---
title: "Sound Waves in Fluids"
book: "University Physics — Year 2"
subject: physics
language: en
chapter: 7
exercises: 12
source: https://one-course.com/books/physics/4/en/chapter/7-sound-waves-in-fluids
---

# Chapter 7 — Sound 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](https://one-course.com/books/physics/4/en/chapter/3-perfect-fluids-euler-and-bernoulli#ch-b2-euler-bernoulli) and [2](https://one-course.com/books/physics/4/en/chapter/2-fluid-kinematics#ch-b2-fluid-kinematics) to a fluid at rest disturbed by a small pressure ripple, and finds the [d’Alembert equation](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#thm-b2-waves-on-strings-equation) again: the [speed of sound](#thm-b2-sound-waves-equation), 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](https://one-course.com/books/physics/4/en/chapter/2-fluid-kinematics#def-b2-fluid-kinematics-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](https://one-course.com/books/physics/4/en/chapter/3-perfect-fluids-euler-and-bernoulli#thm-b2-euler-bernoulli-euler), 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](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#thm-b2-waves-on-strings-equation) for the [overpressure](#def-b2-sound-waves-approximation),

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

(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](#thm-b2-sound-waves-equation) 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](https://one-course.com/books/physics/4/en/chapter/25-heat-conduction#ch-b2-heat-conduction)), nothing compared with a wavelength — so the transformation is adiabatic and reversible, and $\gamma$ appears. The [speed of sound](#thm-b2-sound-waves-equation) 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).](https://one-course.com/images/onecourse/chapters/physics-4/b2-sound-waves/fig-7bacfd66f4ad.svg)

*Left: a slab of fluid displaced by $\xi(x, t)$ and pushed by the pressures on its faces; its compression sets the [overpressure](#def-b2-sound-waves-approximation). 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](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#prop-b2-waves-on-strings-chain)).*

## 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](#def-b2-sound-waves-approximation), and

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

where $Z$ is the *[acoustic impedance](#prop-b2-sound-waves-plane)* 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.](https://one-course.com/images/onecourse/chapters/physics-4/b2-sound-waves/fig-4b7cdb0c2bb5.svg)

![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.](https://one-course.com/images/onecourse/chapters/physics-4/b2-sound-waves/fig-12ce0ecf8b38.svg)

*Left: in a plane progressive sound wave the [overpressure](#def-b2-sound-waves-approximation) and the fluid velocity are in phase, and the displacement lags by a quarter period. Right: the [decibel](#def-b2-sound-waves-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](#prop-b2-sound-waves-spherical)

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

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](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#ch-b2-maxwell-equations)), so $u = rp$ obeys the one-dimensional [wave equation](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#thm-b2-waves-on-strings-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](#def-b2-sound-waves-decibel) 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.](https://one-course.com/images/onecourse/chapters/physics-4/b2-sound-waves/fig-1bbdcb9d5aaf.svg)

*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](#rem-b2-sound-waves-mach), 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](#rem-b2-sound-waves-mach)*); the pressure jump carried by the cone is the [sonic boom](#rem-b2-sound-waves-mach), 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](#def-b2-sound-waves-decibel) $L$: $I = I_0\,10^{L/10}$; 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](#thm-b2-sound-waves-equation) 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 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 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 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 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 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 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 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](#def-b2-sound-waves-approximation) is zero at an open end and the velocity is zero at a closed end. (a) [Standing waves](https://one-course.com/books/physics/4/en/chapter/6-waves-on-strings-and-rods-the-dalembert-equation#prop-b2-waves-on-strings-modes) 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 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 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](#def-b2-sound-waves-approximation) 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 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](#rem-b2-sound-waves-mach).* 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 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 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](https://one-course.com/books/physics/4/en/chapter/4-viscous-flows#def-b2-viscous-flows-reynolds), 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.](https://one-course.com/images/onecourse/chapters/physics-4/b2-sound-waves/img-0db9cd7d5b28.jpg)

*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.**

1. [Speed of sound](#thm-b2-sound-waves-equation) at $15{}^{\circ}\mathrm{C}$ and at $30{}^{\circ}\mathrm{C}$ ; justify the rule "three seconds per kilometre".
2. Thunder arrives $4.0\,\mathrm{s}$ after the flash: distance of the strike. The thunder lasts several seconds although the flash is instantaneous: why?
3. Near the strike the level reaches $120\,\mathrm{dB}$ : pressure amplitude and velocity amplitude of the air.
4. Energy received per square metre of wall during a thunderclap of $2\,\mathrm{s}$ at $120\,\mathrm{dB}$ .
5. 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?
6. 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).
7. 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.

8. Acoustic power; intensity and level at $10\,\mathrm{m}$ .
9. Pressure amplitude there; velocity and displacement amplitudes at $100\,\mathrm{Hz}$ .
10. Level at $1\,\mathrm{m}$ (is it safe?) and distance at which the music falls to $60\,\mathrm{dB}$ .
11. 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.
12. 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](https://one-course.com/books/physics/4/en/chapter/15-reflection-and-transmission-at-interfaces#ch-b2-wave-interfaces)); tissue attenuates sound by about $0.5\,\mathrm{dB}$ per centimetre and per megahertz.

13. Wavelength in tissue at $3\,\mathrm{MHz}$ , $5\,\mathrm{MHz}$ , $10\,\mathrm{MHz}$ ; the smallest detail each can resolve is about one wavelength.
14. 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.
15. 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?
16. 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](#def-b2-sound-waves-decibel) .
17. Round-trip attenuation in [decibels](#def-b2-sound-waves-decibel) 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.
18. 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?
19. 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.**

20. 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 $2^{1/12}$ ).
21. The listener is in a car going at $15\,\mathrm{m}/\mathrm{s}$ toward the approaching ambulance: frequency heard.
22. At what instant does a listener on the pavement hear exactly $700\,\mathrm{Hz}$ ?
23. The ambulance’s wavefronts ahead of it: wavelength; speed of the sound relative to the ambulance ahead and behind.
24. 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).
25. 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 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.