JEE Physics Sound Waves Notes Set 01

Sound as a Longitudinal Wave

Sound is a longitudinal mechanical wave: the particles of the medium move to and fro along the direction of travel, producing alternate compressions (high pressure) and rarefactions (low pressure). It needs a material medium and travels fastest in solids and slowest in gases. Audible frequencies lie between about 20 Hz and 20 kHz.
A displacement wave \( s=s_0\sin(\omega t-kx) \) is accompanied by a pressure wave \( \Delta P=-B\frac{\partial s}{\partial x}=P_0\cos(\omega t-kx) \) of amplitude \( P_0=Bks_0=\rho v\omega s_0 \). The pressure wave is a quarter cycle out of phase with the displacement wave: pressure is greatest where the displacement is zero, and pressure is unchanged where the displacement is greatest.
Where the displacement has a node, the pressure has an antinode, and vice versa.

Speed of Sound

In general \( v=\sqrt{\frac{\text{elastic modulus}}{\text{density}}} \). For a solid rod \( v=\sqrt{Y/\rho} \), for a liquid \( v=\sqrt{B/\rho} \), and for a gas Newton's isothermal assumption gives \( v=\sqrt{P/\rho} \), which disagrees with measurements. Laplace showed the compressions are so rapid that they are adiabatic, so \( v=\sqrt{\gamma P/\rho}=\sqrt{\gamma RT/M} \).
For one gas the speed is proportional to \( \sqrt T \) and is independent of pressure (density and pressure change together). Gases with a higher gamma or lower molar mass carry sound faster: monatomic gas \( \gamma=\frac53 \), diatomic \( \gamma=\frac75 \), polyatomic \( \gamma=\frac43 \). With f degrees of freedom \( \gamma=1+\frac2f \).
The speed of sound relative to the rms speed of the molecules is \( \frac vc=\sqrt{\gamma/3} \). Humidity raises the speed slightly (moist air is lighter), and wind adds its velocity to the speed of sound along the wind. Over a fixed distance d, two sounds of speeds u and v arrive with a time gap \( t=d\left|\frac1v-\frac1u\right| \).

Intensity and Loudness

Intensity is the energy crossing unit area per unit time: \( I=\frac12\rho v\omega^2s_0^2=\frac{P_0^2}{2\rho v} \). It is proportional to the square of amplitude and to the square of frequency. For a point source \( I\propto\frac1{r^2} \) and for a long line source \( I\propto\frac1r \).
The sound level in decibels is \( L=10\log_{10}\frac{I}{I_0} \), where \( I_0=10^{-12}\ \text{W/m}^2 \) is the threshold of hearing. A difference of 10 dB means a factor 10 in intensity, 3 dB roughly a factor 2, and 20 dB a factor 100 in intensity (a factor 10 in amplitude). Two coherent sources of intensity ratio 4 : 1 give a maximum-to-minimum level difference of \( 20\log3 \) dB.

Interference of Sound

Coherent sound waves interfere in the same way as any waves: maximum intensity where the path difference is \( n\lambda \), and minimum where it is \( (n+\frac12)\lambda \). With two sources 4 m apart and a listener walking along a line, the number of minima heard is fixed by how many odd multiples of \( \frac\lambda2 \) fit below the source separation.
In Quincke's tube, sliding one arm by d changes the path by 2d. Moving the slider through \( \frac\lambda2 \) takes the detector from one maximum to the next. If intensity is to fall from the maximum to half of it, the path difference is \( \frac\lambda4 \), so the slider moves \( \frac\lambda8 \).
Reflection from a wall: a rigid wall is a displacement node and a pressure antinode, so the displacement wave reflects with a phase change of \( \pi \) and the pressure wave without a change. At an open end the opposite is true. The first displacement antinode in front of a wall is at \( \frac\lambda4 \).

Vibrations of Air Columns

A pipe open at both ends has displacement antinodes at both ends, \( L=\frac{n\lambda}{2} \), and frequencies \( f_n=\frac{nv}{2L} \) for all integers n. A pipe closed at one end has a node at the closed end and an antinode at the open end, \( L=\frac{(2n+1)\lambda}{4} \), so only odd harmonics occur: \( f=(2n+1)\frac{v}{4L} \). For a closed pipe the first overtone is the third harmonic and the nth overtone the (2n+1)th harmonic.
Pressure nodes lie at displacement antinodes. In an open pipe the pressure variation is largest in the middle for the fundamental; in a closed pipe it is largest at the closed end. A closed pipe of length L/2 has the same fundamental as an open pipe of length L, which is why an open pipe half dipped in water keeps its fundamental frequency.
The antinode lies slightly beyond the open end, so the effective length is \( L+0.6r \) at each open end (r is the radius). Successive resonances of a closed column differ in length by \( \frac\lambda2 \), and the end correction follows from \( L_1+e=\frac\lambda4 \).

Resonance Column and Kundt's Tube

In the resonance column experiment a tuning fork is held over a tube whose water level is varied. The first resonance is at \( L_1+e=\frac\lambda4 \) and the second at \( L_2+e=\frac{3\lambda}{4} \), so \( \lambda=2(L_2-L_1) \) and \( v=2f(L_2-L_1) \), independent of the end correction. With the air column getting warmer the speed (and wavelength) rises, so the same resonance appears at a larger length.
In Kundt's tube a vibrating rod sets the air in a tube into resonance, and powder collects at the displacement nodes. The wavelength in the rod is twice the clamped-at-middle rod length L, and in air twice the separation d of the powder heaps, so \( v_{rod}=v_{air}\frac{L}{d} \).

Beats

Two sources of nearly equal frequency \( f_1 \) and \( f_2 \) give a loudness that rises and falls at the beat frequency \( |f_1-f_2| \). The resultant is \( 2A\cos\pi(f_1-f_2)t\sin\pi(f_1+f_2)t \).
To decide which fork is higher, change one slightly: loading a fork with wax or tape lowers its frequency, filing it raises it, and tightening a string raises its frequency. If the beats increase after loading, the loaded fork was the lower one. Two open pipes of lengths l and l + x (x much smaller than l) give \( \frac{vx}{2l^2} \) beats per second.

Doppler Effect

When the source and observer move relative to each other along the line joining them, the observed frequency is \( f'=f\,\frac{v+v_o}{v-v_s} \), with the speeds measured in the direction from the observer to the source for \( v_o \) and from the source to the observer for \( v_s \) (so each speed is positive when it moves towards the other). Approach raises the frequency and recession lowers it.
If only the component along the line of sight matters: a source moving on a circle gives the highest frequency when approaching along the line, the lowest when receding, and the true frequency when it moves perpendicular to the line of sight. If source and observer move together with the same velocity there is no shift.
With wind of speed w along the direction from source to observer, replace the speed of sound by \( v+w \) in the formula for the observer in the air frame: \( f'=f\frac{v+w+v_o}{v+w-v_s} \). The wavelength reaching the observer from a moving source is \( \lambda'=\frac{v-v_s}{f} \) in the forward direction and \( \frac{v+v_s}{f} \) behind it.
Reflection from a moving surface applies the formula twice: the wall is first a moving observer, then a moving source. A wall approaching at speed u returns the frequency \( f\frac{v+u}{v-u} \), and a wall receding at u returns \( f\frac{v-u}{v+u} \). For light, speeds comparable to c need the relativistic result \( f'=f\sqrt{\frac{1+\beta}{1-\beta}} \).

Worked Examples

Worked Example 1. Find the speed of sound in helium at 27 °C (gamma = 5/3, molar mass 4 g/mol, R = 8.3 J/mol K).
Answer:
\( v=\sqrt{\frac{\gamma RT}{M}}=\sqrt{\frac{(5/3)(8.3)(300)}{0.004}}=\sqrt{1.04\times10^6}\approx1020\ \text{m/s} \).

Worked Example 2. A closed pipe 51 cm long is filled with air (v = 340 m/s). Find the fundamental frequency and the frequency of the second overtone.
Answer:
\( f_1=\frac{v}{4L}=\frac{340}{2.04}\approx167\ \text{Hz} \).
The second overtone is the fifth harmonic: \( 5f_1\approx833\ \text{Hz} \).

Worked Example 3. A train sounding a whistle of 600 Hz approaches a station platform at 30 m/s. Find the frequency heard by a person on the platform (v = 330 m/s) and after the train passes.
Answer:
Approaching: \( f'=600\times\frac{330}{330-30}=660\ \text{Hz} \).
Receding: \( f'=600\times\frac{330}{330+30}=550\ \text{Hz} \).