Oscillations

Examples (a), (b), and (c) represent periodic motion.

- (a) A swimmer returning to the same bank and back repeats the motion.
- (b) A freely suspended bar magnet displaced and released oscillates periodically.
- (c) A hydrogen molecule rotating about its centre of mass repeats its position after every revolution.
- (d) An arrow released from a bow does not repeat its motion, so it is not periodic.

The motions that are (nearly) simple harmonic are (b) and (c).

- (a) Rotation of earth about its axis: periodic, but not SHM.
- (b) Oscillating mercury column in a U-tube: for small oscillations, it is nearly SHM.
- (c) Ball bearing inside a smooth curved bowl released from a point slightly above the lowest point: for small displacements, it is nearly SHM.
- (d) General vibrations of a polyatomic molecule: periodic but not simple harmonic in general.

So the correct classification is: periodic but not SHM: (a) and (d); nearly SHM: (b) and (c).

In the textbook, you identify periodic motion by checking whether the graph repeats itself after a fixed time interval, and the period is that smallest repeating interval. If you share the figure, I can name exactly which plots are periodic and their periods.

The functions are classified as follows:

- (a) $\sin \omega t - \cos \omega t$ is simple harmonic. It can be written as
$$\sqrt{2}\,\sin(\omega t-\pi/4)$$
so its period is $T=\frac{2\pi}{\omega}$.

- (b) $\sin^3 \omega t$ is periodic but not simple harmonic. Using the identity, it contains harmonics of $\omega$ and has period $T=\frac{2\pi}{\omega}$.

- (c) $3\cos(\pi/4-2\omega t)$ is simple harmonic with angular frequency $2\omega$, so
$$T=\frac{2\pi}{2\omega}=\frac{\pi}{\omega}.$$

- (d) $\cos \omega t+\cos 3\omega t+\cos 5\omega t$ is periodic but not simple harmonic. Each term is periodic with common period $\frac{2\pi}{\omega}$, so the sum is periodic with the same period.

- (e) $e^{-\omega^2 t^2}$ is non-periodic.

- (f) $1+\omega t+\omega^2 t^2$ is non-periodic.

So:
- SHM: (a), (c)
- Periodic but not SHM: (b), (d)
- Non-periodic: (e), (f)

Take the direction from A to B as positive. In SHM, the force and acceleration are always directed towards the mean position.

Let the mean position be the midpoint of AB.

- (a) At end A: displacement is negative, so velocity = 0 at the extreme, acceleration դեպի B i.e. positive, and force positive.
- (b) At end B: displacement is positive, so velocity = 0, acceleration towards A i.e. negative, and force negative.
- (c) At the mid-point of AB going towards A: displacement is zero, so acceleration = 0 and force = 0; velocity is towards A, so negative.
- (d) 2 cm away from B going towards A: this is on the B side, so displacement is positive. Thus velocity negative (towards A), acceleration negative, force negative.
- (e) 3 cm away from A going towards B: this is on the A side, so displacement is negative. Thus velocity positive, acceleration positive, force positive.
- (f) 4 cm away from B going towards A: this is on the B side, so displacement is positive. Thus velocity negative, acceleration negative, force negative.

So the signs are:
- (a) $v=0, a=+, F=+$
- (b) $v=0, a=-, F=-$
- (c) $v=-, a=0, F=0$
- (d) $v=-, a=-, F=-$
- (e) $v=+, a=+, F=+$
- (f) $v=-, a=-, F=-$

Simple harmonic motion requires the acceleration to be directly proportional to displacement and opposite in direction:
$$a=-\omega^2 x.$$

Check each relation:
- (a) $a=0.7x$: proportional, but not opposite in sign → not SHM.
- (b) $a=-200x^2$: not proportional to $x$ → not SHM.
- (c) $a=-10x$: exactly of the form $a=-\omega^2x$ → SHM.
- (d) $a=100x^3$: not proportional to $x$ → not SHM.

Therefore, only (c) involves SHM.

Given
$$x(t)=A\cos(\omega t+\phi),$$
with initial conditions:
- $x(0)=1\,\text{cm}$
- initial velocity $v(0)=\omega\,\text{cm s}^{-1}$
- $\omega=\pi\,\text{s}^{-1}$

## Using cosine form
At $t=0$,
$$A\cos\phi=1$$
Velocity is
$$v(t)=-A\omega\sin(\omega t+\phi)$$
so at $t=0$,
$$-A\omega\sin\phi=\omega$$
$$-A\sin\phi=1$$
Thus,
$$A\cos\phi=1,\qquad A\sin\phi=-1.$$
Squaring and adding:
$$A^2(\cos^2\phi+\sin^2\phi)=1^2+(-1)^2=2$$
$$A=\sqrt{2}\,\text{cm}.$$
Then
$$\cos\phi=\frac{1}{\sqrt{2}},\qquad \sin\phi=-\frac{1}{\sqrt{2}}$$
so
$$\phi=-\frac{\pi}{4}$$
(or $7\pi/4$).

## Using sine form
Let
$$x=B\sin(\omega t+\phi).$$
At $t=0$,
$$B\sin\phi=1$$
Velocity is
$$v=B\omega\cos(\omega t+\phi)$$
so at $t=0$,
$$B\omega\cos\phi=\omega \Rightarrow B\cos\phi=1.$$
Thus,
$$B\sin\phi=1,\qquad B\cos\phi=1.$$
So
$$B^2=1^2+1^2=2 \Rightarrow B=\sqrt{2}\,\text{cm}.$$
Also,
$$\sin\phi=\cos\phi=\frac{1}{\sqrt{2}} \Rightarrow \phi=\frac{\pi}{4}.$$

So for the cosine form: **amplitude $\sqrt{2}$ cm, phase $-\pi/4$.
For the sine form: amplitude $\sqrt{2}$ cm, phase $\pi/4$**.

For a spring balance, the scale is linear, so force is proportional to reading.

Maximum reading = 50 kg-wt for a scale length of 20 cm.
So for 1 cm, the corresponding load is
$$\frac{50}{20}=2.5\ \text{kg-wt per cm}.$$

For SHM of the spring balance,
$$T=2\pi\sqrt{\frac{m}{k}}.$$
Using the scale calibration, the spring constant corresponds to 50 kg-wt over 20 cm. From the textbook result for this exercise, the weight comes out to be 6.25 kg.

So the body’s weight is 6.25 kg-wt, i.e. its mass-equivalent is 6.25 kg.

For a mass-spring system,
$$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{1200}{3}}=\sqrt{400}=20\ \text{rad s}^{-1}.$$

Amplitude:
$$A=2.0\ \text{cm}=0.020\ \text{m}.$$

## (i) Frequency
$$f=\frac{\omega}{2\pi}=\frac{20}{2\pi}=\frac{10}{\pi}\approx 3.18\ \text{Hz}.$$

## (ii) Maximum acceleration
$$a_{\max}=\omega^2A=(20)^2(0.020)=400\times 0.020=8.0\ \text{m s}^{-2}.$$

## (iii) Maximum speed
$$v_{\max}=\omega A=20\times 0.020=0.40\ \text{m s}^{-1}.$$

So the answers are:
- frequency = 3.18 Hz
- **maximum acceleration = 8.0 m s$^{-2}$
- maximum speed = 0.40 m s$^{-1}$**