Chapter 15 — Waves

Waves transport energy and information without transporting matter. From a ripple on water, to the sound of your voice, to a photon from a distant star, waves are everywhere. This chapter restricts itself to mechanical waves in elastic media, where particles oscillate about fixed positions and the disturbance propagates outward.

We will write down the travelling-wave equation, compute wave speeds in strings and air, develop the principle of superposition, study reflection and standing waves, derive the beat phenomenon, and end with the Doppler effect in its full vector glory.

Concept Map

• 15.1 Transverse and longitudinal waves; mechanical vs electromagnetic.
• 15.2 Wave parameters — $\lambda$, $T$, $f$, $v$; $v = f\lambda$.
• 15.3 The travelling-wave equation $y(x,t)=A\sin(kx-\omega t+\phi)$.
• 15.4 Speed of a transverse wave on a string: $v=\sqrt{T/\mu}$.
• 15.5 Speed of a longitudinal wave in solid/liquid/gas; Newton's formula, Laplace correction.
• 15.6 Principle of superposition.
• 15.7 Reflection at fixed and free ends.
• 15.8 Standing waves on a stretched string — harmonics.
• 15.9 Standing waves in air columns — open and closed organ pipes.
• 15.10 Beats — derivation of beat frequency.
• 15.11 Doppler effect for sound — full formula and applications.

15.1 Transverse and Longitudinal Waves

Definitions

• Transverse wave. Particles oscillate perpendicular to the direction of propagation. Example: a string wave, light, ripples on water (approximately).
• Longitudinal wave. Particles oscillate parallel to the direction of propagation. Example: sound in air, compression waves in a spring (slinky).

A solid supports both kinds of waves; a fluid (gas or non-viscous liquid) supports only longitudinal waves because it cannot sustain a shear stress.

Mechanical vs Electromagnetic

• Mechanical waves need a material medium (sound, strings, water).
• Electromagnetic waves do not (light, radio, X-rays). They propagate in vacuum at $c=3\times 10^8\ \text{m/s}$.

Worked Example

Which type of wave is the P-wave in an earthquake? S-wave?

P (primary) waves are longitudinal — they travel through both solids and the molten outer core, hence faster. S (secondary) waves are transverse and cannot pass the liquid outer core. The shadow zone confirms this.

Pitfalls

• Surface water waves are neither purely transverse nor purely longitudinal — particles trace ellipses.
• A "longitudinal" wave still has crests and troughs — just in pressure rather than displacement.

15.2 Wave Parameters

The Quartet $\lambda, T, f, v$

• Wavelength $\lambda$ — distance over which the wave shape repeats (m).
• Period $T$ — time for one full oscillation (s).
• Frequency $f = 1/T$ — oscillations per second (Hz).
• Wave speed $v$ — speed at which the wave shape moves (m/s).

Master Relation

A wave covers a distance $\lambda$ in time $T$, hence $v = \lambda/T = f\lambda$.

Angular Counterparts

• Angular frequency $\omega = 2\pi f = 2\pi/T$ (rad/s).
• Wave number $k = 2\pi/\lambda$ (rad/m).

Then

Worked Example

A wave of frequency 500 Hz has wavelength 0.66 m in air. Find $v$, $\omega$, $k$.

$v = (500)(0.66) = 330\ \text{m/s}$; $\omega = 2\pi(500)\approx 3142\ \text{rad/s}$; $k = 2\pi/0.66\approx 9.52\ \text{rad/m}$.

Pitfalls

• $v$ is a property of the medium; $f$ is a property of the source. When a wave changes medium, $f$ stays the same and $\lambda$ changes.
• Hz means 1/s, not "cycles per second" in modern SI — but the meaning is the same.

15.3 The Travelling-Wave Equation

Form

A wave moving with speed $v$ in the $+x$-direction is described by a function

i.e., the shape at time $t$ is the shape at $t=0$ shifted by $vt$. The simplest such function is a sinusoid:

with $\omega = vk$.

A wave moving in the $-x$-direction is $y(x,t)=A\sin(kx+\omega t+\phi)$.

Deriving the Form

A snapshot at $t=0$ is $y(x,0)=A\sin(kx+\phi)$. As $t$ increases, the pattern shifts by $vt$ to the right, so replace $x\to x-vt$:

Particle Velocity vs Wave Velocity

The wave speed $v=\omega/k$ is constant. The particle velocity of a string element is

with maximum $A\omega$. These are very different quantities: wave speed of m/s, particle speed (often) cm/s.

The Linear Wave Equation

$y(x,t)$ satisfies

— the second-order linear wave equation. Any function of the form $f(x\pm vt)$ is a solution.

Worked Example

Given $y = 0.02\sin(50x - 200t)$ in SI units, find $A$, $\lambda$, $T$, $v$, direction.

$A = 0.02\ \text{m}$, $k=50\ \text{rad/m}\Rightarrow\lambda=2\pi/50\approx 0.126\ \text{m}$, $\omega=200\Rightarrow T=2\pi/200\approx 0.0314\ \text{s}$, $v=\omega/k=4\ \text{m/s}$, direction $+x$.

Pitfalls

• The sign between $kx$ and $\omega t$ sets the direction. $(kx-\omega t)$ travels in $+x$; $(kx+\omega t)$ travels in $-x$.
• Wave equation requires the same units in $kx$ and $\omega t$ — both are radians.

15.4 Speed of a Transverse Wave on a String

Setup and Derivation Outline

Consider a string of linear mass density $\mu$ (kg/m) under tension $T$. Pluck the string — a small bump of length $\Delta\ell$ moves along the string at speed $v$.

Switch to the frame of the bump. In this frame, the string elements move backward through the bump at speed $v$, following a small circular arc of radius $R$ at the top of the bump.

The centripetal force on a string element of length $\Delta\ell = R\Delta\theta$ comes from the two tensions at the ends of the element:

Setting this equal to $\mu\,\Delta\ell\,(v^2/R) = \mu R\Delta\theta(v^2/R) = \mu v^2\Delta\theta$:

Worked Example

A 2.0 m long string has mass 4 g and is under 100 N tension. Find $v$.

$\mu = 0.004/2.0 = 0.002\ \text{kg/m}$. $v = \sqrt{100/0.002}\approx 224\ \text{m/s}$.

Pitfalls

• $T$ is the tension in the string (N), not the period.
• $\mu = m/L$ — linear mass density, not volume density.
• The wave speed is independent of frequency (for an ideal string), but depends on tension and density.

15.5 Speed of a Longitudinal Wave

General Formula

For a longitudinal wave in any medium, the speed is

where $B$ is the bulk modulus, $Y$ Young's modulus, $\eta$ shear modulus, and $\rho$ density.

In a Gas — Newton's Formula

Newton assumed sound propagation is isothermal: $B_T = P$. So

For air at STP: $v_\text{Newton} = \sqrt{1.013\times 10^5/1.29}\approx 280\ \text{m/s}$ — but the measured value is $\sim 332\ \text{m/s}$. Newton's formula is wrong by 15%.

Laplace Correction

Laplace recognised that sound oscillations are too fast for heat exchange — they are essentially adiabatic. For an adiabatic process, $B_S = \gamma P$. Hence

For air ($\gamma=1.4$, $M=28.8\ \text{g/mol}$, $T=273$ K):

Dependence on Conditions

• $v \propto \sqrt T$ — sound is faster on hot days.
• $v$ is independent of pressure at fixed $T$ (since $P/\rho\propto T$).
• $v$ depends on molecular mass: hydrogen sound is much faster than air sound.
• Humidity slightly increases $v$ (water vapour is lighter than $N_2$/$O_2$).

Worked Example

Find the speed of sound in helium at 300 K. ($\gamma_\text{He}=5/3$, $M=4\ \text{g/mol}$.)

$v = \sqrt{(5/3)(8.314)(300)/0.004}\approx 1019\ \text{m/s}$.

Pitfalls

• Newton's formula gives a result; only the Laplace correction matches experiment.
• $v$ is independent of pressure at fixed $T$ — a common trap question.
• $v$ does depend on temperature; $\sqrt{T(\text{K})}$, not Celsius.

15.6 Principle of Superposition

Statement

When two or more waves traverse the same medium, the resultant displacement at any point is the algebraic (vector) sum of the displacements due to each wave separately.

This holds because the underlying wave equation is linear (no $y^2$ or higher).

Consequences

• Interference — coherent superposition gives stable patterns of constructive ($\Delta\phi = 2n\pi$) and destructive ($\Delta\phi = (2n+1)\pi$) interference.
• Beats — superposition of two slightly different frequencies (15.10).
• Standing waves — superposition of a wave with its reflection (15.8).

Worked Example

Two waves $y_1 = a\sin(\omega t - kx)$, $y_2 = a\sin(\omega t - kx + \phi)$ superpose. Find the resultant amplitude.

amplitude $A_\text{net} = 2a\vert \cos(\phi/2)\vert$. Max when $\phi=2n\pi$, zero when $\phi=(2n+1)\pi$.

Pitfalls

• Superposition holds for small amplitudes (linear regime). At large amplitudes, shock waves form.
• The intensities don't add; the amplitudes (with phase) do. $I\propto A^2$.

15.7 Reflection of Waves

Two Cases

When a wave on a string hits a boundary, it partially reflects (and partially transmits, if the boundary isn't perfect).

• Fixed end (rigid boundary). The wall cannot move. The string element must satisfy $y = 0$ at the wall. This forces the reflected wave to have opposite sign — a phase change of $\pi$ (i.e., half a wavelength).

If $y_\text{inc} = A\sin(kx-\omega t)$ hits a fixed end at $x=L$, the reflected wave is

• Free end. No constraint on $y$. The reflected wave has the same sign — no phase change.

Reflection of Sound at a Closed/Open Pipe

• Closed end of a pipe (rigid wall) → reflects with no phase change in pressure but $\pi$ phase change in displacement.
• Open end → reflects with no phase change in displacement but $\pi$ phase change in pressure.

This sounds confusing but is forced by the boundary conditions: at a closed end, displacement = 0 (node) and pressure is max (antinode). At an open end, pressure = atmospheric (node) and displacement is free (antinode).

Worked Example

A pulse travels along a string and hits a fixed end. Sketch the reflected pulse.

The reflected pulse is inverted (turned upside down) compared to the incident pulse.

Pitfalls

• "Phase change of $\pi$" means a sign flip — not a $90^\circ$ shift.
• A free end of a string usually means a light ring on a smooth rod — not a hanging end.
• For sound, the boundary condition is in pressure at a closed end and in displacement at an open end; do not mix the two.

15.8 Standing Waves on a Stretched String

Formation

Superpose an incident and a reflected wave on a string of length $L$ fixed at both ends:

(depending on sign conventions). The resultant is

This is a standing wave — the spatial and temporal parts factor. Nodes (zero amplitude) occur where $\sin(kx)=0$, antinodes (max) where $\sin(kx)=\pm 1$.

Boundary Conditions for a String Fixed at Both Ends

Nodes at $x=0$ and $x=L$:

So $k_n = n\pi/L$, $\lambda_n = 2L/n$, and

• $f_1 = v/(2L)$ — fundamental or first harmonic.
• $f_2 = 2f_1$ — second harmonic / first overtone.
• $f_3 = 3f_1$ — third harmonic / second overtone.

All integer multiples of $f_1$ are present.

Worked Example

A guitar string 65 cm long has $\mu = 4\times 10^{-4}\ \text{kg/m}$ and is tuned to 330 Hz (fundamental). Find the tension.

$v = 2Lf_1 = 2(0.65)(330) = 429\ \text{m/s}$. $T = \mu v^2 = (4\times 10^{-4})(429)^2\approx 73.6\ \text{N}$.

Pitfalls

• "Harmonic" = integer multiple of $f_1$ including $f_1$.
• "Overtone" = above-fundamental frequencies. First overtone = second harmonic for a string.
• Doubling $T$ multiplies $f_n$ by $\sqrt 2$, not by 2.

15.9 Standing Waves in Air Columns

Closed Organ Pipe (One End Closed)

A closed end is a displacement node, an open end is a displacement antinode. For a pipe of length $L$ closed at one end:

Allowed wavelengths satisfy $L = (2n-1)\lambda/4$, i.e.,

So $f_1, 3f_1, 5f_1,\dots$ — only odd harmonics are present.

Open Organ Pipe (Both Ends Open)

Both ends are displacement antinodes. $L = n\lambda/2$:

All harmonics are present.

Comparison

For the same length $L$:

• Open pipe fundamental: $v/(2L)$.
• Closed pipe fundamental: $v/(4L)$ — an octave lower (half the frequency).

End Correction

In practice, the antinode at an open end lies a small distance $e\approx 0.6r$ (where $r$ is the pipe radius) outside the open end. Effective length:

This correction is most important for short, wide pipes.

Worked Example

A closed pipe sounds its fundamental at 256 Hz. Find $L$. (Speed of sound 340 m/s.)

$L = v/(4f_1) = 340/(4\cdot 256) \approx 0.332\ \text{m}.$

Pitfalls

• Closed pipe has only odd harmonics, but they are still labelled $n=1,3,5,\dots$ — be careful with $n$.
• Fundamental of a closed pipe is half that of an open pipe of the same length, not twice.
• End correction is sometimes ignored in school problems but expected for precision.

15.10 Beats

Formation

Superpose two sound waves of nearly equal frequencies $f_1, f_2$ at the same point:

Use the sum-to-product identity:

The result is a wave at the average frequency $(f_1+f_2)/2$ whose amplitude is slowly modulated at $(f_1-f_2)/2$. The intensity, $\propto y^2$, is modulated at $\vert f_1-f_2\vert$.

Beat Frequency

You hear $\vert f_1-f_2\vert$ amplitude maxima per second.

Use in Tuning

A piano tuner listens for beats between a tuning fork and the piano string. As the string is tuned closer to the fork, the beat frequency drops to zero.

Worked Example

Two tuning forks of 256 Hz and 260 Hz are sounded together. How many beats per second?

$f_\text{beat} = \vert 260-256\vert = 4$ beats/s.

Pitfalls

• The audible beat frequency is $\vert f_1-f_2\vert$, not $(f_1-f_2)/2$.
• Beats are most easily heard when $\vert f_1-f_2\vert < 10$ Hz; above that they merge into a single rough tone.
• Loading one fork with wax lowers its frequency. If beats decreased after loading, the loaded fork was the higher one.

15.11 Doppler Effect for Sound

The General Formula

For an observer (subscript O) and a source (subscript S) moving in a medium where sound speed is $v$:

with the convention:

• $v_O$ is positive if the observer moves toward the source, negative if away.
• $v_S$ is positive if the source moves toward the observer, negative if away.

(In NCERT notation, sometimes the form $f' = f(v-v_O)/(v-v_S)$ with a different sign convention is used; what matters is consistency.)

Special Cases

• Source moving, observer at rest: $f' = f\,\dfrac{v}{v - v_S}$.
• Observer moving, source at rest: $f' = f\,\dfrac{v + v_O}{v}$.
• Both at rest: $f' = f$. Of course.

If source moves toward observer at rest, $f' > f$ — pitch rises (approaching ambulance siren). If source moves away, $f' < f$ — pitch falls.

Derivation Sketch (source moving toward stationary observer)

Source emits frequency $f$. In one period $T = 1/f$, the source moves $v_S T$ closer to the observer. The next wavefront is launched $v_S T$ closer, so successive crests are separated by $\lambda' = (v - v_S)T = (v - v_S)/f$. The observer (at rest in the medium) sees the wave move at $v$, so:

Effect of Wind

If wind has speed $w$ from source to observer (taken positive), replace $v$ everywhere by $v+w$. Wind affects both $v_O$ and $v_S$ measurements unless they are measured with respect to the medium.

Applications

• Radar — bouncing EM waves off cars: $\Delta f/f = 2v/c$ (for cars approaching).
• Sonar — same idea with sound underwater.
• Astronomy — redshift of galaxies, expansion of the universe.
• Medical ultrasound — measure blood-flow velocity.

Worked Example

A police car emits a 1000 Hz siren and approaches a stationary observer at 30 m/s. Sound speed 340 m/s. Find the frequency heard.

After the car passes:

Pitfalls

• The Doppler formula for sound is asymmetric between source and observer motion (because there's a medium). For light there is a single relativistic formula.
• Always be explicit about your sign convention. A wrong sign turns $1097$ Hz into something nonsensical.
• The formula assumes motion along the line joining source and observer. For oblique motion, take components.

Solved Problems

Problem 1. A 1.2 m long string fixed at both ends vibrates in its third harmonic at 480 Hz. Find $v$ and the wavelength of the fundamental.

Solution. $f_3 = 3v/(2L)\Rightarrow v = 2Lf_3/3 = 2(1.2)(480)/3 = 384$ m/s. $\lambda_1 = 2L = 2.4$ m.

Problem 2. A wave $y = 5\sin(0.5x - 200t)$ cm with $x$ in cm and $t$ in seconds. Find $v$ and the maximum particle speed.

Solution. $v = \omega/k = 200/0.5 = 400$ cm/s = 4 m/s. $v_{p,\text{max}}=A\omega = 5\cdot 200 = 1000$ cm/s = 10 m/s.

Problem 3. A pipe closed at one end resonates at 256 Hz and 768 Hz with no resonance in between. Identify the harmonics and find $L$ (use $v=340$ m/s).

Solution. Closed pipe gives only odd harmonics. 256 Hz = fundamental, 768 Hz = third harmonic. $L = v/(4f_1) = 340/1024 \approx 0.332$ m.

Problem 4. Two open organ pipes of lengths 50 cm and 51 cm produce how many beats per second? ($v=340$ m/s)

Solution. $f_1 = 340/(2\cdot 0.50) = 340\ \text{Hz}$, $f_2=340/(2\cdot 0.51)\approx 333.3\ \text{Hz}$. Beats $\approx 6.7$ per second.

Problem 5. A train moves at 30 m/s toward a stationary observer, sounding a 480 Hz whistle. Speed of sound 340 m/s. Find the frequency heard while approaching and while receding.

Solution. Approach: $f' = 480\cdot 340/(340-30) = 526$ Hz. Recede: $f'=480\cdot 340/(340+30)=441$ Hz.

Problem 6. Find the speed of sound in nitrogen at 300 K. ($\gamma=7/5$, $M=28$ g/mol.)

Solution. $v = \sqrt{(1.4)(8.314)(300)/0.028} \approx 353$ m/s.

Problem 7. A string of length 60 cm and mass 6 g is under 36 N tension. Find the lowest two natural frequencies.

Solution. $\mu = 0.006/0.6 = 0.01\ \text{kg/m}$. $v = \sqrt{36/0.01}=60$ m/s. $f_1 = 60/(2\cdot 0.6) = 50$ Hz; $f_2 = 100$ Hz.

JEE / NEET Edge Cases

• Phase difference $\Delta\phi$ vs path difference $\Delta x$. $\Delta\phi = (2\pi/\lambda)\Delta x = k\Delta x$.
• Intensity from amplitude. $I\propto A^2 f^2$ for a sound wave at fixed $\rho$ and $v$.
• Quincke's tube and resonance tube. Adjust path lengths until constructive/destructive interference; useful for measuring $v$ of sound.
• Beats from loaded forks. Loading by wax lowers the frequency of a fork; filing raises it.
• Doppler with reflection. A wall acts as a moving observer that then re-emits — apply the formula twice.
• Doppler for light. Use the relativistic formula $f'=f\sqrt{(1+\beta)/(1-\beta)}$ for approach; not the sound formula.
• String with one end fixed, one free. Behaves like a closed pipe — only odd harmonics; $f_n = (2n-1)v/(4L)$.
• Sonometer. $f = (1/2L)\sqrt{T/\mu}$; law of length, law of tension, law of mass per unit length.
• Group velocity vs phase velocity. $v_p = \omega/k$, $v_g = d\omega/dk$ — in non-dispersive media they coincide.

Quick Recap

• Travelling wave: $y = A\sin(kx\mp\omega t+\phi)$. Sign in $kx\mp\omega t$ sets direction.
• $v=f\lambda=\omega/k$.
• Transverse on string: $v=\sqrt{T/\mu}$.
• Sound in gas: $v=\sqrt{\gamma RT/M}$ (Newton + Laplace).
• Stretched string both ends fixed: $f_n=n v/(2L)$, all harmonics.
• Closed pipe: $f_n=(2n-1)v/(4L)$, odd harmonics only.
• Open pipe: $f_n = nv/(2L)$, all harmonics.
• Beats: $f_\text{beat} = \vert f_1-f_2\vert$.
• Doppler (sound): $f' = f(v+v_O)/(v-v_S)$ with the right sign convention.

Formula Sheet