Oscillations and Waves — Practice Questions with Answers

Simple harmonic motion is a periodic motion in which the restoring force (and hence acceleration) on a particle is always directed towards a fixed equilibrium point, and is directly proportional to the particle's displacement from that point.

Defining equation: $\dfrac{d^2x}{dt^2} = -\omega^2x$, where $\omega$ is the angular frequency. The general solution is $x = A\sin(\omega t + \phi)$, where $A$ is amplitude and $\phi$ is the initial phase.

For $x = A\sin(\omega t)$, velocity $v = A\omega\cos(\omega t)$.

Kinetic energy: $KE = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2A^2\cos^2(\omega t)$

Potential energy: $PE = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2A^2\sin^2(\omega t)$ (using $k=m\omega^2$)

Total energy: $E = KE+PE = \frac{1}{2}m\omega^2A^2[\cos^2(\omega t)+\sin^2(\omega t)] = \frac{1}{2}m\omega^2A^2$ — a constant, independent of time.

Transverse waves: Particle vibration is perpendicular to the direction of wave propagation. Example: waves on a stretched string, electromagnetic waves.

Longitudinal waves: Particle vibration is parallel to (along) the direction of wave propagation, involving compressions and rarefactions. Example: sound waves in air.

Resonance occurs when a system is driven by a periodic external force whose frequency matches (or is very close to) the system's own natural frequency of oscillation — this results in a dramatic increase in the amplitude of oscillation.

Example: Pushing a child on a swing at exactly the swing's natural frequency produces increasingly large swings with minimal effort, while pushing at a mismatched frequency produces little effect — another example is a singer's voice shattering a glass by matching the glass's natural resonant frequency.

Beats are periodic variations in the loudness of sound heard when two sound waves of slightly different frequencies superpose — the amplitude of the resultant wave periodically rises and falls.

For two waves of frequencies $f_1$ and $f_2$, superposition gives a resultant with amplitude modulated at frequency $|f_1-f_2|/2$; since loudness depends on the square of amplitude, the perceived beat frequency (loud-soft-loud cycles per second) is:

$f_{beat} = |f_1 - f_2|$

A damped oscillation is one in which the amplitude progressively decreases over time due to dissipative forces (like friction or air resistance) that continuously remove energy from the oscillating system.

As damping acts, both amplitude and total mechanical energy decrease over time — typically exponentially — until the oscillation eventually stops entirely, with all the initial mechanical energy converted to heat via the dissipative (damping) force.

The principle of superposition states that when two or more waves overlap at a point, the resultant displacement at that point is the vector (algebraic) sum of the individual displacements each wave would produce alone.

Constructive interference occurs when waves arrive in phase (crest meets crest), so their displacements add, producing a larger resultant amplitude. Destructive interference occurs when waves arrive out of phase (crest meets trough), so their displacements partially or fully cancel, producing a smaller (or zero) resultant amplitude.

A standing wave forms when two identical waves of the same frequency and amplitude travel in opposite directions along the same medium (often due to reflection) and superpose — this produces a pattern that appears to "stand still," with fixed points of zero displacement (nodes) and maximum displacement (antinodes), rather than a travelling wave pattern.

The distance between two successive nodes (or two successive antinodes) is $\lambda/2$, half the wavelength.

For a pendulum bob displaced through small angle $\theta$, the restoring tangential force is $-mg\sin\theta \approx -mg\theta$ (for small $\theta$, in radians).

Since $\theta = x/l$ (arc length $x$ over pendulum length $l$): $F = -mg\dfrac{x}{l}$

Comparing with SHM's $F=-kx$ (with $k = mg/l$): $\omega^2 = k/m = g/l$

$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{l}{g}}$

The speed of sound in a gas is given by $v = \sqrt{\gamma RT/M}$ (or equivalently $v = \sqrt{\gamma P/\rho}$), showing direct dependence on the square root of absolute temperature.

As temperature increases, gas molecules move faster on average (higher RMS speed) — since sound propagation fundamentally relies on molecular collisions transmitting the disturbance through the medium, faster-moving molecules transmit this disturbance more quickly, increasing the speed of sound with increasing temperature.

(a) Angular frequency: $\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2}\text{ rad/s} \approx 1.57\text{ rad/s}$

(b) Maximum velocity: $v_{max} = A\omega = 0.05\times1.57 \approx 0.0785\text{ m/s} \approx 7.85\text{ cm/s}$

(c) Maximum acceleration: $a_{max} = A\omega^2 = 0.05\times(1.57)^2 = 0.05\times2.47 \approx \mathbf{0.123\text{ m/s}^2}$

Time period on Earth: $T = 2\pi\sqrt{l/g} = 2\pi\sqrt{1/9.8} = 2\pi\sqrt{0.102} \approx 2\pi(0.319) \approx 2.01\text{ s}$

For same period on Moon: $2\pi\sqrt{l_E/g_E} = 2\pi\sqrt{l_M/g_M} \Rightarrow \dfrac{l_M}{l_E} = \dfrac{g_M}{g_E}$

$l_M = l_E\times\dfrac{g_M}{g_E} = 1\times\dfrac{1.63}{9.8} \approx \mathbf{0.166\text{ m} \approx 16.6\text{ cm}}$

Frequency: $f = \dfrac{v}{\lambda} = \dfrac{340}{0.5} = 680\text{ Hz}$

Time period: $T = \dfrac{1}{f} = \dfrac{1}{680} \approx \mathbf{1.47\times10^{-3}\text{ s} = 1.47\text{ ms}}$

Beat frequency: $f_{beat} = |f_1-f_2| = |260-256| = 4\text{ Hz}$

Time between successive maxima: $t = \dfrac{1}{f_{beat}} = \dfrac{1}{4} = \mathbf{0.25\text{ s}}$

For a closed pipe, the fundamental has a node at the closed end and antinode at the open end, so $l = \lambda/4 \Rightarrow \lambda = 4l$

$\lambda = 4\times0.5 = 2\text{ m}$

$f = \dfrac{v}{\lambda} = \dfrac{340}{2} = \mathbf{170\text{ Hz}}$

Velocity in SHM: $v = \omega\sqrt{A^2-x^2}$. Given $v = v_{max}/2 = \omega A/2$:

$\dfrac{\omega A}{2} = \omega\sqrt{A^2-x^2}$

$\dfrac{A}{2} = \sqrt{A^2-x^2} \Rightarrow \dfrac{A^2}{4} = A^2-x^2 \Rightarrow x^2 = A^2 - \dfrac{A^2}{4} = \dfrac{3A^2}{4}$

Given $x = 3\text{ cm}$: $9 = \dfrac{3A^2}{4} \Rightarrow A^2 = 12 \Rightarrow A = \sqrt{12} \approx \mathbf{3.46\text{ cm}}$

For an observer moving towards a stationary source, apparent frequency: $f' = f\left(\dfrac{v+v_o}{v}\right)$

$f' = 500\left(\dfrac{340+20}{340}\right) = 500\times\dfrac{360}{340}$

$f' = 500\times1.0588 \approx \mathbf{529.4\text{ Hz}}$

(a) Angular frequency: $\omega = \sqrt{k/m} = \sqrt{200/2} = \sqrt{100} = 10\text{ rad/s}$

(b) Time period: $T = \dfrac{2\pi}{\omega} = \dfrac{2\pi}{10} \approx \mathbf{0.628\text{ s}}$

For a string fixed at both ends, the $n$th harmonic satisfies $l = n\dfrac{\lambda}{2}$, so $\lambda = \dfrac{2l}{n}$

For the third harmonic, $n=3$: $\lambda = \dfrac{2\times1}{3} = \mathbf{0.667\text{ m} \approx 66.7\text{ cm}}$

$PE = \frac{1}{2}kx^2$, Total energy $E = \frac{1}{2}kA^2$. So $\dfrac{PE}{E} = \dfrac{x^2}{A^2}$

$\dfrac{PE}{E} = \dfrac{6^2}{10^2} = \dfrac{36}{100} = 0.36$

Since $KE + PE = E$: $\dfrac{KE}{E} = 1 - \dfrac{PE}{E} = 1 - 0.36 = \mathbf{0.64}$

Kinetic energy is 64% of the total energy at this displacement.