A wedding procession with a brass band and a loudspeaker truck passes down a lane. The glass panes of the houses rattle as it goes by. No air has blown from the loudspeaker to the windows. Yet something pushed on the glass and made it move. The sound brought energy to the window.
A sound wave carries energy
The chapter Propagation of Sound Waves described sound as a longitudinal wave in air. Each layer of air swings to and fro along the direction of travel, with a displacement \(s(x, t)\). The layers crowd into compressions and spread into rarefactions. The chapter Propagating Waves on a String found the energy carried by a wave on a string. This lesson finds the energy carried by a sound wave in the same way.
Every slab of air that takes part in the wave has energy in two forms. The first is kinetic energy, because the slab moves to and fro. The second is potential energy, because the slab is squeezed or stretched. We find each in turn, and then compare them.
A slab of air
Take a tube of air with a cross-section of area \(A\). With no sound, the air has density \(\rho\) and pressure \(P_0\). Mark a thin slab of this air between the planes \(x\) and \(x + dx\). Its volume is \(A\,dx\), so its mass is \(dm = \rho A\,dx\).
When the sound passes, the left face of the slab moves by \(s\). The right face moves by a slightly different amount, \(s + ds\), as Fig 1.1 shows. The whole slab moves with the particle velocity \(\partial s/\partial t\). This is the speed of the air itself, not the speed of the wave.
Keep two speeds apart. The particle velocity \(\partial s/\partial t\) is the to-and-fro speed of one slab of air, and it changes from moment to moment. The wave speed \(v = \sqrt{B/\rho}\) is the speed of the pattern, about \(340\,\text{m/s}\) in air, and the medium fixes it.
\[ dK = \tfrac12\,(dm)\left(\dfrac{\partial s}{\partial t}\right)^2 \]
Why. the kinetic energy of a body of mass \(dm\) moving at the speed \(\partial s/\partial t\).
\[ dm = \rho A\,dx \]
Why. the mass of the slab is its density times its volume at rest.
\[ dK = \tfrac12\,(\rho A\,dx)\left(\dfrac{\partial s}{\partial t}\right)^2 \]
Why. put the mass into step 1. This is the kinetic energy of the slab.
\[ \dfrac{dK}{dV} = \tfrac12\,\rho\left(\dfrac{\partial s}{\partial t}\right)^2 \]
Read it. divide by the volume \(dV = A\,dx\) of the slab. This is the kinetic energy per unit volume at the point \(x\).
For a wave on a string the natural measure was energy per unit length. Air fills space, so the natural measure is energy per unit volume. It is called the energy density and has the unit J/m³.
In a loud sound, the air at some point moves with a particle velocity of \(0.05\,\text{m/s}\) at one instant. Take the density of air as \(1.2\,\text{kg/m}^3\). What is the kinetic energy per unit volume there, in mJ/m³?
Potential energy
A squeezed slab of air pushes back, like a compressed spring. So work must be done to squeeze it. That work is stored as potential energy. A stretched slab stores energy in the same way, because work was done to pull it apart against the air around it.
First find how much the slab is squeezed. Its thickness changes from \(dx\) to \(dx + ds\). Here \(ds = (\partial s/\partial x)\,dx\) is the difference between the displacements of its two faces. So its volume changes by \(\Delta V = A\,ds = A(\partial s/\partial x)\,dx\). Divide by the volume \(V = A\,dx\) to find the fractional change.
\[ \frac{\Delta V}{V} = \frac{\partial s}{\partial x} \]
The bulk modulus \(B\) of the air links this fractional change to the extra pressure \(\Delta P\) in the slab: \(B = -\Delta P/(\Delta V/V)\). So \(\Delta P = -B\,\partial s/\partial x\), the result of the chapter Propagation of Sound Waves. For air, \(B = \rho v^2 = 1.2 \times 340^2 \approx 1.39 \times 10^5\,\text{Pa}\).
Now squeeze the slab slowly from its volume at rest, \(V\), to its final volume \(V + \Delta V\). Let \(\delta V\) be the change in volume at some stage on the way. The extra pressure at that stage is \(\delta P = -B\,\delta V/V\). It grows in step with the squeeze, from zero to \(\Delta P\), just as the force of a spring grows with its compression.
\[ dU = -\int_0^{\Delta V} \delta P\; d(\delta V) \]
Why. the work done on a gas squeezed by \(d(\delta V)\) is minus the pressure times \(d(\delta V)\). Only the extra pressure of the wave counts here (see the caution below).
\[ dU = \int_0^{\Delta V} \dfrac{B}{V}\,\delta V\; d(\delta V) \]
Why. put in \(\delta P = -B\,\delta V/V\). The two minus signs cancel.
\[ dU = \dfrac{B\,(\Delta V)^2}{2V} \]
Why. the integral of \(\delta V\) from \(0\) to \(\Delta V\) is \((\Delta V)^2/2\). This has the shape of \(\tfrac12 kx^2\) for a spring.
\[ \Delta V = -\dfrac{V\,\Delta P}{B} \]
Why. the definition of the bulk modulus, \(B = -\Delta P/(\Delta V/V)\), solved for \(\Delta V\).
\[ dU = \dfrac{V\,(\Delta P)^2}{2B} \]
Why. put \(\Delta V\) into step 3. One factor \(V\) cancels.
\[ \dfrac{dU}{dV} = \dfrac{(\Delta P)^2}{2B} = \tfrac12\,B\left(\dfrac{\partial s}{\partial x}\right)^2 \]
Read it. divide by the volume of the slab. The second form uses \(\Delta P = -B\,\partial s/\partial x\).
The steady pressure \(P_0\) of the air also does work on the slab as it is squeezed. But the slab is squeezed and then stretched by the same amount in every cycle. So the work of \(P_0\) adds to zero on average. The potential energy of the wave is only the extra energy stored by the extra pressure. In the same way, the energy of a string was only the energy of its extra stretch.
The formula has a square of \(\Delta P\). So a compression and a rarefaction with the same size of \(\Delta P\) store the same energy. A slab stores energy whether it is squeezed or stretched, as a spring does.
At one instant, slab P of a sound wave has an extra pressure of \(+0.5\,\text{Pa}\), and slab Q has \(-0.5\,\text{Pa}\). Which stores more potential energy per unit volume?
At some point in a sound wave in air the extra pressure is \(1.0\,\text{Pa}\). Take \(B = 1.39 \times 10^5\,\text{Pa}\). What is the potential energy per unit volume there, in μJ/m³?
The two energies are equal
For a wave on a string, the kinetic and the potential energy of every piece were equal. The same holds for sound. The proof needs two results you already have. The first is the relation of the chapter Introduction to Waves, \(\partial s/\partial t = -v\,\partial s/\partial x\), for every wave \(s = f(x - vt)\). The second is the speed of sound, \(v^2 = B/\rho\).
\[ \left(\dfrac{\partial s}{\partial t}\right)^2 = v^2\left(\dfrac{\partial s}{\partial x}\right)^2 \]
Why. square \(\partial s/\partial t = -v\,\partial s/\partial x\). The minus sign disappears.
\[ \tfrac12\,\rho\left(\dfrac{\partial s}{\partial t}\right)^2 = \tfrac12\,\rho v^2\left(\dfrac{\partial s}{\partial x}\right)^2 \]
Why. multiply both sides by \(\tfrac12\rho\). The left side is now the kinetic energy per unit volume.
?
Your turn. which fact turns the right side into the potential energy per unit volume?
\[ \tfrac12\,\rho\left(\dfrac{\partial s}{\partial t}\right)^2 = \tfrac12\,B\left(\dfrac{\partial s}{\partial x}\right)^2 \]
Read it. put \(\rho v^2 = B\) into step 2. The right side is \((\Delta P)^2/(2B)\), the potential energy per unit volume.
In a sound wave travelling one way, the kinetic and the potential energy of every small slab of air are equal at every instant. The total energy density is twice the kinetic energy density, \(u = \rho(\partial s/\partial t)^2\).
The same argument shows a useful link between pressure and motion. Put \(\partial s/\partial x = -(1/v)\,\partial s/\partial t\) into \(\Delta P = -B\,\partial s/\partial x\), and use \(B/v = \rho v\). Then \(\Delta P = \rho v\,\partial s/\partial t\). The extra pressure of a slab is in step with its particle velocity. The air moves forward fastest where it is squeezed most.
A harmonic sound wave
Now apply these results to a harmonic sound wave, \(s = s_{max}\cos(kx - \omega t)\). Here \(s_{max}\) is the displacement amplitude, \(k\) the angular wave number and \(\omega\) the angular frequency. The two derivatives both contain the same sine.
\[ \begin{aligned} \frac{\partial s}{\partial t} &= \omega s_{max}\sin(kx - \omega t) \\ \Delta P = -B\frac{\partial s}{\partial x} &= Bk\,s_{max}\sin(kx - \omega t) \end{aligned} \]
The second line is the pressure wave of the chapter Propagation of Sound Waves. Put the first line into the kinetic energy density, and double it for the total. The energy density is then \(u = \rho\omega^2 s_{max}^2\sin^2(kx - \omega t)\).
A harmonic sound wave travels to the right in a tube of air. At one instant, where is the energy per unit volume greatest?
Press play on the figure above. The compressions C and the rarefactions R travel to the right at the wave speed. The humps of energy under them travel too, half a wavelength apart. The painted particles only swing to and fro. Each passes on its energy and stays near home.
On a string the energy sat where the string crossed its rest line, where the displacement was zero. The sound wave agrees. At a compression or a rarefaction the displacement \(s\) is zero, and the air moves fastest through its rest position.
The average energy density
At a fixed point the energy density rises and falls twice in every period. Its average over a period is steady. The average of \(\sin^2\) over one period is \(\tfrac12\), exactly as the average of \(\cos^2\) was for a string. So the average energy density is half the peak value.
The formula is the twin of \(\tfrac12\mu\omega^2A^2\) for a string. The mass per unit length \(\mu\) becomes the mass per unit volume \(\rho\), and the amplitude \(A\) becomes \(s_{max}\). The energy grows as the square of the amplitude and as the square of the frequency.
A sound of frequency \(500\,\text{Hz}\) travels through air of density \(1.2\,\text{kg/m}^3\) at \(340\,\text{m/s}\). Its displacement amplitude is \(1.0 \times 10^{-6}\,\text{m}\). Find (a) the average energy density, (b) the peak energy density, and (c) the largest fractional change in the volume of the air.
(a) The angular frequency is \(\omega = 2\pi f = 2\pi \times 500 = 3142\,\text{rad/s}\). So \(\langle u\rangle = \tfrac12\rho\omega^2 s_{max}^2 = \tfrac12(1.2)(3142)^2(1.0 \times 10^{-6})^2 = 5.9 \times 10^{-6}\,\text{J/m}^3\).
(b) The peak value is twice the average, \(\rho\omega^2 s_{max}^2 = 1.2 \times 10^{-5}\,\text{J/m}^3\). It occurs at the compressions and the rarefactions.
(c) The fractional change in volume is \(\partial s/\partial x\), whose largest size is \(k s_{max}\). Here \(k = \omega/v = 3142/340 = 9.24\,\text{rad/m}\), so \(k s_{max} = 9.2 \times 10^{-6}\).
Even this loud sound squeezes the air by less than one part in a hundred thousand. The energy it carries is small, but the ear is very sensitive.
A second sound in the same air has the same displacement amplitude but twice the frequency. How does its average energy density compare?
A sound of frequency \(200\,\text{Hz}\) in air of density \(1.2\,\text{kg/m}^3\) has a displacement amplitude of \(5.0 \times 10^{-6}\,\text{m}\). What is its average energy density, in μJ/m³?
Now return to the rattling windows. Every cubic metre of air near the band holds energy, half as motion and half as squeeze. That energy sits in the compressions and rarefactions and travels with them. When they reach a window, the extra pressure pushes and pulls on the glass, and the glass shakes.
- Using the wave speed \(v\) in the kinetic energy of a slab.The slab moves with the particle velocity \(\partial s/\partial t\). The wave speed enters only through \(\partial s/\partial t = -v\,\partial s/\partial x\).
- Thinking a rarefaction holds less energy than a compression.The potential energy goes as \((\Delta P)^2\). Equal squeezing and stretching store equal energy.
- Placing the energy where the displacement is largest.There the air is at rest for an instant and has its normal density. The energy sits where \(s = 0\).
- Counting the work done by the steady air pressure \(P_0\).It adds to zero over a cycle. Only the extra pressure \(\Delta P\) stores the energy of the wave.