Sound Waves in an Ideal Gas

Consider a uniform ideal gas of equilibrium mass density $\rho$ and equilibrium pressure $p$. Let us investigate the longitudinal oscillations of such a gas. Of course, these oscillations are usually referred to as sound waves. Generally speaking, a sound wave in an ideal gas oscillates sufficiently rapidly that heat is unable to flow fast enough to smooth out any temperature perturbations generated by the wave. Under these circumstances, the gas obeys the adiabatic gas law,

$$p\,V^\gamma = {\rm constant},$$ (315)

where $p$ is the pressure, $V$ the volume, and $\gamma$ the ratio of specific heats (i.e., the ratio of the gas's specific heat at constant pressure to its specific heat at constant volume). This ratio is approximately $1.4$ for ordinary air.

Consider a sound wave in a column of gas of cross-sectional area $A$. Let $x$ measure distance along the column. Suppose that the wave generates an $x$-directed displacement of the column, $\psi(x,t)$. Consider a small section of the column lying between $x-\delta x/2$ and $x+\delta x/2$. The change in volume of the section is $\delta V = A\,[\psi(x+\delta x/2,t)-\psi(x-\delta x/2,t)]$. Hence, the relative change in volume, which is assumed to be small, is

$$\frac{\delta V}{V} = \frac{A\,[\psi(x+\delta x/2,t)-\psi(x-\delta x/2,t)]}{A\,\delta x}.$$ (316)

In the limit $\delta x\rightarrow 0$, this becomes

$$\frac{\delta V(x,t)}{V} = \frac{\partial\psi(x,t)}{\partial x}.$$ (317)

The pressure perturbation $\delta p(x,t)$ associated with the volume perturbation $\delta V(x,t)$ follows from (315), which yields

$$(p+\delta p)\,(V+\delta V)^\gamma = p\,V^\gamma,$$ (318)

or

$$(1+\delta p/p)\,(1+\delta V/V)^\gamma \simeq 1+ \delta p/p + \gamma\,\delta V/V=1,$$ (319)

giving

$$\delta p = - \gamma\,p\,\frac{\delta V}{V}= -\gamma\,p\,\frac{\partial\psi}{\partial x},$$ (320)

where use has been made of (317).

Consider a section of the gas column lying between $x-\delta x/2$ and $x+\delta x/2$. The mass of this section is $\rho\,A\,\delta x$. The $x$-directed force acting on its left boundary is $A\,[p+\delta p(x-\delta x/2,t)]$, whereas the $x$-directed force acting on its right boundary is $-A\,[p+\delta p(x+\delta x/2,t)]$. Finally, the average longitudinal (i.e., $x$-directed) acceleration of the section is $\partial^2\psi(x,t)/\partial t^2$. Thus, the section's longitudinal equation of motion is written

$$\rho\,A\,\delta x\,\frac{\partial^2\psi(x,t)}{\partial t^2} ... ...left[\delta p(x+\delta x/2,t)-\delta p(x-\delta x/2,t)\right].$$ (321)

In the limit $\delta x\rightarrow 0$, this equation reduces to

$$\rho\,\frac{\partial^2\psi(x,t)}{\partial t^2} = - \frac{\partial\delta p(x,t)}{\partial x}.$$ (322)

Finally, (320) yields

$$\frac{\partial^2\psi}{\partial t^2} = c^2\,\frac{\partial^2\psi}{\partial x^2},$$ (323)

where $c=\sqrt{\gamma\,p/\rho}$ is a constant with the dimensions of velocity, which turns out to be the sound speed in the gas (see Section 7.1).

Figure 33: First three normal modes of an organ pipe.

As an example, suppose that a standing wave is excited in a uniform organ pipe of length $l$. Let the closed end of the pipe lie at $x=0$, and the open end at $x=l$. The standing wave satisfies the wave equation (323), where $c$ represents the speed of sound in air. The boundary conditions are that $\psi(0,t)=0$--i.e., there is zero longitudinal displacement of the air at the closed end of the pipe--and $\partial\psi(l,t)/\partial x=0$--i.e., there is zero pressure perturbation at the open end of the pipe (since the small pressure perturbation in the pipe is not intense enough to modify the pressure of the air external to the pipe). Let us write the displacement pattern associated with the standing wave in the form

$$\psi(x,t) = A\,\sin(k\,x)\,\cos(\omega\,t-\phi),$$ (324)

where $A>0$, $k>0$, $\omega>0$, and $\phi$ are constants. This expression automatically satisfies the boundary condition $\psi(0,t)=0$. The other boundary condition is satisfied provided

$$\cos(k\,l) = 0,$$ (325)

which yields

$$k\,l = (n-1/2)\,\pi,$$ (326)

where the mode number $n$ is a positive integer. Equations (323) and (324) yield the dispersion relation

$$\omega = k\,c.$$ (327)

Hence, the $n$th normal mode has a wavelength

$$\lambda_n =\frac{4\,l}{2n-1},$$ (328)

and an oscillation frequency (in Hertz)

$$f_n = (2n-1)\,f_1,$$ (329)

where $f_1=c/4\,l$ is the frequency of the fundamental harmonic (i.e., the normal mode with the lowest oscillation frequency). Figure 33 shows the characteristic displacement patterns (which are pictured as transverse displacements, for the sake of clarity) and oscillation frequencies of the pipe's first three normal modes (i.e., $n=1, 2$, and 3). It can be seen that the modes all have a node at the closed end of the pipe, and an anti-node at the open end. The fundamental harmonic has a wavelength which is four times the length of the pipe. The first overtone harmonic has a wavelength which is $4/3$rds the length of the pipe, and a frequency which is three times that of the fundamental. Finally, the second overtone has a wavelength which is $4/5$ths the length of the pipe, and a frequency which is five times that of the fundamental. By contrast, the normal modes of a guitar string have nodes at either end of the string. See Figure 23. Thus, as is easily demonstrated, the fundamental harmonic has a wavelength which is twice the length of the string. The first overtone harmonic has a wavelength which is the length of the string, and a frequency which is twice that of the fundamental. Finally, the second overtone harmonic has a wavelength which is $2/3$rds the length of the string, and a frequency which is three times that of the fundamental.