

At $$t = 0$$ the mass is at the maximum compressed position. Compressing the spring moves the mass in the negative ($$-x$$) direction, so $$x = -A$$ at $$t = 0$$. Hence

$$x(t) = -A\cos\omega t = -2\cos 20t$$

(here $$x$$ is in cm and $$t$$ in s).

Comparison of the three functions. Writing each as a cosine: (a) $$2\sin 20t = 2\cos\!\left(20t - \dfrac{\pi}{2}\right)$$, (b) $$2\cos 20t$$, (c) $$-2\cos 20t = 2\cos(20t + \pi)$$.

All three have the same amplitude ($$2 \, \mathrm{cm}$$) and the same angular frequency / frequency ($$\omega = 20 \, \mathrm{rad\,s^{-1}}$$). They differ only in the initial phase — namely $$-\dfrac{\pi}{2}$$, $$0$$ and $$\pi$$ respectively.

To read off the reference circle, convert the function to the standard form $$x = A\cos(\omega t + \phi)$$ (this represents the $$x$$-projection of a particle rotating anticlockwise).

Use the identity $$-\sin\theta = \cos\!\left(\theta + \dfrac{\pi}{2}\right)$$:

$$x = -2\sin\!\left(3t + \dfrac{\pi}{3}\right) = 2\cos\!\left(3t + \dfrac{\pi}{3} + \dfrac{\pi}{2}\right) = 2\cos\!\left(3t + \dfrac{5\pi}{6}\right)$$

Comparing with $$x = A\cos(\omega t + \phi)$$:

• Radius of the circle (amplitude): $$A = 2 \, \mathrm{cm}$$.
• Angular speed of the rotating particle: $$\omega = 3 \, \mathrm{rad\,s^{-1}}$$.
• Initial phase: $$\phi = \dfrac{5\pi}{6} = 150°$$.

So at $$t = 0$$ the radius vector OP points at $$150°$$, measured anticlockwise from the positive $$x$$-axis. Its $$x$$-projection at $$t = 0$$ is $$x(0) = 2\cos 150° = -\sqrt{3} \approx -1.73 \, \mathrm{cm}$$.

Since cosine is an even function, $$\cos(-\theta) = \cos\theta$$, so

$$x = \cos\!\left(\dfrac{\pi}{6} - t\right) = \cos\!\left(t - \dfrac{\pi}{6}\right)$$

Comparing with the standard form $$x = A\cos(\omega t + \phi)$$:

• Radius of the circle (amplitude): $$A = 1 \, \mathrm{cm}$$.
• Angular speed of the rotating particle: $$\omega = 1 \, \mathrm{rad\,s^{-1}}$$.
• Initial phase: $$\phi = -\dfrac{\pi}{6} = -30°$$.

So at $$t = 0$$ the radius vector OP points at $$-30°$$ (i.e. $$30°$$ below the positive $$x$$-axis). Its $$x$$-projection at $$t = 0$$ is $$x(0) = \cos(-30°) = \dfrac{\sqrt{3}}{2} \approx 0.87 \, \mathrm{cm}$$.

Convert the sine to a cosine using $$\sin\theta = \cos\!\left(\theta - \dfrac{\pi}{2}\right)$$:

$$x = 3\sin\!\left(2\pi t + \dfrac{\pi}{4}\right) = 3\cos\!\left(2\pi t + \dfrac{\pi}{4} - \dfrac{\pi}{2}\right) = 3\cos\!\left(2\pi t - \dfrac{\pi}{4}\right)$$

Comparing with the standard form $$x = A\cos(\omega t + \phi)$$:

• Radius of the circle (amplitude): $$A = 3 \, \mathrm{cm}$$.
• Angular speed of the rotating particle: $$\omega = 2\pi \, \mathrm{rad\,s^{-1}}$$.
• Initial phase: $$\phi = -\dfrac{\pi}{4} = -45°$$.

So at $$t = 0$$ the radius vector OP points at $$-45°$$ from the positive $$x$$-axis. Its $$x$$-projection at $$t = 0$$ is $$x(0) = 3\cos(-45°) \approx 2.12 \, \mathrm{cm}$$.

The function $$x = 2\cos\pi t$$ is already in the standard form $$x = A\cos(\omega t + \phi)$$.

Comparing directly:

• Radius of the circle (amplitude): $$A = 2 \, \mathrm{cm}$$.
• Angular speed of the rotating particle: $$\omega = \pi \, \mathrm{rad\,s^{-1}}$$.
• Initial phase: $$\phi = 0$$.

So at $$t = 0$$ the radius vector OP lies along the positive $$x$$-axis, and the $$x$$-projection is $$x(0) = 2\cos 0 = 2 \, \mathrm{cm}$$.

Figure (a): One end of the spring is clamped to a rigid wall and a force $$F$$ is applied at the free end. In equilibrium the wall supplies an equal and opposite reaction $$F$$. The spring is therefore under a tension $$F$$, so by Hooke's law its extension is

$$x_a = \dfrac{F}{k}$$

Figure (b): Both ends of the spring are free, and each end is pulled outward by a force $$F$$. These two equal and opposite forces keep the spring in equilibrium. The tension at every cross-section of the spring is still $$F$$ (it is not $$2F$$ — a stretched spring pulled at both ends by $$F$$ has tension $$F$$ throughout, just as in case (a) where the wall plays the role of the second $$F$$). Hence its extension is

$$x_b = \dfrac{F}{k}$$

Therefore the maximum extension of the spring is the same in both cases, equal to $$\dfrac{F}{k}$$.

Case (a): a single mass $$m$$ attached to a spring of force constant $$k$$ whose other end is fixed. This is the standard linear oscillator, for which the restoring force is $$F = -kx$$ and the period is

$$T_a = 2\pi\sqrt{\dfrac{m}{k}}$$

Case (b): two equal masses $$m$$ attached to the two ends of the same spring. By symmetry, the mid-point of the spring does not move at all — it behaves like a fixed point.

Each mass therefore behaves as if it were attached to only half of the spring. A spring of half the length is stiffer: halving the length doubles the force constant, so each half has force constant $$2k$$.

Hence each mass oscillates on an effective spring of constant $$2k$$, giving

$$T_b = 2\pi\sqrt{\dfrac{m}{2k}}$$

(The same result follows by treating it as a two-body problem with reduced mass $$\mu = \dfrac{m \cdot m}{m + m} = \dfrac{m}{2}$$, so that $$T_b = 2\pi\sqrt{\dfrac{\mu}{k}} = 2\pi\sqrt{\dfrac{m}{2k}}$$.)