Oscillations and Waves

A particle in SHM of amplitude $A$. At what displacement are its kinetic and potential energies equal?

Almost everyone says $A/2$. It is $\frac{A}{\sqrt{2}}$.

$U=\frac{1}{2}k{x}^2$ and $E=\frac{1}{2}k{A}^2$. Setting $U=\frac{1}{2}E$:

$\frac{1}{2}k{x}^2=\frac{1}{2}\left(\frac{1}{2}k{A}^2\right)\Rightarrow x=\frac{A}{\sqrt{2}}$

Energy goes as $x^2$, not $x$. Half the energy sits at $1/\sqrt{2}$ of the amplitude, not half of it.

That squared relationship runs through the whole chapter — and the other thread is phase, which decides everything from standing-wave shapes to whether two sounds reinforce or cancel.

Scope note. The 2023 NTA revision removed damped, forced and free oscillations, resonance, and the Doppler effect in sound from JEE Main, and they stay out for 2026. All four remain in JEE Advanced — section 10 covers them and flags them.

1. What makes motion simple harmonic

$F=-kx⟺a=-{\omega }^{2}x$

The minus sign is the entire physical content: the force always points back toward equilibrium, which is what makes the motion oscillate rather than run away.

The defining test. Acceleration proportional to displacement and oppositely directed. If asked to prove a motion is SHM, this is what you show.

Almost any smooth potential well looks parabolic near its minimum, so almost every oscillation is approximately SHM for small enough amplitude. That is why SHM turns up in so many unrelated systems.

2. The solution

$x=A\sin (\omega t+\varphi )$

Starting from an extreme gives a cosine; starting from equilibrium gives a sine.

Periodic, but not necessarily simple harmonic

Every SHM is periodic. Not every periodic motion is SHM — the extra requirement is $a\propto -x$, and questions exploit the gap constantly.

To find the period of a combination, look for the smallest $T$ that returns every term to its starting value.

The last row is worth a moment: two perfectly periodic motions can add to something that never repeats at all, because the ratio of their periods is irrational.

Illustration 1

Which of these represent SHM, and what is the period of each? (a) $\sin \omega t+\cos \omega t$, (b) ${\sin }^2\omega t$, (c) $\sin \omega t+\sin 2\omega t$.

(a) Combine into a single sinusoid:

$\sin \omega t+\cos \omega t=\sqrt{2}\sin \left(\omega t+\frac{\pi }{4}\right)$

A single sine of angular frequency $\omega$, so SHM with $T=2\pi /\omega$, amplitude $\sqrt{2}$.

(b) Use the double-angle identity, which is the whole trick:

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

SHM — but about the centre $y=\frac{1}{2}$, and at angular frequency $2\omega$, so $T=\pi /\omega$. Half the period you would guess from the $\omega$ in the expression.

(c) Differentiate twice and the two terms pick up different factors:

$a=-{\omega }^2\sin \omega t-4{\omega }^2\sin 2\omega t$

That is not a multiple of $x$, so it is periodic but not SHM. The period is still $2\pi /\omega$, since the second term completes two whole cycles in that time.

Illustration 2

A particle moves as $x=5\sin \left(4\pi t+\frac{\pi }{6}\right)$ cm. Find the amplitude, period, frequency, maximum speed and maximum acceleration, and the position and velocity at $t=0$.

Read the constants straight off: $A=5$ cm, $\omega =4\pi$ rad/s, $\varphi =\pi /6$.

$T=\frac{2\pi }{\omega }=0.5\text{s},f=2\text{Hz}$

$v_{max}=A\omega =5(4\pi )=62.8\text{cm/s}$

$a_{max}=A{\omega }^2=5(4\pi {)}^2=790{\text{cm/s}}^2$

At $t=0$: $x=5\sin \frac{\pi }{6}=2.5\text{cm}$, and $v=A\omega \cos \frac{\pi }{6}=62.8(0.866)=54.4\text{cm/s}$

Check the velocity independently: $v=\omega \sqrt{A^2-x^2}=4\pi \sqrt{25-6.25}=12.57(4.33)=54.4$ cm/s. The two routes agree, and the second never needed the time at all.

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{k}}$

Period does not depend on amplitude. This is isochronism, and it is why pendulum clocks work — the clock keeps time even as the swing decays.

3. Velocity, acceleration and phase

$v=A\omega \cos (\omega t+\varphi ),a=-A{\omega }^2\sin (\omega t+\varphi )=-{\omega }^{2}x$

Eliminating $t$ gives the form most numericals actually need:

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

At the extremes the particle is momentarily at rest but under maximum force. At equilibrium it moves fastest but feels no force at all.

Acceleration is exactly antiphase with displacement — which is just $a=-{\omega }^{2}x$ read as a statement about phase.

SHM is the shadow of circular motion

Take a point moving at constant angular speed $\omega$ round a circle of radius $A$ — the reference circle — and project it onto a diameter. The projection is exactly $x=A\sin (\omega t+\varphi )$.

This is not an analogy; it is an identity, and it hands you three things at once:

• $\varphi$ is a real angle on the circle, which is why phase differences are quoted in radians.
• $v_{max}=A\omega$ is just the speed round the circle, and $a_{max}=A{\omega }^2$ its centripetal acceleration.
• $v=\omega \sqrt{A^2-x^2}$ falls out of Pythagoras on the reference circle, with no calculus at all.

4. Energy

$U=\frac{1}{2}k{x}^2,K=\frac{1}{2}m{\omega }^2(A^2-x^2),E=\frac{1}{2}k{A}^2=\frac{1}{2}m{\omega }^2{A}^2$

$E\propto A^2$ — double the amplitude, quadruple the energy.

• $K$ and $U$ each oscillate at twice the frequency of the motion, because the particle crosses equilibrium twice per cycle.
• Averaged over a cycle, $⟨K⟩=⟨U⟩=E/2$.
• They are instantaneously equal at $x=A/\sqrt{2}$, not $A/2$.

Illustration 3

A 0.2 kg particle oscillates with amplitude $0.1\text{m}$ and period $0.4\text{s}$. Find the total energy, and the kinetic and potential energies at $x=0.05\text{m}$.

$\omega =\frac{2\pi }{0.4}=15.7\text{rad/s}$, so ${\omega }^2=246.7$

$E=\frac{1}{2}m{\omega }^2{A}^2=\frac{1}{2}(0.2)(246.7)(0.01)=0.247\text{J}$

At $x=0.05=A/2$:

$U=\frac{1}{2}m{\omega }^2{x}^2=\frac{1}{2}(0.2)(246.7)(0.0025)=0.0617\text{J}$

$K=E-U=0.185\text{J}$

Read off the ratio: at $x=A/2$ the split is $K:U=3:1$, not $1:1$. The potential energy takes only a quarter of the total, because it goes as $x^2$ and $(1/2{)}^2=1/4$. Equality arrives later, at $A/\sqrt{2}$, exactly as the opening question said.

5. Springs and pendulums

$T=2\pi \sqrt{\frac{m}{k}}$

Holds whether the spring is horizontal or vertical — a vertical spring only shifts the equilibrium position, it does not change the period.

Simple pendulum

The restoring force is $-mg\sin \theta$, which is not proportional to displacement — so strictly this is not SHM. The small-angle approximation $\sin \theta \approx \theta$ rescues it:

$T=2\pi \sqrt{\frac{L}{g}}$

• Independent of the bob's mass and of the amplitude — which is what made the pendulum the basis of timekeeping.
• Good to about 1% up to $10°$; beyond that a large-amplitude pendulum runs slow.
• A seconds pendulum has $T=2$ s, so $L\approx 0.994$ m — near enough to a metre to use as an anchor.
• In a lift accelerating up at $a$, replace $g$ by $g+a$. In free fall $g_{eff}=0$ and the pendulum stops oscillating entirely.

Illustration 4

A seconds pendulum ($T=2$ s) is carried to the Moon, where $g$ is one sixth of Earth's. Find its new period, and the length that would restore a 2-second beat there.

$T\propto \frac{1}{\sqrt{g}}$ at fixed length, so

$T_{moon}=2\sqrt{6}=4.90\text{s}$

A clock driven by this pendulum would run at less than half speed.

To recover $T=2$ s the length must fall by the same factor as $g$:

$L_{moon}=\frac{L_{earth}}{6}=\frac{0.994}{6}=0.166\text{m}$

The general point: only the ratio $L/g$ matters. Neither the bob's mass nor the amplitude appears anywhere, which is exactly why a pendulum was a usable clock centuries before anyone could measure $g$ accurately.

Illustration 5

Two springs $k_1=200$ and $k_2=300\text{N/m}$ carry a 6 kg mass. Find the period in series and in parallel.

Parallel: $k=500\text{N/m}$, $T=2\pi \sqrt{6/500}=2\pi (0.110)\approx 0.69\text{s}$

Series: $k=\frac{200(300)}{500}=120\text{N/m}$, $T=2\pi \sqrt{6/120}=2\pi (0.224)\approx 1.40\text{s}$

Series is softer, so it is slower — by almost exactly a factor of 2 here. Note the series stiffness is below both individual values, and the parallel one is above both.

6. Waves

A wave transports energy and momentum through a medium without transporting the medium itself. Each particle oscillates about its own fixed position while the disturbance travels.

Transverse mechanical waves need rigidity, because restoring a sideways displacement is a shear. Fluids have zero rigidity modulus — so sound in air, water or any gas is longitudinal only.

Seismologists use exactly this: S-waves fail to cross the Earth's outer core, which is how we know it is liquid.

$v=f\lambda$

Wave speed is a property of the medium, not the source. Change the source frequency and the wavelength changes to compensate; $v$ does not move.

Newton used the isothermal bulk modulus for sound and got an answer ~15% too low. Laplace corrected it: the compressions are too rapid for heat to flow, so they are adiabatic — hence the $\gamma$.

Sound speed in a gas depends on temperature but not pressure, since $P/\rho$ is fixed at a given $T$ by the gas law. It goes as $\sqrt{T}$.

Illustration 6

(a) A string of mass $4\text{g}$ and length $80\text{cm}$ is stretched to a tension of $100\text{N}$. Find the wave speed. (b) The speed of sound in air is $331\text{m/s}$ at $0°\text{C}$. Find it at $100°\text{C}$.

(a) Linear density first — the commonest slip is using the total mass:

$\mu =\frac{4\times 10^{-3}}{0.8}=5\times 10^{-3}\text{kg/m}$

$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{100}{0.005}}=\sqrt{2\times 10^4}=141\text{m/s}$

(b) $v\propto \sqrt{T}$ with $T$ absolute:

$v_{100}=331\sqrt{\frac{373}{273}}=331(1.169)=387\text{m/s}$

Note what did not appear: the pressure. Raising the pressure raises the density in exact proportion, so $P/\rho$ and hence the speed are untouched. A sound wave travels no faster at the bottom of a mine than at the top, other than through the temperature difference.

7. Progressive waves and standing waves

$y=A\sin (\omega t-kx),k=\frac{2\pi }{\lambda }$

Minus sign $\Rightarrow$ travelling in $+x$. Plus sign reverses it.

Two speeds live in that one equation and they are not the same thing:

$v_{wave}=\frac{\omega }{k}\text{versus}{v}_{particle}=\frac{\partial y}{\partial t}=A\omega \cos (\omega t-kx)$

The wave speed is fixed by the medium; the particle speed depends on where in its cycle the particle happens to be, and peaks at $A\omega$.

Illustration 7

A wave is described by $y=0.02\sin (300t-5x)$ in SI units. Find the amplitude, wavelength, frequency and wave speed, and the maximum speed of any particle of the medium.

Match against $y=A\sin (\omega t-kx)$:

$v_{particle,max}=A\omega =0.02(300)=6\text{m/s}$

Read the two together: the disturbance races along at 60 m/s while no particle of the medium ever exceeds 6 m/s — a tenth as fast, and only sideways. That gap is the clearest statement of what a wave actually is: the pattern travels, the matter does not.

Cross-check: $v=f\lambda =47.7(1.26)=60$ m/s, matching $\omega /k$.

Reflection. At a fixed end there is a phase reversal of $\pi$ — the end cannot move, so the reflected wave must cancel the incident one there. At a free end, no phase change.

Two identical waves travelling oppositely superpose to:

$y=2A\sin (kx)\cos (\omega t)$

The $x$ and $t$ dependence have separated. That is the mathematical signature of a standing wave: every particle oscillates at the same frequency, with an amplitude $2A\sin (kx)$ fixed by where it is.

• Nodes $\lambda /2$ apart; antinodes midway, also $\lambda /2$ apart. Node to nearest antinode: $\lambda /4$.
• A standing wave carries no net energy — equal flow both ways. That is the clearest difference from a progressive wave.
• All particles between adjacent nodes are in phase; across a node they are antiphase. There is no progressive phase lag as in a travelling wave.

8. Strings and pipes

$f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu }}\text{(string)},f_n=\frac{(2n-1)v}{4L}\text{(closed pipe)}$

A closed pipe's fundamental is half an open pipe's of the same length — which is why short closed tubes give low notes. The missing even harmonics change the timbre, giving the closed pipe its hollow tone.

An end correction of about $0.6r$ must be added at each open end, because the antinode sits slightly outside the tube.

Illustration 8

An open pipe of length 50 cm and a closed pipe of length 25 cm. Take $v=340$ m/s. Compare their fundamentals and list the first three harmonics of each.

Open, $L=0.5$: $f_1=\frac{340}{2(0.5)}=340$ Hz. Harmonics 340, 680, 1020 Hz.

Closed, $L=0.25$: $f_1=\frac{340}{4(0.25)}=340$ Hz. Harmonics 340, 1020, 1700 Hz.

Same fundamental — but the closed pipe skips 680 Hz entirely, because only odd multiples survive. Two pipes at identical pitch that sound completely different.

9. Beats

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

Two waves of slightly different frequency superpose, and the resultant amplitude rises and falls. The ear resolves beats only up to about 10 per second.

Beats are how instruments are tuned — a tuner adjusts a string until the beats slow and vanish, which is far more sensitive than judging pitch directly.

Beats give you the size of the difference, never the sign. To settle which frequency is higher, change one deliberately: loading a tuning fork with wax lowers its frequency, and whether the beat rate rises or falls tells you which side you were on.

Illustration 9

Two open organ pipes of lengths $100\text{cm}$ and $101\text{cm}$ are sounded together at their fundamentals. Take $v=330\text{m/s}$. Find the beat frequency.

$f_1=\frac{v}{2L_1}=\frac{330}{2.00}=165.0\text{Hz}$

$f_2=\frac{v}{2L_2}=\frac{330}{2.02}=163.4\text{Hz}$

$f_{beat}=|165.0-163.4|=1.6\text{Hz}$

Read it back: a 1% difference in length gives a 1% difference in frequency, which is roughly 1.6 Hz here — slow enough to count on your fingers. This is exactly the sensitivity that makes beats the standard tuning method: a mismatch far too small to hear as a pitch difference becomes an obvious throb.

Illustration 10

A stretched string vibrates in 3 loops. The tension is changed so that at the same frequency it vibrates in 2 loops. Find the ratio of the new tension to the old.

$f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$, with $f$, $L$ and $\mu$ all unchanged, so $n\sqrt{T}$ is constant:

$3\sqrt{T_1}=2\sqrt{T_2}\Rightarrow \frac{T_2}{T_1}=\frac{9}{4}$

The tension must be raised by a factor of 2.25.

That makes sense: fewer loops means a longer wavelength, and to keep the frequency fixed the wave must travel faster — which needs more tension.

10. Beyond JEE Main — Advanced only

Removed from Main in 2023, still in the Advanced syllabus.

• Damped oscillations — resistive forces decay the amplitude exponentially, with the frequency shifted slightly below the undamped value.
• Forced oscillations — the system settles at the driver's frequency, not its own.
• Resonance — driving frequency matches the natural frequency; amplitude is limited only by damping.
• Doppler effect — $f^{\prime }=f\frac{v\pm v_o}{v\mp v_s}$, signs chosen so that approach raises the frequency.

Summary

• SHM is defined by $a=-{\omega }^{2}x$: restoring force proportional to displacement. That is what you prove.
• $A$ comes from how it was started; $\omega$ from the system itself. Period is independent of amplitude.
• Every SHM is periodic; not every periodic motion is SHM. ${\sin }^2\omega t$ is SHM, at frequency $2\omega$; $\sin \omega t+\sin 2\omega t$ is not SHM at all.
• SHM is the projection of uniform circular motion, which is where $\varphi$, $A\omega$ and $A{\omega }^2$ all come from.
• $v=\omega \sqrt{A^2-x^2}$ is the form most numericals need.
• $v$ leads $x$ by $\pi /2$; $a$ is exactly antiphase with $x$.
• $E=\frac{1}{2}k{A}^2\propto A^2$. $K=U$ at $x=A/\sqrt{2}$, not $A/2$. Both oscillate at $2f$.
• Springs: parallel adds $k$, series adds $1/k$. Cutting into $n$ pieces gives $nk$ each.
• Pendulum: $T=2\pi \sqrt{L/g}$, independent of mass and amplitude. In free fall it stops.
• Transverse waves need rigidity — sound in fluids is longitudinal only.
• $v=f\lambda$; the medium sets $v$, never the source. Sound goes as $\sqrt{T}$, not with pressure.
• Wave speed $\omega /k$ and particle speed $A\omega$ are different quantities — the pattern travels, the matter only jiggles.
• Standing wave $y=2A\sin kx\cos \omega t$ — $x$ and $t$ separate, no net energy transported.
• String and open pipe: all harmonics, $f_1=v/2L$. Closed pipe: odd only, $f_1=v/4L$.
• $f_{beat}=|f_1-f_2|$, and beats never tell you which is higher.