Sound waves

Small-amplitude sound is a linear disturbance of the medium, which means that we can write, for instance, $\rho =\rho _{0}+\delta \rho ({\bf r},t)$ and keep only terms linear in $\delta \rho$. We will also assume that ${\bf v}$ itself is small, which still allows us to treat sound in drifting fluids by going to a frame moving with the fluid (this works as long as the sound's wavelength is smaller than the size of the drift). The effect of gravity is included to leading order by letting the undisturbed density $\rho _{0}$ vary with altitude. Neglecting viscosity, the Navier-Stokes equation reduces to the linearized Euler equation  

$$\rho _{0}\,\frac{\partial {\bf v}}{\partial t}={\bf -\nabla }p$$ (10)

and the linearized continuity equation is simply  

$$\frac{\partial \rho }{\partial t}+\rho _{0}{\bf \nabla }\cdot {\bf v}=0$$ (11)

To determine everything, we need an equation relating pressure changes to density changes. Assuming that these changes occur either adiabatically or isothermally, we can write, from eq.(5.9),

$$\frac{\partial p}{\partial x}=\frac{B}{\rho }\frac{\partial \rho }{\partial x}$$

where x is any coordinate. In vector notation  

$${\bf \nabla }p=\frac{B}{\rho }{\bf \nabla }\rho$$ (12)

The ratio $B/\rho$ can be taken as constant because we have assumed small density changes; for a liquid, B and $\rho$ vary very little anyhow.

From equations (5.10), (5.11), and (5.12), it is follows that r, p, and v obey the wave equation. Let us derive, for instance, the equation for r. From (5.10) and (5.12) we get

$$\rho _{0}\,\frac{\partial {\bf v}}{\partial t}=-\frac{B}{\rho _{0}}{\bf \nabla }\rho$$

and taking the divergence of both sides:

$$\rho _{0}\,\frac{\partial }{\partial t}\,{\bf \nabla }\cdot {\bf v}=-\frac{B}{\rho _{0}}\nabla ^{2}\rho$$

On the other hand, the time derivative of (5.11) gives  

$$\frac{\partial ^{2}\rho }{\partial t^{2}}+\rho _{0}\frac{\partial }{\partial t}{\bf \nabla }\cdot {\bf v}=0$$ (13)

and from the last two equations it follows that

$$\frac{\partial ^{2}\rho }{\partial t^{2}}=\frac{B}{\rho _{0}}\nabla ^{2}\rho$$ (14)

As is well known, this equation has plane wave solutions of the form $\rho =\rho _{0}+\delta \rho ,$ with

$$\delta \rho =\vert A\vert\,\cos (kx-\omega t+\alpha )$$

where

• |A| is the amplitude, i.e., the maximum value of $\delta \rho$
• k is the wave number, positive for a wave propagating in the positive x direction
• $\omega$ is the angular frequency, always taken to be positive. It is related to k by $\omega ^{2}=c^{2}k^{2},$ where (dropping the subscript on $\rho _{0}$)

  $$c=\sqrt{\frac{B}{\rho }}\quad \quad \quad {\rm is\;the\;speed\;of\;sound}$$

• $\alpha$ is the phase of the wave at x=0, t=0.

Intuition suggests that one should use B_S for low frequencies and B_T for high frequency, with the crossover at $\omega =Dk^{2},$ or $\omega =c^{2}/D,$ where D is the thermal diffusivity (more about this later). For audible sound in air, $\omega /2\pi$ is between 40 Hz and 20kHz (decreasing with age), while $c^{2}/D=5\times 10^{9}$Hz. Thus in practice one should use B_S even for ultrasound, and one finds

$$c_{{\rm air}}=\sqrt{\frac{\gamma p}{\rho }}=\sqrt{\frac{\gam... ...340\sqrt{\frac{T}{293\,{\rm K}}}\,\frac{{\rm m}}{{\rm \sec }}$$

The speed of sound in a liquid is higher than in the corresponding vapor, because B increases more than r does. For water at STP, c is around 1500 m/sec, depending somewhat on purity.