University Physics — Year 2 · Bachelor Year 2

7Sound Waves in Fluids

Count the seconds between the flash and the thunder and divide by three: that is the distance in kilometres. Sound crosses air at a third of a kilometre per second, water at a kilometre and a half, and an ultrasound scanner times the echoes from inside the body to a millionth of a second to draw its picture. This chapter applies the fluid mechanics of Chapters 3 and 2 to a fluid at rest disturbed by a small pressure ripple, and finds the d’Alembert equation again: the speed of sound, what carries the energy, how loud is loud, how sound spreads in space, and why a siren changes pitch as it passes.

7.1 The acoustic approximation

Definition 7.1 (Acoustic perturbation)

A fluid at rest, of uniform pressure $P_0$ and density $\rho_0$ , is disturbed: $P = P_0 + p(M, t)$ , $\rho = \rho_0 + \rho_1(M, t)$ , velocity $\vect v(M, t)$ . The acoustic approximation keeps only the first order in the small quantities $p$ (the overpressure, or acoustic pressure), $\rho_1$ and $\vect v$ : $|p| \ll P_0$ , $|\rho_1| \ll \rho_0$ , $v \ll c$ , and the transformation undergone by each fluid particle is so fast that it exchanges no heat: it is isentropic, with the compressibility $\chi_S = \frac1\rho\bigl(\frac{\partial\rho}{\partial P}\bigr)_S$ . Gravity is neglected.

Theorem 7.2 (Sound waves)

To first order, Euler’s equation, mass conservation and the isentropic law read

\rho_0\frac{\partial\vect v}{\partial t} = -\operatorname{\vect{grad}}p , \qquad \frac{\partial\rho_1}{\partial t} + \rho_0\operatorname{div}\vect v = 0 , \qquad \rho_1 = \rho_0\chi_Sp ,

and together they give the wave equation for the overpressure,

\Delta p = \frac1{c^2}\frac{\partial^2p}{\partial t^2} , \qquad c = \frac1{\sqrt{\rho_0\chi_S}} ,

(the same for $\rho_1$ and for $\vect v$ ). For a perfect gas, $\chi_S = 1/\gamma P_0$ and

c = \sqrt{\frac{\gamma P_0}{\rho_0}} = \sqrt{\frac{\gamma RT}{M}} :

$343\,\mathrm{m}/\mathrm{s}$ in air at $20{}^{\circ}\mathrm{C}$ , rising by $0.6\,\mathrm{m}/\mathrm{s}$ per kelvin; $1000\,\mathrm{m}/\mathrm{s}$ in helium; $1480\,\mathrm{m}/\mathrm{s}$ in water ( $\chi_S = 4.5 \times 10^{-10}\,\mathrm{Pa}^{-1}$ ).

Proof. Euler: the convective term $\rho(\vect v\cdot\operatorname{\vect{grad}})\vect v$ is second order, $\rho\,\partial_t\vect v \approx \rho_0\,\partial_t\vect v$ , and $\operatorname{\vect{grad}}P = \operatorname{\vect{grad}}p$ . Mass: $\partial_t\rho + \operatorname{div}(\rho\vect v) \approx \partial_t\rho_1 + \rho_0\operatorname{div}\vect v$ . Isentropic law: $\dd\rho = \rho\chi_S\,\dd P$ to first order. Take the divergence of the first equation and the time derivative of the second: $\rho_0\partial_t\operatorname{div}\vect v = -\Delta p = -\partial_t^2\rho_1 = -\rho_0\chi_S \partial_t^2p$ . For the perfect gas, $PV^\gamma$ constant along an isentropic transformation gives $\dd\rho/\rho = \dd P/\gamma P$ , hence $\chi_S = 1/\gamma P$ , and $P_0/\rho_0 = RT/M$ . ∎

Remark 7.3 (Why isentropic)

Newton computed the speed of sound with the isothermal compressibility, $\sqrt{P_0/\rho_0} = 290\,\mathrm{m}/\mathrm{s}$ , $15\%$ too low; Laplace saw that the compressions are too fast for heat to flow between the warm crests and the cool troughs — over one period, heat diffuses a few micrometres (Chapter 25), nothing compared with a wavelength — so the transformation is adiabatic and reversible, and $\gamma$ appears. The speed of sound is thus a direct measurement of $\gamma$ : $1.40$ for air, $1.67$ for argon.

Left: a slab of fluid displaced by (x, t) and pushed by the pressures on its faces; its compression sets the overpressure. Right: a plane sound wave — alternate zones of compression and rarefaction moving at c, the fluid itself oscillating back and forth along the direction of propagation (a longitudinal wave). — Left: a slab of fluid displaced by $\xi(x, t)$ and pushed by the pressures on its faces; its compression sets the overpressure. Right: a plane sound wave — alternate zones of compression and rarefaction moving at $c$ , the fluid itself oscillating back and forth along the direction of propagation (a longitudinal wave).

7.2 Plane waves, impedance, intensity

Proposition 7.4 (Plane progressive sound wave)

For a plane wave $p = f(x - ct)$ travelling toward $+x$ , the fluid velocity is along $x$ , in phase with the overpressure, and

p = Z\,v , \qquad Z = \rho_0c = \sqrt{\rho_0/\chi_S} ,

where $Z$ is the acoustic impedance of the medium ( $\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ ): $410\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ for air, $1.5 \times 10^{6}\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ for water, $\sim1.6 \times 10^{6}\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ for soft tissue, $4 \times 10^{7}\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ for steel. The wave carries the energy density and the intensity (power per unit area, toward $+x$ )

e = \tfrac12\rho_0v^2 + \tfrac12\chi_Sp^2 , \qquad I = p\,v ,

which obey $\partial_te + \partial_xI = 0$ ; for the progressive wave the two halves of $e$ are equal and $I = p^2/Z = Zv^2$ . A sinusoidal wave of pressure amplitude $p_m$ has the mean intensity $\langle I\rangle = p_m^2/2Z$ .

Proof. For $p = f(x - ct)$ , $\rho_0\partial_tv = -\partial_xp = f'/c\cdot c\dots$ : precisely, $\rho_0\partial_tv = -f'(x - ct)$ , and $v = f(x - ct)/\rho_0c$ satisfies it. The potential term of $e$ is the work stored in compressing the fluid (an isentropic compression of unit volume by $\dd V/V = -\chi_S\,\dd p$ stores $\int p\,\chi_S\dd p = \tfrac12\chi_Sp^2$ ); $I$ is the work rate of the pressure force $pS$ on the fluid ahead, moving at $v$ , per unit area. Balance: $\partial_te = \rho_0v\partial_tv + \chi_Sp\partial_tp = -v\partial_xp - p\partial_xv = -\partial_x(pv)$ by the two linearized equations. ∎

Definition 7.5 (Sound level)

The sound level of a wave of mean intensity $I$ is

L = 10\log_{10}\frac I{I_0} \ \text{dB} , \qquad I_0 = 1 \times 10^{-12}\,\mathrm{W}/\mathrm{m}^{2}

— the threshold of hearing at $1\,\mathrm{kHz}$ , corresponding in air to the pressure amplitude $p_0 \approx 2 \times 10^{-5}\,\mathrm{Pa}$ (rms) $= 2 \times 10^{-10}$ atmospheres. Doubling the intensity adds $3\,\mathrm{dB}$ ; ten times, $10\,\mathrm{dB}$ ; a hundred times the pressure amplitude, $40\,\mathrm{dB}$ . Quiet room $30\,\mathrm{dB}$ , conversation $60\,\mathrm{dB}$ , busy street $80\,\mathrm{dB}$ , rock concert $110\,\mathrm{dB}$ , pain $120\,\mathrm{dB}$ ( $I = 1\,\mathrm{W}/\mathrm{m}^{2}$ , $p_m = 29\,\mathrm{Pa}$ ).

Example 7.6 (How small a sound is)

At the threshold, $1\,\mathrm{kHz}$ : $p_m = 2.8 \times 10^{-5}\,\mathrm{Pa}$ , $v_m = p_m/Z = 7 \times 10^{-8}\,\mathrm{m}/\mathrm{s}$ , and the displacement amplitude $\xi_m = v_m/\omega = 1 \times 10^{-11}\,\mathrm{m}$ — a tenth of an atomic diameter: the eardrum detects motions smaller than an atom. At $120\,\mathrm{dB}$ , $\xi_m = 11\,\text{µ}\mathrm{m}$ , still invisible, and $p_m/P_0 = 3 \times 10^{-4}$ : even the loudest sounds are tiny perturbations, which is why the linear theory works so well.

Left: in a plane progressive sound wave the overpressure and the fluid velocity are in phase, and the displacement lags by a quarter period. Right: the decibel scale — each 20\, dB step multiplies the intensity by a hundred and the pressure amplitude by ten. — Left: in a plane progressive sound wave the overpressure and the fluid velocity are in phase, and the displacement lags by a quarter period. Right: the decibel scale — each $20\,\mathrm{dB}$ step multiplies the intensity by a hundred and the pressure amplitude by ten.

7.3 Spherical waves

Proposition 7.7 (Spherical waves and the inverse-square law)

A small source radiating equally in all directions produces the spherical wave

p(r, t) = \frac{A}{r}\,f\Bigl(t - \frac rc\Bigr) ,

whose amplitude falls as $1/r$ and whose intensity falls as $1/r^2$ : for a source of acoustic power $\mathcal P$ , $I = \mathcal P/4\pi r^2$ , i.e. $-6\,\mathrm{dB}$ each time the distance doubles. Far from the source the wave is locally plane, with $v = p/Z$ radial.

Proof. In spherical coordinates, for a function of $r$ alone, $\Delta p = \frac1r\frac{\partial^2(rp)}{\partial r^2}$ (admitted; Chapter 11), so $u = rp$ obeys the one-dimensional wave equation $\partial_r^2u = \partial_t^2u/c^2$ , whence $u = Af(t - r/c)$ for the outgoing wave. The power crossing a sphere of radius $r$ is $4\pi r^2I$ , constant if nothing is absorbed. ∎

Example 7.8 (A siren, a conversation)

A $10\,\mathrm{W}$ siren: $I = 10/4\pi r^2$ , $0.8\,\mathrm{W}/\mathrm{m}^{2}$ at $1\,\mathrm{m}$ ( $119\,\mathrm{dB}$ , pain), $79\,\mathrm{dB}$ at $100\,\mathrm{m}$ , $59\,\mathrm{dB}$ at $1\,\mathrm{km}$ — still audible over the city. A voice radiates about $10\,\text{µ}\mathrm{W}$ : $60\,\mathrm{dB}$ at $1\,\mathrm{m}$ , and $40\,\mathrm{dB}$ at $10\,\mathrm{m}$ . Air absorbs sound too (more at high frequency, a few decibels per hundred metres at $10\,\mathrm{kHz}$ ), which is why distant thunder rumbles low.

7.4 The Doppler effect

Theorem 7.9 (Doppler effect)

A source emits at frequency $f$ and moves at speed $v_s$ along the line joining it to an observer who moves at $v_o$ , both speeds counted positive when they approach each other; the sound travels at $c$ in the air at rest. The observer receives the frequency

f' = f\,\frac{c + v_o}{c - v_s} .

A moving source crowds its wavefronts ahead of it ( $\lambda' = (c - v_s)/f$ ) and spreads them behind; a moving observer meets wavefronts at the rate $(c + v_o)/\lambda$ . For $v_s, v_o \ll c$ both give $\Delta f/f \approx (v_s + v_o)/c$ .

Proof. Moving source: the wavefronts emitted at $t$ and $t + 1/f$ are separated by $c/f - v_s/f$ in the direction of motion, so the wavelength ahead is $\lambda' = (c - v_s)/f$ and the frequency received by a fixed observer is $c/\lambda'$ . Moving observer: the wavelength is $\lambda = c/f$ but fronts pass at the relative speed $c + v_o$ . Combine. ∎

Left: the wavefronts of a moving source crowd ahead and spread behind — higher pitch approaching, lower receding. Right: a source faster than sound leaves its wavefronts behind; their envelope is the Mach cone, heard on the ground as a boom.

Example 7.10 (Sirens, radar guns, red cells)

An ambulance siren at $700\,\mathrm{Hz}$ passing at $25\,\mathrm{m}/\mathrm{s}$ : $755\,\mathrm{Hz}$ approaching, $651\,\mathrm{Hz}$ receding — a drop of a sixth of an octave, the familiar "nee-naw" dip. A police radar gun measures the same effect on a $24\,\mathrm{GHz}$ radio wave reflected by a car (the wave is shifted twice, by the car receiving and re-emitting): $\Delta f = 2fv/c = 4.8\,\mathrm{kHz}$ at $30\,\mathrm{m}/\mathrm{s}$ . An ultrasound probe at $5\,\mathrm{MHz}$ aimed at an artery hears the blood cells at $\Delta f = 2fv\cos\theta/c \approx 1.6\,\mathrm{kHz}$ for $0.5\,\mathrm{m}/\mathrm{s}$ at $60{}^{\circ}$ : an audible whistle whose pitch is the blood speed. And the spectral lines of a receding galaxy are shifted red by $\Delta\lambda/\lambda = v/c$ — the formula holds for light at low speed, with the relativistic correction beyond.

Remark 7.11 (Supersonic)

When $v_s > c$ the formula breaks down ( $f' < 0$ ahead): no sound precedes the source. The wavefronts emitted along the path have a conical envelope of half-angle $\theta$ with $\sin\theta = c/v_s$ (the Mach cone); the pressure jump carried by the cone is the sonic boom, which reaches a listener on the ground only after the aircraft has passed — at Mach $2$ and $10\,\mathrm{km}$ of altitude, $\theta = 30{}^{\circ}$ and the boom arrives $10\,\text{km}/\tan\theta/v_s \approx 25\,\mathrm{s}$ after the overhead passage. A bullet’s crack and the tip of a whip are the same cone.

Method 7.12 (Orders of magnitude in acoustics)

Given a sound level $L$ : $I = I_0\,10^{L/10}$ ; then $p_m = \sqrt{2ZI}$ , $v_m = p_m/Z$ , $\xi_m = v_m/\omega$ ; the power of a source is $4\pi r^2I$ if it radiates in all directions (twice less over a hard floor, which reflects into a half-space). Going from $r_1$ to $r_2$ changes $L$ by $20\log_{10}(r_1/r_2)$ . Adding $n$ incoherent equal sources adds $10\log_{10}n$ ; two coherent sources in phase at a point add $6\,\mathrm{dB}$ .

7.5 Exercises

Exercise 7.1 ★

Speed of sound in air ( $\gamma = 1.40$ , $M = 29\,\mathrm{g}/\mathrm{mol}$ ) at $0{}^{\circ}\mathrm{C}$ , $20{}^{\circ}\mathrm{C}$ and at $-50{}^{\circ}\mathrm{C}$ ( $10\,\mathrm{km}$ altitude); in helium ( $\gamma = 1.67$ , $M = 4\,\mathrm{g}/\mathrm{mol}$ ) and carbon dioxide ( $\gamma = 1.30$ , $M = 44\,\mathrm{g}/\mathrm{mol}$ ) at $20{}^{\circ}\mathrm{C}$ . Why does a voice sound high-pitched after breathing helium?

Solution

Solution of Exercise 7.1.

$c = \sqrt{\gamma RT/M}$ : $331\,\mathrm{m}/\mathrm{s}$ , $343\,\mathrm{m}/\mathrm{s}$ , $299\,\mathrm{m}/\mathrm{s}$ ; helium $1010\,\mathrm{m}/\mathrm{s}$ ; CO $_2$ $268\,\mathrm{m}/\mathrm{s}$ . The vocal cords vibrate at the same frequency, but the resonances of the vocal tract (which shape the timbre) scale with $c$ and move up by a factor three: the voice sounds high-pitched.

Exercise 7.2 ★

A conversation at $60\,\mathrm{dB}$ , $500\,\mathrm{Hz}$ : intensity, pressure amplitude, velocity amplitude and displacement amplitude in air ( $Z = 410\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ ). Same for $120\,\mathrm{dB}$ . Level of two people talking at once; of a hundred.

Solution

Solution of Exercise 7.2.

$60\,\mathrm{dB}$ : $I = 1 \times 10^{-6}\,\mathrm{W}/\mathrm{m}^{2}$ , $p_m = \sqrt{2ZI} = 29\,\mathrm{mPa}$ , $v_m = p_m/Z = 7 \times 10^{-5}\,\mathrm{m}/\mathrm{s}$ , $\xi_m = v_m/\omega = 22\,\mathrm{nm}$ . $120\,\mathrm{dB}$ : $1\,\mathrm{W}/\mathrm{m}^{2}$ , $29\,\mathrm{Pa}$ , $7\,\mathrm{cm}/\mathrm{s}$ , $22\,\text{µ}\mathrm{m}$ . Two: $63\,\mathrm{dB}$ ; a hundred: $80\,\mathrm{dB}$ .

Exercise 7.3 ★

A loudspeaker radiates $1.0\,\mathrm{W}$ of sound equally in all directions. Intensity and level at $1\,\mathrm{m}$ , $10\,\mathrm{m}$ , $100\,\mathrm{m}$ ; distance at which the level falls to $60\,\mathrm{dB}$ ; power needed for $100\,\mathrm{dB}$ at $30\,\mathrm{m}$ (an open-air concert).

Solution

Solution of Exercise 7.3.

$I = 1/4\pi r^2$ : $0.080\,\mathrm{W}/\mathrm{m}^{2}$ ( $109\,\mathrm{dB}$ ), $89\,\mathrm{dB}$ , $69\,\mathrm{dB}$ ; $I = 10^{-6}$ at $r = \sqrt{1/4\pi10^{-6}} = 280\,\mathrm{m}$ ; $100\,\mathrm{dB}$ at $30\,\mathrm{m}$ needs $4\pi \times 900 \times 10^{-2} = 110\,\mathrm{W}$ of sound.

Exercise 7.4 ★

Thunder is heard $4.5\,\mathrm{s}$ after the flash: distance. A ship’s horn echoes off a cliff after $3.0\,\mathrm{s}$ : distance. Sonar: an echo from the sea floor returns after $2.4\,\mathrm{s}$ ( $c = 1500\,\mathrm{m}/\mathrm{s}$ ): depth. A $5\,\mathrm{MHz}$ ultrasound echo returns after $130\,\text{µ}\mathrm{s}$ in tissue ( $1540\,\mathrm{m}/\mathrm{s}$ ): depth.

Solution

Solution of Exercise 7.4.

$4.5 \times 340 = 1.5\,\mathrm{km}$ ; $340 \times 1.5 = 510\,\mathrm{m}$ ; $1500 \times 1.2 = 1.8\,\mathrm{km}$ ; $1540 \times 65 \times 10^{-6} = 10\,\mathrm{cm}$ .

Exercise 7.5 ★★

A train horn at $500\,\mathrm{Hz}$ ; $c = 340\,\mathrm{m}/\mathrm{s}$ . (a) Frequency heard by a person on the platform as the train approaches at $40\,\mathrm{m}/\mathrm{s}$ , and after it has passed. (b) The person is on a train moving toward the horn at $40\,\mathrm{m}/\mathrm{s}$ , the horn at rest; then both moving toward each other at $40\,\mathrm{m}/\mathrm{s}$ : compare with a relative speed of $80\,\mathrm{m}/\mathrm{s}$ and one body at rest. (c) Frequency heard by the driver of the horn’s train from the echo off a wall ahead.

Solution

Solution of Exercise 7.5.

(a) $500 \times 340/300 = 567\,\mathrm{Hz}$ ; $500 \times 340/380 = 447\,\mathrm{Hz}$ . (b) $500 \times 380/340 = 559\,\mathrm{Hz}$ ; both: $500 \times 380/300 = 633\,\mathrm{Hz}$ , against $654\,\mathrm{Hz}$ (source alone at $80\,\mathrm{m}/\mathrm{s}$ ) or $618\,\mathrm{Hz}$ (observer alone): the air, not the relative speed, sets the result. (c) The wall receives $567\,\mathrm{Hz}$ and re-emits it at rest; the driver approaches at $40\,\mathrm{m}/\mathrm{s}$ : $567 \times 380/340 = 633\,\mathrm{Hz}$ .

Exercise 7.6 ★★

For a pressure amplitude of $1.0\,\mathrm{Pa}$ at $1\,\mathrm{kHz}$ , compute the velocity and displacement amplitudes, the intensity and the level in air, in water ( $Z = 1.5 \times 10^{6}\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ ) and in steel ( $4 \times 10^{7}\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ ). Same pressure, very different intensities: comment. (Underwater levels are quoted relative to $1\,\text{µ}\mathrm{Pa}$ : what is $1\,\mathrm{Pa}$ on that scale?)

Solution

Solution of Exercise 7.6.

$v_m = p_m/Z$ : $2.4\,\mathrm{mm}/\mathrm{s}$ , $0.67\,\text{µ}\mathrm{m}/\mathrm{s}$ , $25\,\mathrm{nm}/\mathrm{s}$ ; $\xi_m = v_m/\omega$ : $390\,\mathrm{nm}$ , $0.11\,\mathrm{nm}$ , $4\,\mathrm{pm}$ ; $I = p_m^2/2Z$ : $1.2 \times 10^{-3}\,\mathrm{W}/\mathrm{m}^{2}$ ( $91\,\mathrm{dB}$ ), $3.3 \times 10^{-7}\,\mathrm{W}/\mathrm{m}^{2}$ ( $55\,\mathrm{dB}$ ), $1.3 \times 10^{-8}\,\mathrm{W}/\mathrm{m}^{2}$ ( $41\,\mathrm{dB}$ ). A stiff medium barely moves under a given pressure and carries little power. $1\,\mathrm{Pa}$ is $120\,\mathrm{dB}$ re $1\,\text{µ}\mathrm{Pa}$ .

Exercise 7.7 ★★

(a) Newton’s isothermal speed $\sqrt{P_0/\rho_0}$ for air at $20{}^{\circ}\mathrm{C}$ ( $\rho_0 = 1.20\,\mathrm{kg}/\mathrm{m}^{3}$ ) and Laplace’s correction. (b) Measured: $343\,\mathrm{m}/\mathrm{s}$ in air, $323\,\mathrm{m}/\mathrm{s}$ in argon ( $M = 40\,\mathrm{g}/\mathrm{mol}$ ) at $20{}^{\circ}\mathrm{C}$ : deduce $\gamma$ for each; interpret with the number of degrees of freedom. (c) Estimate how far heat diffuses in one period at $1\,\mathrm{kHz}$ (thermal diffusivity of air $2 \times 10^{-5}\,\mathrm{m}^{2}/\mathrm{s}$ ) and compare with the wavelength.

Solution

Solution of Exercise 7.7.

(a) $\sqrt{1.013 \times 10^5/1.2} = 290\,\mathrm{m}/\mathrm{s}$ ; $\times\sqrt{1.4}$ : $343\,\mathrm{m}/\mathrm{s}$ . (b) $\gamma = c^2M/RT$ : $1.40$ for air, $1.71$ for argon ( $5/3$ within the rounding): three translational degrees of freedom ( $\gamma = 1 + 2/3$ ) against five for a diatomic gas ( $1 + 2/5$ ). (c) $\sqrt{D/f} = \sqrt{2 \times 10^{-8}} = 0.14\,\mathrm{mm} \ll \lambda = 34\,\mathrm{cm}$ : no heat flows between crests and troughs.

Exercise 7.8 ★★

Organ pipes. In a pipe the overpressure is zero at an open end and the velocity is zero at a closed end. (a) Standing waves in a pipe open at both ends: show the modes are $f_n = nc/2L$ ; in a pipe closed at one end, $f_n = (2n - 1)c/4L$ . (b) Lengths for a $440\,\mathrm{Hz}$ fundamental in each case. (c) Which harmonics does each pipe contain? (d) The real open end sits about $0.6$ radius beyond the pipe (end correction): for a $4\,\mathrm{cm}$ pipe, the correction on $L$ and on $f$ .

Solution

Solution of Exercise 7.8.

(a) Open–open: $p = p_m\sin kx\cos\omega t$ with $kL = n\pi$ ; closed–open: $p = p_m\cos kx\cos\omega t$ (velocity node, pressure antinode at the closed end) with $kL = (2n - 1)\pi/2$ . (b) $L = c/2f = 39\,\mathrm{cm}$ ; $c/4f = 19.5\,\mathrm{cm}$ . (c) All harmonics; odd harmonics only (the clarinet’s hollow sound). (d) $0.6 \times 2 = 1.2\,\mathrm{cm}$ per open end: $L_{\text{eff}} = 41.4\,\mathrm{cm}$ , $f = 414\,\mathrm{Hz}$ , $6\%$ flat — the pipe must be cut to $36.6\,\mathrm{cm}$ .

Exercise 7.9 ★★

A $2.0\,\mathrm{kW}$ electric siren converts $20\%$ of its power into sound radiated into a half-space (it sits on a roof). (a) Intensity, level and pressure amplitude at $100\,\mathrm{m}$ . (b) Distance of the pain threshold ( $120\,\mathrm{dB}$ ); of the $70\,\mathrm{dB}$ contour where it stops being alarming. (c) Air absorption adds $1\,\mathrm{dB}$ per $100\,\mathrm{m}$ at its frequency: distance of the $70\,\mathrm{dB}$ contour now (solve numerically). (d) How many such sirens cover a city of $10\,\mathrm{km}$ radius?

Solution

Solution of Exercise 7.9.

$\mathcal P = 400\,\mathrm{W}$ , $I = \mathcal P/2\pi r^2$ . (a) $6.4 \times 10^{-3}\,\mathrm{W}/\mathrm{m}^{2}$ , $98\,\mathrm{dB}$ , $p_m = 2.3\,\mathrm{Pa}$ . (b) $r = \sqrt{400/2\pi} = 8\,\mathrm{m}$ ; $70\,\mathrm{dB}$ at $2.5\,\mathrm{km}$ . (c) $98 - 20\log(r/100) - (r - 100)/100 = 70$ : $r \approx 950\,\mathrm{m}$ . (d) $\pi(10)^2/\pi(0.95)^2 \approx 110$ sirens.

Exercise 7.10 ★★★

Energy of a standing sound wave. In a closed tube of length $L$ the overpressure is $p = p_m\cos(kx)\cos(\omega t)$ , $k = n\pi/L$ . (a) Find $v(x, t)$ from the linearized Euler equation. (b) Kinetic and potential energy densities; show that they oscillate in quadrature in time and are shifted by a quarter wavelength in space. (c) Total energy in the tube (section $S$ ); check it is constant. (d) Mean intensity: zero everywhere — and yet the energy moves: describe how.

Solution

Solution of Exercise 7.10.

(a) $\rho_0\partial_tv = p_mk\sin kx\cos\omega t$ : $v = (p_m/Z)\sin kx\sin\omega t$ . (b) $e_k = \tfrac12\chi_Sp_m^2\sin^2kx\sin^2\omega t$ , $e_p = \tfrac12\chi_Sp_m^2\cos^2kx\cos^2\omega t$ : one is maximal when the other vanishes, in time and in space. (c) $E = S\int_0^L(e_k + e_p)\dd x = \tfrac14\chi_Sp_m^2SL$ , constant. (d) $I = pv = (p_m^2/4Z)\sin2kx\sin2\omega t$ : zero mean, but twice per period energy flows from the pressure antinodes to the velocity antinodes and back.

Exercise 7.11 ★★★

Sonic boom. An aircraft flies horizontally at $v = 1.6\,c$ at altitude $h = 12\,\mathrm{km}$ , $c = 300\,\mathrm{m}/\mathrm{s}$ there. (a) Derive the Mach angle from the envelope of the spherical wavefronts. (b) Time between the overhead passage and the boom on the ground (neglect the variation of $c$ with altitude). (c) Width on the ground of the strip that hears the boom as the aircraft flies by, if the cone’s trace is heard within $\pm30\,\mathrm{km}$ of the track: why is supersonic flight over land banned? (d) A rifle bullet at $800\,\mathrm{m}/\mathrm{s}$ : Mach angle; what an observer standing $10\,\mathrm{m}$ from the trajectory hears, and in what order.

Solution

Solution of Exercise 7.11.

(a) The front emitted at $t = 0$ has radius $ct$ when the source is $vt$ away: the tangent cone has $\sin\theta = c/v$ . (b) $\theta = 39{}^{\circ}$ ; the cone reaches the ground under the aircraft when it is $h/\tan\theta = 15\,\mathrm{km}$ further: $15000/480 = 31\,\mathrm{s}$ after the overhead passage. (c) A $60\,\mathrm{km}$ -wide carpet along the whole route hears the boom. (d) $\sin\theta = 340/800$ , $\theta = 25{}^{\circ}$ ; the crack of the cone arrives first, the muzzle blast, travelling at $c$ from the gun, later.

Exercise 7.12 ★★★

Doppler ultrasound. A probe emits at $f = 5.0\,\mathrm{MHz}$ into tissue ( $c = 1540\,\mathrm{m}/\mathrm{s}$ ); red cells move at $v$ along a vessel making the angle $\theta$ with the beam. (a) Frequency received by a cell (moving observer). (b) It re-emits (scatters) this frequency as a moving source: frequency back at the probe; show $\Delta f = 2fv\cos\theta/c$ for $v \ll c$ . (c) $\Delta f$ for $v = 0.50\,\mathrm{m}/\mathrm{s}$ , $\theta = 60{}^{\circ}$ ; for $\theta = 90{}^{\circ}$ — what does the operator do? (d) The probe can resolve $50\,\mathrm{Hz}$ : velocity resolution. Why is the Doppler signal made audible rather than displayed only?

Solution

Solution of Exercise 7.12.

(a) $f_1 = f(c + v\cos\theta)/c$ . (b) $f_2 = f_1c/(c - v\cos\theta) = f(c + v\cos\theta) /(c - v\cos\theta) \approx f(1 + 2v\cos\theta/c)$ . (c) $2 \times 5 \times 10^6 \times 0.5 \times 0.5/1540 = 1.6\,\mathrm{kHz}$ ; zero at $90{}^{\circ}$ : tilt the probe. (d) $\Delta v = 50 \times 1540/ (2 \times 5 \times 10^6 \times 0.5) = 1.5\,\mathrm{cm}/\mathrm{s}$ . The ear follows pitch and timbre in real time: a smooth whistle means laminar flow, a hiss means turbulence behind a narrowing.

An ultrasound scanner: a few megahertz of sound, timed echoes and the impedance mismatches between tissues draw the image; the gel removes the film of air that would reflect everything.

7.6 Problem: From thunder to the echograph

Problem 7.1

Weekend problem — sound in the open air, in a concert hall, inside the body and from a passing siren

Air: $\gamma = 1.40$ , $M = 29\,\mathrm{g}/\mathrm{mol}$ , $Z = 410\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ at $15{}^{\circ}\mathrm{C}$ ; $R = 8.31\,\mathrm{J}/(\mathrm{mol}\,\mathrm{K})$ .

Part I — The storm.

Speed of sound at $15{}^{\circ}\mathrm{C}$ and at $30{}^{\circ}\mathrm{C}$ ; justify the rule "three seconds per kilometre".
Thunder arrives $4.0\,\mathrm{s}$ after the flash: distance of the strike. The thunder lasts several seconds although the flash is instantaneous: why?
Near the strike the level reaches $120\,\mathrm{dB}$ : pressure amplitude and velocity amplitude of the air.
Energy received per square metre of wall during a thunderclap of $2\,\mathrm{s}$ at $120\,\mathrm{dB}$ .
On a clear night the ground cools and the air is colder below than above. Sound rays obey Snell’s law between layers, $\sin i/c =$ const: do rays launched upward bend down or up? Why are distant sounds heard so well at night and over water?
By day, the reverse: a listener at $3\,\mathrm{km}$ hears nothing. Estimate the angle by which a horizontal ray has bent after $3\,\mathrm{km}$ if $c$ falls by $1\,\%$ per $500\,\mathrm{m}$ of altitude (treat the bending as uniform: the ray is a circle).
Air absorbs high frequencies far more than low ones: why does distant thunder rumble while a nearby strike cracks?

Part II — The concert. A loudspeaker receives $100\,\mathrm{W}$ of electrical power and converts $2\%$ of it into sound, radiated equally into the half-space in front of it.

Acoustic power; intensity and level at $10\,\mathrm{m}$ .
Pressure amplitude there; velocity and displacement amplitudes at $100\,\mathrm{Hz}$ .
Level at $1\,\mathrm{m}$ (is it safe?) and distance at which the music falls to $60\,\mathrm{dB}$ .
Ten identical loudspeakers spread around the stage, not in phase: level at $10\,\mathrm{m}$ . Two loudspeakers fed in phase, at the same distance from a listener on their axis: level there.
The hall (volume $10\,000\,\mathrm{m}^{3}$ ) has $2000\,\mathrm{m}^{2}$ of walls and audience that absorb $30\%$ of the sound hitting them: estimate the acoustic energy stored in the hall at equilibrium and the reverberation time (time for the level to drop $60\,\mathrm{dB}$ after the music stops), using the balance between the power emitted and the power absorbed ( $I_{\text{wall}} \approx ce/4$ for a diffuse field, admitted).

Part III — The echograph. Soft tissue: $c = 1540\,\mathrm{m}/\mathrm{s}$ , $Z = 1.6 \times 10^{6}\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ ; bone $Z = 7 \times 10^{6}\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ ; air $400\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ . At normal incidence the pressure amplitude reflected at an interface is $r = (Z_2 - Z_1)/(Z_2 + Z_1)$ times the incident one (derived in Chapter 15); tissue attenuates sound by about $0.5\,\mathrm{dB}$ per centimetre and per megahertz.

Wavelength in tissue at $3\,\mathrm{MHz}$ , $5\,\mathrm{MHz}$ , $10\,\mathrm{MHz}$ ; the smallest detail each can resolve is about one wavelength.
Echo delay from an organ $12\,\mathrm{cm}$ deep; maximum rate at which pulses can be sent without confusing echoes; time to build an image of $200$ lines.
Fraction of the pressure, and of the energy, reflected at a tissue–air interface; at a tissue–bone interface; between fat ( $Z = 1.38 \times 10^{6}\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ ) and muscle ( $1.70 \times 10^{6}\,\mathrm{kg}\,\mathrm{m}^{-2}\,\mathrm{s}^{-1}$ ). Why the gel, and why bones cast shadows?
At $5\,\mathrm{MHz}$ the beam crosses $2\,\mathrm{cm}$ of fat before meeting muscle: fraction of the emitted energy that comes back to the probe from that interface (reflection and round-trip attenuation), in decibels.
Round-trip attenuation in decibels for the organ of question 12 at each of the three frequencies; if the scanner can handle $80\,\mathrm{dB}$ of loss, which frequencies work? State the resolution versus depth trade-off.
The probe emits pulses of $1\,\text{µ}\mathrm{s}$ at $5\,\mathrm{MHz}$ : how many wavelengths long is a pulse, and how does that limit the depth resolution?
Doppler mode at $5\,\mathrm{MHz}$ on an artery at $0.40\,\mathrm{m}/\mathrm{s}$ , beam at $60{}^{\circ}$ : shift; what the operator hears.

Part IV — The siren.

An ambulance passes at $90\,\mathrm{km}/\mathrm{h}$ with a $700\,\mathrm{Hz}$ siren: frequencies heard approaching and receding; the interval between them in semitones (a semitone is a ratio $2^{1/12}$ ).
The listener is in a car going at $15\,\mathrm{m}/\mathrm{s}$ toward the approaching ambulance: frequency heard.
At what instant does a listener on the pavement hear exactly $700\,\mathrm{Hz}$ ?
The ambulance’s wavefronts ahead of it: wavelength; speed of the sound relative to the ambulance ahead and behind.
A radar gun at $24\,\mathrm{GHz}$ reads a $3.2\,\mathrm{kHz}$ shift from a car: speed of the car (the wave is shifted twice).
Sum up: the six speeds of this problem (sound in air, in tissue, the ambulance, the car, the red cells, light) and the one formula that handles them all.

Solution

Solution of Problem 7.1.

1. $c = \sqrt{1.4 \times 8.31 \times 288/0.029} = 340\,\mathrm{m}/\mathrm{s}$ ; $349\,\mathrm{m}/\mathrm{s}$ at $30{}^{\circ}\mathrm{C}$ . $1\,\mathrm{km}$ in $2.9\,\mathrm{s}$ .

2. $1.4\,\mathrm{km}$ . The channel is kilometres long; the sound of its different parts arrives over seconds, with echoes from clouds and ground.

3. $I = 1\,\mathrm{W}/\mathrm{m}^{2}$ : $p_m = \sqrt{2ZI} = 29\,\mathrm{Pa}$ , $v_m = 7\,\mathrm{cm}/\mathrm{s}$ .

4. $It = 2\,\mathrm{J}/\mathrm{m}^{2}$ .

5. $c$ grows upward; $\sin i/c$ constant makes $i$ grow, so the ray bends toward the horizontal and back down: sound is channelled along the ground and carries far. Over water the lowest air is cooled by the water: the same.

6. The rays bend upward, leaving a shadow zone. Radius of curvature $R = c/(\dd c/\dd z) = 50\,\mathrm{km}$ : after $3\,\mathrm{km}$ the ray has turned by $3/50 = 0.06\,\mathrm{rad} = 3.4{}^{\circ}$ and risen $90\,\mathrm{m}$ : the listener is below it.

7. Absorption grows as $f^2$ : the crack’s high frequencies are gone after a few kilometres, the low rumble remains.

8. $\mathcal P = 2\,\mathrm{W}$ ; $I = 2/2\pi r^2 = 3.2 \times 10^{-3}\,\mathrm{W}/\mathrm{m}^{2}$ at $10\,\mathrm{m}$ : $95\,\mathrm{dB}$ .

9. $p_m = \sqrt{2 \times 410 \times 3.2 \times 10^{-3}} = 1.6\,\mathrm{Pa}$ ; $v_m = 3.9\,\mathrm{mm}/\mathrm{s}$ ; $\xi_m = v_m/2\pi f = 6\,\text{µ}\mathrm{m}$ .

10. $+20\,\mathrm{dB}$ : $115\,\mathrm{dB}$ at $1\,\mathrm{m}$ — damaging within minutes. $60\,\mathrm{dB}$ : $r = \sqrt{2/2\pi10^{-6}} = 560\,\mathrm{m}$ .

11. Ten incoherent: $+10\,\mathrm{dB}$ , $105\,\mathrm{dB}$ . Two in phase: the pressure doubles, $+6\,\mathrm{dB}$ , $101\,\mathrm{dB}$ .

12. Balance $\mathcal P = \alpha A\,ce/4$ : $e = 8/(0.3 \times 2000 \times 340) = 3.9 \times 10^{-5}\,\mathrm{J}/\mathrm{m}^{3}$ , $E = eV = 0.4\,\mathrm{J}$ . After the stop, $\dd E/\dd t = -(\alpha Ac /4V)E$ : $\tau = 4V/\alpha Ac = 0.20\,\mathrm{s}$ , and $60\,\mathrm{dB}$ takes $\tau\ln10^6 = 2.7\,\mathrm{s}$ (Sabine’s reverberation time).

13. $\lambda = c/f$ : $0.51\,\mathrm{mm}$ , $0.31\,\mathrm{mm}$ , $0.15\,\mathrm{mm}$ .

14. $2 \times 0.12/1540 = 156\,\text{µ}\mathrm{s}$ ; $6.4\,\mathrm{kHz}$ ; $200$ lines in $31\,\mathrm{ms}$ (thirty images a second).

15. Tissue–air: $r = -0.9995$ , $99.9\%$ of the energy reflected — hence the gel, which removes the air film. Tissue–bone: $r = 0.63$ , $40\%$ of the energy; the rest is absorbed by the bone, which casts a shadow. Fat–muscle: $r = 0.10$ , $1\%$ of the energy: the faint echoes the image is made of.

16. Reflected fraction $1.1\%$ ( $-20\,\mathrm{dB}$ ); round trip $4\,\mathrm{cm}$ at $5\,\mathrm{MHz}$ : $-10\,\mathrm{dB}$ ; in all $-30\,\mathrm{dB}$ , one thousandth.

17. Round trip $24\,\mathrm{cm}$ : $36\,\mathrm{dB}$ , $60\,\mathrm{dB}$ , $120\,\mathrm{dB}$ : $3\,\mathrm{MHz}$ and $5\,\mathrm{MHz}$ work, $10\,\mathrm{MHz}$ does not — deep organs are imaged at low frequency with coarse resolution, superficial ones at high frequency finely.

18. Five wavelengths, $1.5\,\mathrm{mm}$ long: two interfaces closer than about $0.8\,\mathrm{mm}$ (half the pulse) give overlapping echoes.

19. $2 \times 5 \times 10^6 \times 0.4 \times 0.5/1540 = 1.3\,\mathrm{kHz}$ : a whistle whose pitch rises and falls with each heartbeat.

20. $v = 25\,\mathrm{m}/\mathrm{s}$ : $700 \times 340/315 = 756\,\mathrm{Hz}$ , $700 \times 340/365 = 652\,\mathrm{Hz}$ ; ratio $1.16$ , $12\log_21.16 = 2.6$ semitones.

21. $700 \times 355/315 = 789\,\mathrm{Hz}$ .

22. At closest approach, when the ambulance’s velocity is perpendicular to the line of sight (strictly, when the sound now arriving was emitted at that point).

23. $\lambda' = (340 - 25)/700 = 0.45\,\mathrm{m}$ (against $0.49\,\mathrm{m}$ ); the sound recedes from the ambulance at $315\,\mathrm{m}/\mathrm{s}$ ahead and $365\,\mathrm{m}/\mathrm{s}$ behind.

24. $v = c\,\Delta f/2f = 3 \times 10^8 \times 3200/4.8 \times 10^{10} = 20\,\mathrm{m}/\mathrm{s} = 72\,\mathrm{km}/\mathrm{h}$ .

25. $340\,\mathrm{m}/\mathrm{s}$ , $1540\,\mathrm{m}/\mathrm{s}$ , $25\,\mathrm{m}/\mathrm{s}$ , $20\,\mathrm{m}/\mathrm{s}$ , $0.4\,\mathrm{m}/\mathrm{s}$ , $3 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ : $f' = f(c + v_o)/(c - v_s)$ , applied once or twice, and for light at low speed.

Terms defined in this chapter

See all 393 terms in the glossary