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


1 Answer 1


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.


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 $$dE_\text{k}=\frac{1}{2}\rho dV v^2 = \frac{1}{2}\rho\omega^2 s^2_0\cos^2(kx-\omega t).$$ 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 $$dE_\text{p} = -\frac{1}{2}p\frac{\partial s}{\partial x}dV = \frac{1}{2}\rho c^2k^2s_0^2\cos^2(kx-\omega t).$$ By definition, we know that $k^2=\omega^2/c^2$, and so we may then write $$dE_\text{p} = \frac{1}{2}\rho \omega^2s_0^2\cos^2(kx-\omega t).$$

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 $$\frac{\omega}{2\pi}\int_0^{2\pi/\omega}\left[\frac{dE_\text{k}}{dV} + \frac{dE_\text{p}}{dV}\right]dt = \frac{\rho\omega^3s_0^2}{2\pi}\int_0^{2\pi/\omega}\left[\frac{1 + \cos(2kx-2\omega t)}{2}\right]dt = \frac{1}{2}\rho\omega^2s_0^2.$$

  • $\begingroup$ so i got it right when i was deriving it, thanks man i was going crazy over this. $\endgroup$
    – user695849
    Jun 1, 2022 at 14:39

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