Unit 8: Oscillations & Waves

Oscillations and Waves together carry 6–10 % weight in JEE-Main. The chapter is highly formulaic but contains some of the trickiest problems on JEE-Advanced — particularly Doppler-shift problems with multiple reflections, beats with frequency-dependent intensities, and combinations of two springs with a moving constraint.

Expected pattern:

• JEE-Main: 1 SHM problem (time period of a non-obvious system), 1 wave equation / velocity question, 1 Doppler or beats numerical.
• JEE-Advanced: A multi-step problem combining SHM with energy conservation, or a Doppler effect with reflection.

The single most useful organising idea is: any system with $F = -kx$ executes SHM with $\omega = \sqrt{k/m}$. Whether $x$ is a length, an angle, a charge or a current, this is the only thing that matters.

Concept Map

OSCILLATIONS
│
├── SHM kinematics
│     ├── x(t) = A sin(ωt + φ)
│     ├── v² = ω²(A² − x²)
│     └── a = -ω² x
│
├── Energy:  E = ½mω²A²  (KE + PE = const)
│
├── Springs
│     ├── Single mass, vertical:  T = 2π√(m/k)
│     ├── Series: 1/k_eq = 1/k₁ + 1/k₂
│     ├── Parallel: k_eq = k₁ + k₂
│     └── Two-mass spring system → reduced mass μ
│
├── Pendulums
│     ├── Simple: T = 2π√(L/g)
│     ├── Physical: T = 2π√(I/mgd)
│     └── Torsional: T = 2π√(I/κ)
│
└── Damped & Forced
       ├── ẍ + γẋ + ω₀²x = 0  → underdamped, critical, overdamped
       └── Resonance: |x|_max at ω = √(ω₀² − γ²/2)
│
WAVES
│
├── Wave equation:  ∂²y/∂t² = v² ∂²y/∂x²
│
├── Travelling: y = A sin(kx ∓ ωt)
│
├── String: v = √(T/μ)
│
├── Sound: v = √(γP/ρ) (Laplace); v ∝ √T
│
├── Reflection: free (no phase flip) vs fixed (π flip)
│
├── Standing waves
│     ├── String fixed-fixed: f_n = nv/(2L)
│     ├── Pipe open-open: f_n = nv/(2L)
│     └── Pipe closed-open: f_n = (2n−1)v/(4L)
│
├── Beats: f_beat = |f₁ − f₂|
│
└── Doppler effect
       f' = f · (v ± v_o)/(v ∓ v_s)

Topic 1: Simple Harmonic Motion (SHM)

Sub-topic A: Defining Equation and Solutions

A particle executes SHM if its acceleration is always directed toward a fixed point and proportional to its displacement from that point:

General solution:

with $\omega = \sqrt{k/m}$, period $T = 2\pi/\omega$, frequency $f = 1/T$.

Sub-topic B: Kinematics of SHM

From $x = A\sin(\omega t + \phi)$:

• Velocity: $v = A\omega \cos(\omega t + \phi)$
• Acceleration: $a = -A\omega^2 \sin(\omega t + \phi) = -\omega^2 x$

Useful position-velocity relation (eliminating $t$):

Maxima:

• $\vert v\vert _{\max} = A\omega$ at $x = 0$
• $\vert a\vert _{\max} = A\omega^2$ at $x = \pm A$

Sub-topic C: Energy of an SHM

Let $x = A\sin(\omega t + \phi)$, with spring (restoring) force $F = -kx$, $k = m\omega^2$.

• KE $= \tfrac{1}{2}m v^2 = \tfrac{1}{2} m \omega^2 (A^2 - x^2)$
• PE $= \tfrac{1}{2} k x^2 = \tfrac{1}{2} m\omega^2 x^2$
• Total energy $E = \tfrac{1}{2} m \omega^2 A^2$ — constant, independent of $x$ or $t$.

Time-averages:

Sub-topic D: SHM as Projection of UCM (Phasor Picture)

A particle going round a circle of radius $A$ with angular velocity $\omega$, projected onto a diameter, performs SHM along that diameter. This is the geometric origin of phasors, and the cleanest way to remember phase relationships:

• Velocity leads displacement by $\pi/2$
• Acceleration leads displacement by $\pi$

Sub-topic E: Phase and Phase Difference

Phase of an SHM at time $t$ is $\omega t + \phi$. Two oscillators with phase difference $\Delta\phi$:

• $\Delta\phi = 0$: in phase
• $\Delta\phi = \pi$: anti-phase
• $\Delta\phi = \pi/2$: quadrature

Worked Problem 1

A particle in SHM has amplitude $5$ cm and time period $2$ s. Find its speed when it is $3$ cm from the mean position.

Solution. $\omega = 2\pi/T = \pi$ rad/s. $v = \omega\sqrt{A^2 - x^2} = \pi\sqrt{25-9} = 4\pi$ cm/s $\approx 12.57$ cm/s.

Worked Problem 2

A particle in SHM along the $x$-axis is at $x = 2$ cm with $v = 4$ cm/s. At $x = 3$ cm it has $v = 3$ cm/s. Find the amplitude and angular frequency.

Solution. $v^2 = \omega^2(A^2 - x^2)$. Two equations: $16 = \omega^2(A^2-4),\quad 9 = \omega^2(A^2-9)$ Subtract $\Rightarrow 7 = 5\omega^2 \Rightarrow \omega = \sqrt{7/5}\approx 1.18$ rad/s. From first: $A^2 = 4 + 16/(7/5) = 4 + 80/7 \approx 15.43 \Rightarrow A \approx 3.93$ cm.

Topic 2: Spring Systems

Sub-topic A: Vertical Spring with Gravity

A spring of constant $k$ hung vertically, mass $m$ attached. New equilibrium at $x = mg/k$. About this equilibrium, the dynamics are pure SHM with $\omega = \sqrt{k/m}$ — gravity only shifts equilibrium, doesn't change period.

Sub-topic B: Combinations of Springs

Series: Same force $F$ through each, total extension $x = F/k_1 + F/k_2$:

Parallel: Same extension, forces add:

For a block between two springs (one on each side) attached to walls, both springs work in parallel as the block moves (one compresses, the other stretches): $k_{\text{eq}} = k_1 + k_2$.

Sub-topic C: Spring Between Two Masses — Reduced Mass

Two masses $m_1$ and $m_2$ connected by a spring of constant $k$, no external forces. Let $x_1, x_2$ be positions. Equation of motion:

Let $x = x_2 - x_1 - \ell_0$ (extension). Then

where the reduced mass is

So

Worked Problem 3

A spring of natural length $\ell_0$ and constant $k$ is cut into two pieces in the ratio $1:2$. Find the spring constants of the two pieces.

Solution. For a uniform spring $k\ell$ = const (since $k = AE/\ell$). So $k_1 \cdot \ell/3 = k\ell \Rightarrow k_1 = 3k$. Similarly $k_2 = 3k/2$.

Worked Problem 4 (JEE-Advanced)

A block of mass $m$ rests on a smooth horizontal surface between two walls. Two springs of constants $k$ and $4k$ are attached on either side (natural length, no preload). Find the period of small oscillations.

Solution. Both springs work in parallel: $k_{\text{eq}} = k + 4k = 5k$. $T = 2\pi\sqrt{m/(5k)}$.

Topic 3: Pendulums

Sub-topic A: Simple Pendulum

A point mass $m$ at the end of a massless string of length $L$. The restoring torque about the pivot is $-mgL\sin\theta$, and for small $\theta$, $\sin\theta \approx \theta$:

valid for small oscillations only. For amplitude $\theta_0$, the leading correction is $T \approx 2\pi\sqrt{L/g}(1 + \theta_0^2/16)$ — JEE-Advanced occasionally.

Sub-topic B: Compound (Physical) Pendulum

A rigid body free to rotate about a horizontal axis under gravity. Let $I$ be the moment of inertia about the pivot, $d$ the distance from pivot to centre of mass. Restoring torque: $-mgd\sin\theta \approx -mgd\,\theta$. Newton's rotational equation:

Equivalent simple-pendulum length $L_{\text{eff}} = I/(md)$.

Sub-topic C: Torsional Pendulum

A disk hung by a wire that resists twisting with torque $\tau = -\kappa\theta$. Then

Worked Problem 5

A uniform rod of length $L$ swings as a physical pendulum about a horizontal axis through one end. Find its period.

Solution. $I = mL^2/3$, $d = L/2$.

Worked Problem 6

A simple pendulum has period $T$ on Earth. What is its period inside a freely falling lift?

Solution. Effective $g = 0 \Rightarrow T = \infty$ — the pendulum does not oscillate.

Topic 4: Damped and Forced Oscillations

Sub-topic A: Damped SHM

Add a velocity-dependent friction $-b\dot x$:

where $\gamma = b/(2m)$, $\omega_0 = \sqrt{k/m}$.

Trial solution $x = e^{\lambda t}$: $\lambda = -\gamma \pm \sqrt{\gamma^2 - \omega_0^2}$. Three regimes:

Energy decays as $E \propto e^{-2\gamma t}$ in underdamped case.

Sub-topic B: Forced SHM and Resonance

External driver $F(t) = F_0 \cos\omega t$:

Steady state: $x = A(\omega)\cos(\omega t - \delta)$ with

Maximum amplitude at the resonance frequency:

For small damping $\omega_{\text{res}}\to\omega_0$.

Sub-topic C: Superposition of SHMs

Along the same line (collinear): $x_1 = A_1\sin\omega t$, $x_2 = A_2\sin(\omega t + \phi)$. Result:

Perpendicular (Lissajous): $x = A_1\sin\omega t$, $y = A_2 \sin(\omega t + \phi)$. The figure depends on $\phi$:

• $\phi = 0$: line $y = (A_2/A_1) x$
• $\phi = \pi/2$: ellipse with axes $A_1, A_2$
• $\phi = \pi$: line with negative slope

If frequencies differ ($\omega_1 \ne \omega_2$), figure closes only when $\omega_1:\omega_2$ is rational.

Topic 5: Waves — Travelling Waves

Sub-topic A: One-Dimensional Wave Equation

Any quantity satisfying $\partial^2 y/\partial t^2 = v^2 \partial^2 y/\partial x^2$ represents a wave propagating at speed $v$. General solution (d'Alembert):

First term is right-moving, second left-moving.

Sub-topic B: Sinusoidal Travelling Wave

with wave number $k = 2\pi/\lambda$ and angular frequency $\omega = 2\pi f$. Speed:

Velocity of a particle (at fixed $x$): $v_p = \partial y/\partial t = -A\omega\cos(kx - \omega t)$, max $A\omega$ — not the wave speed.

Sub-topic C: Speed of a Wave on a String

Take a stretched string of tension $T$ and linear mass density $\mu$. Consider a small arc subtending angle $2\theta$ at the centre of curvature, radius $R$, moving with speed $v$ (transverse pulse, in the frame where it is stationary).

Centripetal force on element of mass $\mu(2R\theta) = 2\mu R\theta$:

Sub-topic D: Speed of Sound — Newton's Formula and Laplace Correction

Newton's formula: if sound is an isothermal process, $v = \sqrt{P/\rho}$. For air at STP this gives $280$ m/s — too low.

Laplace correction: sound waves are too fast for heat exchange with surroundings — they propagate adiabatically:

For air ($\gamma = 1.4$), $v \approx 332$ m/s at $0\,^\circ$C, in excellent agreement with experiment.

Effects:

• Temperature: $v \propto \sqrt T$, rises by about $0.6$ m/s per $^\circ$C near room temperature.
• Pressure: at fixed $T$, $P/\rho$ is constant (ideal gas), so $v$ independent of $P$ at fixed $T$.
• Humidity: moist air is less dense ⇒ sound faster in humid air.

Worked Problem 7

A stretched wire of length $1$ m, mass $4$ g, vibrates at $200$ Hz in its fundamental mode. Find the tension.

Solution. $\mu = 4\times 10^{-3}/1 = 4\times 10^{-3}$ kg/m. Fundamental wavelength $\lambda = 2L = 2$ m. $v = f\lambda = 400$ m/s. $T = \mu v^2 = 4\times 10^{-3}\times 1.6\times 10^5 = 640$ N.

Topic 6: Reflection, Transmission and Standing Waves

Sub-topic A: Reflection from Fixed and Free Ends

• Fixed end (rigid): incident pulse inverts on reflection — phase change of $\pi$.
• Free end: reflects without inversion.

Physically: at a fixed boundary the medium cannot move, requiring an inverted reflected wave to cancel; at a free boundary the slope is zero (no constraint on displacement).

Sub-topic B: Standing Waves on a String

Two oppositely directed travelling waves of equal amplitude superpose:

This is a standing wave. Nodes at $\sin kx = 0$, i.e. $x = n\lambda/2$. Antinodes halfway between, at $x = (n+\tfrac{1}{2})\lambda/2$.

For a string fixed at $x = 0$ and $x = L$: $y(L) = 0 \Rightarrow kL = n\pi \Rightarrow \lambda_n = 2L/n$. Frequencies:

All harmonics ($f_1, 2f_1, 3f_1, \ldots$) are present.

Sub-topic C: Standing Waves in Air Columns

Open–open pipe (both ends antinodes): $\lambda_n = 2L/n$, so

All harmonics present.

Closed–open pipe (one end node, the other antinode): only odd harmonics:

Fundamental $f_1 = v/(4L)$ — half of the open pipe of the same length.

Comparison table:

End correction: real pipes have antinodes a bit outside the open end (≈ $0.6\,r$ for a tube of radius $r$). For most JEE problems use the bare formulas unless the question explicitly mentions end correction.

Sub-topic D: Quincke's Tube and Resonance

A common JEE setup. Tuning forks resonate with air columns at specific lengths — knowing two successive resonance lengths $L_1, L_2$ removes end correction:

Worked Problem 8

A pipe closed at one end resonates with a 512 Hz tuning fork at lengths $16.5$ cm and $50.5$ cm. Find the speed of sound and end correction.

Solution. $\lambda/2 = 50.5 - 16.5 = 34$ cm $\Rightarrow \lambda = 68$ cm. $v = f\lambda = 512 \times 0.68 = 348$ m/s. End correction $e$: $\lambda/4 = L_1 + e \Rightarrow 17 = 16.5 + e \Rightarrow e = 0.5$ cm.

Topic 7: Beats

Sub-topic A: Derivation

Two sound waves of nearly equal frequencies $f_1$ and $f_2$ at the same point:

The slowly-varying $\cos$-envelope has frequency $(f_1-f_2)/2$, but intensity $\propto y^2$ has frequency $\vert f_1-f_2\vert$.

Condition: $\vert f_1 - f_2\vert$ small (typically < 10 Hz, audible as throbbing).

Worked Problem 9 (classic trap)

Two tuning forks A and B produce 5 beats per second. When A is loaded with a little wax, the beat frequency becomes 3 per second. If A is 256 Hz, what is the frequency of B?

Solution. Loading wax lowers $f_A$. Beats dropped from 5 → 3 ⇒ originally $f_B > f_A$ (so lowering $f_A$ widened the gap if $f_B$ above — no wait — careful):

If $f_B > f_A$, lowering $f_A$ increases beat: but here beat decreased (5 → 3), so $f_B < f_A$. Originally $f_B = f_A - 5 = 251$ Hz.

Verify: after wax, $f_A$ drops by 2, beat $= \vert 251 - 254\vert = 3$. ✓

Worked Problem 10

A string of frequency $f$ vibrates with a tuning fork of $256$ Hz producing $4$ beats per second. The string's tension is increased, and beats now become $6$/s. Original $f$?

Solution. Increasing tension raises $f_{\text{string}}$. Beats increased from $4 \to 6$ ⇒ $f_{\text{string}} > 256$, so original $f = 260$ Hz.

Topic 8: Doppler Effect

Sub-topic A: General Formula for Sound

Sound is a wave in a medium (air). With speed of sound $v$, observer moving at $v_o$, source moving at $v_s$, both measured along the source-to-observer line:

Sign convention (everything measured along the line from source toward observer):

• $v_o > 0$ if observer moves toward source.
• $v_s > 0$ if source moves toward observer.
• $v_o, v_s < 0$ for opposite directions.

If both move along same line away from each other, $v_o < 0$, $v_s < 0$, signs in formula give a smaller $f'$ — frequency drops.

Sub-topic B: Special Cases

1. Source moving, observer stationary ($v_o = 0$): $f' = f \cdot v/(v - v_s)$. Approaching $\Rightarrow f' > f$; receding $\Rightarrow f' < f$.
2. Observer moving, source stationary ($v_s = 0$): $f' = f \cdot (v + v_o)/v$.
3. Both moving towards each other: $f' = f\cdot(v + v_o)/(v - v_s)$ — maximum upshift.
4. Both moving in same direction (observer chasing source): $f' = f\cdot(v - v_o)/(v - v_s)$ with appropriate signs.

Sub-topic C: Moving Medium (Wind)

If the medium itself moves with velocity $v_m$ (e.g. wind), replace $v \to v + v_m$ (sign per direction):

Sub-topic D: Reflection Doppler (Two-Stage)

Sound from source reflects off a moving wall and comes back to source. Treat as two consecutive Dopplers: wall is observer first, then becomes a source.

Worked Problem 11

A police car moving at $36$ km/h sounds a horn of $400$ Hz. A motorist drives at $54$ km/h directly toward the police car. What frequency does the motorist hear? ($v_{\text{sound}} = 340$ m/s.)

Solution. $v_s = 10$ m/s (toward observer), $v_o = 15$ m/s (toward source). Both positive:

Worked Problem 12 (JEE-Advanced)

A source of $1000$ Hz moves at $10$ m/s toward a wall. A stationary observer is between the source and the wall. Find the beat frequency. ($v_{\text{sound}} = 340$ m/s.)

Solution. Observer hears two waves:

(i) Direct from approaching source: $f_1 = 1000\cdot 340/(340-10) = 1030.3$ Hz.

(ii) Reflected from wall. Wall first acts as observer (stationary): hears $f_{\text{wall}} = 1000\cdot 340/(340-10) = 1030.3$ Hz; then wall re-radiates as stationary source: observer (stationary) hears the same $1030.3$ Hz.

Wait — the reflected wave has the same frequency $1030.3$ Hz as the direct? Yes — because the wall is stationary, and observer is stationary, so no further Doppler shift on reflection.

Beats $= 0$.

(But if observer were behind the source, direct sound would be the receding $f = 1000\cdot 340/(340+10) = 971.4$ Hz, and reflected (still 1030.3 Hz from approaching source toward wall) → beat frequency $\approx 58.9$ Hz.)

Worked Problem 13

Two trains approach each other on parallel tracks at speeds $20$ m/s and $30$ m/s. Train A blows whistle of $400$ Hz. Frequency heard by passenger in train B? ($v = 340$ m/s.)

Solution. $v_o = +30$, $v_s = +20$ (both moving toward each other):

Problem-Solving Heuristics

1. Identify the SHM. Anything with $F = -kx$ (or $\tau = -\kappa\theta$) gives SHM. The hard part is finding $k$ — usually by linearising the restoring force about equilibrium.
2. Use energy conservation. $\tfrac{1}{2}mv^2 + \tfrac{1}{2}kx^2 = \tfrac{1}{2}kA^2$ gives quick answers without integrating.
3. Spring trick: if a spring of constant $k$ is cut into $n$ equal pieces, each piece has constant $nk$.
4. For two-mass + spring problems: use reduced mass $\mu = m_1m_2/(m_1+m_2)$.
5. For pendulums: derive period from torque, $T = 2\pi\sqrt{I/(mgd)}$ — never confuse $I_{\text{cm}}$ with $I_{\text{pivot}}$ (parallel-axis!).
6. For a wave on a string, $v = \sqrt{T/\mu}$ — increasing tension or decreasing mass per length raises speed.
7. String fundamental: $\lambda = 2L$. Pipe closed: $\lambda = 4L$. Pipe open: $\lambda = 2L$.
8. Closed pipe has only odd harmonics. Fundamental of closed pipe = half of open pipe of same length.
9. Beats: load wax test. Adding mass lowers the frequency of a fork. Compare beat frequency before/after to decide whether $f_{\text{unknown}}$ was above or below the reference.
10. Doppler — use one master formula with signs. Don't try to memorise four formulas. Define "+" along source→observer direction.
11. Doppler with reflection = two-stage Doppler. Wall absorbs at $f_1$, re-emits at $f_1$.
12. Light Doppler is different — uses relativistic formula $f' = f\sqrt{(1-\beta)/(1+\beta)}$, no medium.

Common Traps & Mistakes

• Forgetting the $\pi/2$ phase lead of velocity over position in SHM — affects intensity / wave problems with two SHMs.
• Adding amplitudes algebraically when two SHMs of equal frequency superpose. Use the vector/phasor addition.
• Mass of the spring. For a non-massless spring, effective mass is $m + m_{\text{spring}}/3$ for vertical, simple-pendulum-style oscillation. JEE-Advanced may probe this.
• Hooke's range only. SHM assumes small displacements (linear restoring force). At large amplitude any real system deviates.
• Pipe vs string confusion — they have different fundamentals (depending on boundary conditions). Make sure you draw a node-antinode diagram.
• Direction of $v_s, v_o$ in Doppler. Pick one axis (e.g. "+" = from source to observer) and stick with it.
• Sound vs light Doppler. Sound formula has separate $v_s, v_o$ because medium matters; light depends only on relative velocity.
• Beat frequency vs envelope frequency. The envelope wobbles at $\vert f_1 - f_2\vert /2$, but intensity wobbles at $\vert f_1-f_2\vert$. Always report the intensity beat in JEE.
• In a stretched string, fundamental wavelength is $2L$, not $L$.
• Sound speed in medium scales with $\sqrt T$, not $T$.

Quick Revision Card

• SHM: $\ddot x = -\omega^2 x$, $x = A\sin(\omega t + \phi)$, $v^2 = \omega^2(A^2-x^2)$, $E = \tfrac{1}{2}m\omega^2 A^2$.
• Spring: $T = 2\pi\sqrt{m/k}$. Series $1/k$ adds; parallel $k$ adds.
• Two-mass: $T = 2\pi\sqrt{\mu/k}$, $\mu = m_1m_2/(m_1+m_2)$.
• Pendulum: $T = 2\pi\sqrt{L/g}$; physical: $T = 2\pi\sqrt{I/(mgd)}$.
• Damped: $E \propto e^{-2\gamma t}$, underdamped $\omega' = \sqrt{\omega_0^2-\gamma^2}$.
• Wave: $v = f\lambda$. String $v = \sqrt{T/\mu}$. Sound $v = \sqrt{\gamma P/\rho}$, $v\propto\sqrt T$.
• Standing wave on string fixed-fixed: $f_n = nv/(2L)$.
• Open pipe: $f_n = nv/(2L)$. Closed pipe: $f_n = (2n-1)v/(4L)$ — only odd.
• Beats: $f_{\text{beat}} = \vert f_1 - f_2\vert$.
• Doppler: $f' = f(v + v_o)/(v - v_s)$, signs along source→observer.

Formula Sheet