If the frequency of a sound wave doubles, does its energy quadruple?

Question

Or does it double?

What about its 'intensity' or power?

Acoustic physics is confusing me more than electromagnetic waves did....

Is there a single list somewhere of the equation(s) relating to sound?

Answer 1

Bottom line up front: The energy of an acoustic wave does not depend on the frequency (at least not in a simple way).

Derivation:

The energy contained in an acoustic wave depends somewhat on the geometry and whether it is a standing or propagating wave (or a combination). For simplicity, just consider a planar wave of a single frequency propagating in a homogeneous fluid.

The kinetic energy density of any given particle in motion is simply $$ E_k = \frac{1}{2}\rho v^2, $$ where $\rho$ is the mass density of the fluid and $v$ is the particle velocity. The potential energy density comes from the fact that the fluid is being compressed, like a spring, and may be written as $$ E_p = \frac{1}{2}\beta p^2, $$ where $\beta$ is the linear compressibility (related to a spring constant) and $p$ is the acoustic pressure (difference between the total and ambient pressure). The total energy density of this particle would then be written as $$ E_t = E_k + E_p. $$ Another important piece of the puzzle is the sound intensity, which may be written as (without derivation) $$ I = pv. $$

For a plane propagating wave we may write (again, no derivation but this is standard stuff) $$ p = A\sin(\omega t - kx), $$ where $\omega$ is the angular frequency and $k$ is the wavenumber. From the momentum equation (neglecting absorptive terms) we may then write $$ v = A\sqrt{\frac{\beta}{\rho}}\sin(\omega t-kx). $$ Interestingly, the ratio $p/v\equiv Z$ (called the specific acoustic impedance) is independent of $\omega$. Then the total energy density may be written as $$ E_t = A^2\beta\sin^2(\omega t-kx), $$ and the intensity as $$ I = \frac{A^2}{Z}\sin^2(\omega t-kx). $$ If you average these energetic quantities over a period you obtain $$ \bar E_t = \frac{A^2}{2}\beta, $$ and $$ \bar I = \frac{A^2}{2Z}. $$ Neither of these time-averaged quantities (energy at a point and intensity at a point) depend on the frequency of the wave.

Caveats:

This analysis is focused on constant frequency waves, which does not allow for wave packets. If you look at narrow-band pulses the energy and intensity actually does depend on the frequency, but I am not prepared to make any general statements about what that dependence looks like.

Another point to remember is that my analysis is highly simplified; the real world has many more phenomena that I do not account for. These phenomena will, in general, lead to frequency-dependent energetic terms.

Answer 2

It really depends on the cause of the frequency doubling. If, for example, the sound wave reflects off a surface moving at just the right speed to double the frequency, the intensity will double too, because the moving surface adds energy to the wave. But if the frequency doubles because the wave is absorbed by, e.g., a nonlinear resonator and re-emitted at twice the frequency, (ideally, so there are no losses), there will be no intensity change. If you're looking for an analogy with light waves, in the first case the number of phonons (or ohotons) does not change, while in the second case the number of phonons (or photons) is cut in half. The reason the number of phonons or photons in the second case is cut in half is because this amounts to "upconversion" frequency doubling, in which the energy of two phonons (or photons) is combined to make a new phonon (or photon) having twice the frequency and therefore twice the energy.