# How to derive the energy density of a sound wave?

I am studying sound waves and my book just provides the result for the energy density of a sound wave and I've been searching it online for 3 days, i tried to derive it myself countless time but I can't get the provided result $$\frac{dE}{dV}=\frac{1}{2}\rho\omega^{2}s^{2}$$ where $$s = s_0 \sin(kx-\omega t)$$ is the displacement of a tiny piece of volume from the equilibrium

Notes: I believe that the answer you are trying to obtain is incorrect. First, if you are looking at total energy density, you should not have the factor of 1/2. Alternatively, you could be looking at either the potential or the kinetic energy density, not the sum. Second, I believe that $$s$$ in your solution should be $$s_0\cos(kx-\omega t)$$. Alternatively, you could be referring to the mean total energy density, in which case $$s$$ should be replaced with $$s_0$$, and the factor of 1/2 is now accurate.

Another note is that the expression for $$s$$ you have is a traveling wave. The analysis is different for standing waves, though it follows the same steps.

Derivation:

We will assume, as you have in your problem, that the displacement of the sound wave is given by $$s = s_0\sin(kx-\omega t).$$ Then the particle velocity may be obtained by taking the time derivative: $$v = -\omega s_0\cos(kx-\omega t).$$ We may then use the continuity equation to obtain an expression for the pressure. The continuity equation is given by $$\frac{1}{c^2}\frac{\partial p}{\partial t} + \rho\frac{\partial v}{\partial x} = 0,$$ where $$c$$ is the speed of sound and $$\rho$$ is the ambient mass density, which leads to the solution $$p = -\rho c^2ks_0\cos(kx-\omega t).$$

Now consider a single fluid element of volume $$dV$$. The mass of this element is $$\rho dV$$. The total kinetic energy of this element is then given by $$dE_\text{k}=\frac{1}{2}\rho dV v^2 = \frac{1}{2}\rho\omega^2 s^2_0\cos^2(kx-\omega t)\ dV.$$ The potential energy comes from the fact that the fluid element acts like a spring, and so may be written as one half of the force times the compression. The force is the pressure times the cross-sectional area $$dA$$, and the compression is minus the derivative of the displacement times the element length $$dx$$. Noting that $$dx\ dA=dV$$, we may then write $$dE_\text{p} = -\frac{1}{2}p\frac{\partial s}{\partial x}dV = \frac{1}{2}\rho c^2k^2s_0^2\cos^2(kx-\omega t)\ dV.$$ By definition, we know that $$k^2=\omega^2/c^2$$, and so we may then write $$dE_\text{p} = \frac{1}{2}\rho \omega^2s_0^2\cos^2(kx-\omega t)\ dV.$$

If we divide the kinetic and potential energies by $$dV$$, we obtain expressions for the respective energy densities. The total energy density is then given by $$\frac{dE_\text{k}}{dV} + \frac{dE_\text{p}}{dV} = \rho\omega^2s_0^2\cos^2(kx-\omega t).$$

We may be interested in the mean energy density, in which case we integrate the energy density in time over one period and divide by one period. This process leads to $$\frac{\omega}{2\pi}\int_0^{2\pi/\omega}\left[\frac{dE_\text{k}}{dV} + \frac{dE_\text{p}}{dV}\right]dt = \frac{\rho\omega^3s_0^2}{2\pi}\int_0^{2\pi/\omega}\left[\frac{1 + \cos(2kx-2\omega t)}{2}\right]dt = \frac{1}{2}\rho\omega^2s_0^2.$$

• so i got it right when i was deriving it, thanks man i was going crazy over this. Commented Jun 1, 2022 at 14:39
• Well done, however dV is missing on the right-hand side of the expressions for dEk and dEp. Commented Jan 18 at 18:22