Energy in a sound wave possible error? In a certain text discussing longitudinal waves it stated that for a sound wave, both the kinetic energy and potential energy are maximum at the same time.
I always thought that the potential energy depended on the displacement and the kinetic energy the particle velocity, which put them out of phase such that the sum of the two remained constant. Other texts that I have read also state that the potential and kinetic energy are out of phase.
Which is correct?
Thanks in advance for any information.
 A: As knzhou mentioned, this is true even for progressive waves in a string. Your mistake might be to think about simple harmonic motion instead of harmonic waves.
I will show it for a progressive transverse wave in a string. It is easier to visualize. At the end I will give you a sketch for longitudinal waves.
For a string of density $\mu$ and tension $T$ and supporting a wave $y(x-vt)$ the kinetic energy of an element $dx$ is
$$dK=\frac 12\mu dx\left(\frac{\partial y}{\partial t}\right)^2.$$
For the potential energy we have
$$dU=Tdl,$$
where $k$ is some elastic constant and $dl$ is the stretched amount of the string. A small section $dh$ of the string is the hypotenuse of a right angle triangle with basis $dx$ and height $\frac{\partial y}{\partial x}dx$. Hence the amount stretched is
$$dl=\sqrt{dx^2+\left(\frac{\partial y}{\partial x}\right)^2dx^2}-dx=\frac{dx}{2}\left(\frac{\partial y}{\partial x}\right)^2.$$
In the last equation we neglect higher terms in $\left(\frac{\partial y}{\partial x}\right)$ since we assume small displacements. Then
$$dU=\frac{Tdx}{2}\left(\frac{\partial y}{\partial x}\right)^2.$$
Calling $x-vt\equiv u$, we see that
$$dE=dK+dU=\frac 12\mu v^2\left(\frac{\partial y}{\partial u}\right)^2dx+\frac{T}{2}\left(\frac{\partial y}{\partial u}\right)^2dx.$$
As we can see, the kinetic and potential energy are in phase.
For longitudinal waves what changes is the calculation of the potential energy. 
The tension on an elastic medium is proportional to $\frac{\partial \varphi}{\partial x}$, where $\varphi$ is the longitudinal displacement and $\frac{\partial \varphi}{\partial x}$ is the strain. The work and therefore the potential energy is proportional to 
$$\frac{\partial \varphi}{\partial x}d\varphi=\left(\frac{\partial \varphi}{\partial x}\right)^2 dx.$$
Again you will get 
$$dK\propto dU\propto \left(\frac{\partial \varphi}{\partial x}\right)^2dx.$$
