# How can pressure and velocity fluctuations in acoustic plane waves be in phase and still hold to the B.C. that velocity must be zero at a solid wall?

In linear plane-wave acoustics (no mean flow, small perturbations, etc.), it is often derived that the phase of a traveling pressure fluctuation wave and a fluctuation velocity wave are the same. For example, if the pressure fluctuation is given by:

$$P' = B\cos(kx - \omega t + \phi)$$

Where B is the amplitude, k is the wavenumber, $\omega$ is the angular frequency, and $\phi$ is the phase.

It can also be shown that the momentum equation can be linearized to:

$$\frac{\partial u'}{\partial t} = -\frac{1}{\rho_0}\frac{\partial P'}{\partial x} = \frac{1}{\rho_0}kB\sin(kx - \omega t + \phi)$$

Finally, integrating with time then gives:

$$u' = \frac{k}{\rho_0\omega}B\cos(kx - \omega t + \phi)$$

Thus, pressure fluctuations and velocity fluctuations are in phase for plane waves under the given assumptions. But how is this physically possible since all velocities must go to zero at a solid wall and pressure must be maximized? Is there a way to rigorously justify this? Or is it that the assumptions break down near the wall and the phase difference becomes 180 deg.?

## 3 Answers

I assume you are referring to what happens when such a plane wave is propagating parallel to a solid surface. You are correct that strictly speaking, the acoustic particle velocity, like the mean flow velocity, must obey the no-slip condition at the wall. Viscous effects become important very close to the wall only, such that the acoustic motions are no longer sufficiently described by the classical wave equation, as they are far away from the wall.

Consider a plane wave of the type you specified traveling adjacent to a wall. The linearized momentum and energy equations may be written like so:

$$\bar{\rho}\frac{\partial u'}{\partial t} + \frac{\partial p'}{\partial x} = \mu\frac{\partial^2 u'}{\partial y^2}$$

$$\bar{\rho}c_p\frac{\partial T'}{\partial t} - \frac{\partial p'}{\partial t} = k\frac{\partial^2 T'}{\partial y^2}$$

Recall that in normal boundary layer theory we neglect the pressure gradient normal to the wall on the hypothesis that the boundary layer is thin. So, although the velocity changes considerably throughout the region, the pressure remains approximately at its free-stream value throughout. This is one of the crucial assumptions of boundary layer theory generally, and we have written the acoustic boundary layer equations above under the same assumption.

Following the pressure argument, we may replace the $\partial p'/\partial x$ term with the rate of change of the acoustic momentum due to the wave in the free stream. Likewise, we may set the acoustic pressure to that of the free stream wave in the energy equation:

$$\bar{\rho}\frac{\partial u'}{\partial t} - \bar{\rho}\frac{\partial u'_\infty}{\partial t} = \mu\frac{\partial^2 u'}{\partial y^2}$$

$$\bar{\rho}c_p\frac{\partial T'}{\partial t} - \frac{\partial p'_\infty}{\partial t} = k\frac{\partial^2 T'}{\partial y^2}$$

Now, assuming all quantities vary harmonically as $e^{i\omega t}$, we may develop analytical solutions for the acoustic particle velocity and temperature within the boundary layer:

$$u(y) = u_\infty [1 - e^{-(1+i)y/\delta_u}]e^{i\omega t}$$

$$T(y) = T_\infty [1 - e^{-(1+i)y/\delta_T}]e^{i\omega t}$$

where $\delta_u =\sqrt{2\mu/\bar{\rho}\omega}$ and $\delta_T = \delta_u/\sqrt{Pr}$ are respectively, the boundary layer thicknesses of the velocity and temperature fields. An animation for a free stream wave frequency of 30 Hz and Pr = 1.5 is shown below to better understand these solutions. So indeed, very close to the wall we see lag effects on the velocity which causes it to be out of phase with the velocity (and hence pressure) in the free stream. The velocity vectors are in blue, and the temperature is in red.

If this was not exactly what your question was referring to, I apologize for misinterpreting. The boundary conditions just determine how that quantity reflects. The pressure wave reflects in phase, and the velocity wave reflects out of phase.

What you noticed it that for a forward traveling sound wave, the high pressure fluctuations correspond to forward velocity of the particles. When the wave reflects, the velocity of those high pressure regions must reverse so the wave can go the other way.

• Can you please clarify your answer? If I'm understanding your second paragraph, then the velocity and pressure are not in phase at a solid boundary, and the equations that I listed above does not hold near the boundary. – Kimusubi Dec 15 '15 at 3:08
• The boundary conditions affect the standing wave, which is the total of the incident and reflected waves. The particle velocity of the traveling waves has nodes that move along with the wave. The traveling wave will never have a node at a fixed point. It's only by adding traveling waves going in opposite directions that the fixed nodes appear. Try looking at this professor's explanation: acs.psu.edu/drussell/Demos/superposition/superposition.html – Spirko Dec 15 '15 at 4:31
• The zero normal velocity at the wall B.C. does not depend on standing waves or traveling waves -- it's a matter of physics. The normal velocity at a stationary wall should always be zero. – Kimusubi Dec 16 '15 at 22:39
• The boundary conditions on the solution to a differential equation is usually not meet by individual basis states. – Spirko Dec 16 '15 at 22:56
• Another way of saying it is that the forward wave definitely does not meet the zero velocity condition. That is what makes the reflection happen. – Spirko Dec 16 '15 at 22:57

The uni-directional wave equation deals only with waves travelling in 1 direction. However, as sound suffers reflection, the direction of the wave propagation is reversed. Hence the system no longer follows the earlier governing wave equation. Instead it follows the wave equation with velocity reversed. The phase difference between pressure and velocity is 180 there.

To accommodate both the directions of propagation, the governing equation is a second order wave equation, which allows the phase difference of both 0 and 180