JEE Physics Waves Notes Set 04

Travelling Waves

A wave transfers energy and momentum through a medium without carrying the medium itself along. In a transverse wave the particles vibrate perpendicular to the direction of travel (strings, light); in a longitudinal wave they vibrate along it (sound in fluids). Fluids cannot carry transverse mechanical waves because they have no rigidity.
The basic relations are \( v=f\lambda=\frac{\omega}{k} \), \( \omega=2\pi f \) and \( k=\frac{2\pi}{\lambda} \). The frequency is fixed by the source and does not change when a wave enters another medium; speed and wavelength change together.
A wave moving along +x is \( y=A\sin(\omega t-kx+\phi) \) (same signs of kx and ωt give a wave moving along -x). Particle velocity is \( \frac{\partial y}{\partial t} \), with maximum \( A\omega \); it is smaller than the wave speed when \( A<\frac{\lambda}{2\pi} \). A phase difference \( \Delta\phi \) corresponds to a path difference \( \Delta x=\frac{\lambda}{2\pi}\Delta\phi \).
Any function of the single combination \( x\pm vt \) is a travelling wave; a function such as \( f(x^2-v^2t^2) \) is not.

Speed of Waves

Transverse wave on a string: \( v=\sqrt{T/\mu} \) (tension over mass per unit length). Longitudinal wave: \( v=\sqrt{E/\rho} \) with E the appropriate elastic modulus (Young's modulus for a rod, bulk modulus for a fluid).
For a gas, Newton assumed isothermal compressions and obtained \( \sqrt{P/\rho} \), which disagrees with measurement. Laplace showed that the compressions are adiabatic, so \( v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{\gamma RT}{M}} \). Hence the speed is proportional to \( \sqrt T \), independent of pressure at constant temperature, smaller in heavier gases and nearly independent of frequency. Sound travels fastest in solids, then liquids, then gases.
A hanging rope has tension that grows towards the top, so a pulse travelling up speeds up while its frequency stays unchanged and its wavelength increases in proportion to the speed.

Energy, Intensity and Loudness

The average power carried by a wave on a string is \( P=\tfrac12\mu v\omega^2A^2 \), so it depends on \( A^2f^2 \). For sound, intensity is \( I=\tfrac12\rho v\omega^2A^2 \) (power per unit area). A point source spreads its power over a sphere, so \( I\propto1/r^2 \) and the amplitude falls as \( 1/r \).
Loudness level in decibel: \( \beta=10\log_{10}\frac{I}{I_0} \), with the threshold of hearing \( I_0=10^{-12}\ \text{W/m}^2 \). Each 10 dB is a factor 10 in intensity. Pitch is determined by frequency, quality (timbre) by waveform and loudness by intensity. Audible frequencies lie between 20 Hz and 20 kHz; below this is infrasonic, above is ultrasonic (used in SONAR).

Doppler Effect

When the source or the observer moves along the line joining them, the observed frequency changes. With v the speed of sound, \( v_o \) the speed of the observer and \( v_s \) that of the source:
\( f'=f\,\frac{v+v_o}{v-v_s} \) when the observer moves towards the source (speed \( v_o \)) and the source moves towards the observer (speed \( v_s \)). Reverse the sign of a speed when that motion is away from the other body. In short, any motion that brings them closer raises the frequency and any motion that separates them lowers it.
For a source or observer moving in a circle, the extremes of frequency occur at the points where the velocity is directly along the line of sight. For light and small speeds, \( \frac{\Delta\lambda}{\lambda}\approx\frac{v}{c} \) (red shift for recession).

Example 1. A whistle of frequency 500 Hz is whirled in a horizontal circle of radius 1.2 m at 400 rev/min. Find the range of frequencies heard by a distant listener in the plane of the circle (speed of sound 340 m/s).
Answer:
Speed of the whistle: \( v_s=\omega r=\frac{2\pi\times400}{60}\times1.2\approx50\ \text{m/s} \).
Maximum: \( 500\times\frac{340}{340-50}\approx586\ \text{Hz} \); minimum: \( 500\times\frac{340}{340+50}\approx436\ \text{Hz} \).
Range: 436 Hz to 586 Hz.

Superposition, Interference and Beats

Waves obey the principle of superposition (valid when the amplitudes are small and the medium is linear): the resultant displacement is the sum of the individual displacements. Two waves of amplitudes \( A_1,A_2 \) and phase difference \( \phi \) give \( A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi} \). Maximum intensity (constructive) needs path difference \( n\lambda \); minimum (destructive) needs \( (2n+1)\lambda/2 \). The ratio \( \frac{I_{max}}{I_{min}}=\left(\frac{A_1+A_2}{A_1-A_2}\right)^2 \).
Beats arise from two waves of nearly equal frequencies and the beat frequency is \( |f_1-f_2| \). To find which one is higher, change one frequency in a known way: loading a fork with wax lowers its frequency, and increasing the tension of a string raises it.

Stationary Waves, Strings and Pipes

Two identical waves travelling in opposite directions form a stationary wave, for instance \( y=2A\sin kx\cos\omega t \). Nodes (zero amplitude) and antinodes (maximum amplitude) alternate at spacing \( \lambda/2 \) between adjacent nodes. All particles in one loop vibrate in phase, and they pass the mean position together. A rigid or fixed end reflects with a phase change of \( \pi \) (a node forms there); a free end reflects without change (an antinode forms).
String fixed at both ends: \( f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu}} \) with n = 1, 2, 3, ...; the n-th harmonic has n loops, n antinodes and n + 1 nodes. Frequency varies as \( \frac1L\sqrt{T} \) and, for strings of one material, as \( \frac{1}{L\,r}\sqrt T \) with radius r.
Pipe open at both ends: \( f_n=\frac{nv}{2L} \) (all harmonics). Pipe closed at one end: \( f=\frac{(2m-1)v}{4L} \) (odd harmonics only; the closed end is a displacement node and a pressure antinode). The effective length of an air column exceeds the measured length by the end correction e, and in the resonance-tube experiment \( \lambda=2(L_2-L_1) \) eliminates it.

Example 2. A closed organ pipe has fundamental frequency 1500 Hz. How many overtones can a normal person hear (limit 20 kHz)?
Answer:
The frequencies are \( 1500(2n-1) \), and \( 1500(2n-1)\le20000\Rightarrow2n-1\le13 \).
So the harmonics 1, 3, 5, 7, 9, 11 and 13 are audible; excluding the fundamental, there are 6 overtones.

Example 3. On increasing the tension of a stretched string by 2.5 N, its frequency changes in the ratio 3 : 2. Find the original tension.
Answer:
Since \( f\propto\sqrt T \): \( \frac{T+2.5}{T}=\frac94\Rightarrow T=2\ \text{N} \).
The original tension was 2 N.