Reflection of Sound Wave and Light (Electromagnetic) Waves from a Rigid Boundary

Question

We know that when a light wave (Electromagnetic Waves) reflects from a rigid boundary, the reflected ray suffers a phase difference of "Pi" (180 degrees).

But, in case of a sound wave, if it gets reflected from the rigid boundary, we do not consider a phase difference of "Pi". We consider the reflected wave to be in the same phase.

I understand that we can represent a sound wave in two ways:

1. Pressure Wave. Variation of pressure with time and position.
2. Displacement wave. Variation of Instantaneous Position of a particle of the medium, w.r.t its own equilibrium position.

Now, on reflection of a sound wave from a rigid boundary, the Pressure wave doesn't suffer any phase difference, but displacement wave does by "Pi".

It implies that, if in literature, in case of sound waves, if we do not consider the reflected waves to have suffered any phase difference, then we are giving priority to "Pressure Waves" over "Displacement Waves".

Is there any specific reason why we understand Sound Waves, in the form of Pressure Waves, but Not in the form of Displacement Waves?

Or, I am totally misunderstanding?

Answer 1

The difference is the nature of the boundary condition.

A "perfectly rigid" barrier in acoustics is one that allows no displacement. Therefore it is naturally a point where displacement must go to zero. Which means it is a point where pressure in an acoustic wave will have a maximum, meaning the reflected pressure wave must have the same phase as the incoming wave.

In electromagnetics, you were probably considering a perfect conducting boundary as the equivalent of a rigid one in acoustics. A perfect conducting boundary in EM is one that forces the (transverse) electric field to go to zero. So naturally this produces a null in the electric field, meaning the reflected wave must have its E field phase shifted 180 degrees from the incoming wave. But it will also produce a maximum (anti-node) in the magnetic field.

If you consider transitions between different dielectric materials in EM, you can produce either a shifted or unshifted reflection in the E field wave. Or if you consider transitions between solids with different Young's modulus in acoustics, you can produce either a shifted or unshifted reflection in the pressure wave.

So you can produce either situation in either system (acoustics or EM). But a perfectly conductive boundary in EM is equivalent to a perfectly elastic boundary in acoustics rather than a perfectly rigid one, if you're considering the E field and pressure components of the two waves.

Answer 2

We certainly can understand sound as a displacement wave, and sometimes it is convenient to. In solids, this is actually essential, because the displacement determines the strain of the material which is critical for describing sound in the solid.

What determines the phase shift is the impedance of the two materials on either side of the boundary, as said here: http://hyperphysics.phy-astr.gsu.edu/hbase/Sound/reflec.html "That is, reflections off a lower impedance medium will be reversed in phase." When a sound wave inside a solid reflects off the air solid interface, it DOES experience a phase shift. This page: https://www.acs.psu.edu/drussell/demos/reflect/reflect.html also has a wonderful description of what is going on here. Try to think about the questions posited with the animations for varying impedance.

Beyond this, really all I can tell you to understand this better is to get dirty and solve the wave equation $\frac{d^2f}{dt^2}=c^2\frac{d^2f}{dx^2}$ yourself. You can say that for $x \leq 0$, $c=1$ and for $x > 0$, $c=2$, and require continuity of $f$ and $\frac{df}{dx}$ as your boundary conditions.

Edit: I couldn't help myself, and you may not know how to do this (it is a common exercise in introductory quantum mechanics, Griffiths' QM book does it) so I'm going to.

Plane waves $e^{\pm i(kx-\omega t)}$ are solutions to this equation, with $\omega^2 = k^2c^2$. I'll drop the time dependence because it only obfuscates and isn't relevant: $e^{\pm ikx}$

Let's send in a plane wave from negative infinity with amplitude one and wavenumber $k$, and let the right side have wavenumber $q$: $ f(x) = \begin{cases} e^{ikx} + Be^{-ikx}, & x\leq0 \\ Ce^{iqx}, & x>0 \end{cases}$

At the boundary, $1+B=C$ is required by continuity of $f$ and $k(1-B)=qC$ is required by continuity of $\frac{df}{dx}$. So $B=\frac{k-q}{k+q}$, and $C=\frac{2k}{k+q}$. You can get the reflection and transmission coefficients by $R=|B|^2$, $T=1-R$, but your question is about the phase shift. You get a phase shift in the reflected wave if $B<0$, which happens if $k<q$, which happens if the speed of sound is lower (i.e. the impedance is higher) on the right side. You don't get a phase shift otherwise. There is in neither case a phase shift in the transmitted wave. You can see that all this came from requiring continuity and smoothness while obeying the dispersion relations of the materials.

I hope this helps, cheers!