What is the phase shift incurred by a sound wave as a result of reflection?

Question

While studying waves I read the fact that a sound wave gets shifted by $\pi$ as a result of reflection against a surface. But I am unable to prove that fact.

Assuming the interface to be a node I can prove that there is a phase shift of $\pi$ but speaking generally how do I deduce it? If we do not assume that the interface is a node I saw in some case it's not true.

Am I misunderstanding something here? Does the sound wave manipulate itself so as that the interface becomes node? How does it do that?

Answer 1

A wave e.g $$\sin (kx + \omega t + \phi)$$

when reflected runs in the opposite direction. In other words gets a rotation by $\pi$ or what amounts to the same thing gets a phase shift by $\pi$.

$$\sin (kx + \omega t + \phi + \pi)$$

Tentative proof:

Let's say a wave $\psi \sim e^{i(kx-\omega t)}$

on reflection the wave should propagate opposite (or rotated by $\pi$)

i.e $\psi \sim -e^{i(kx-\omega t)}=e^{i\pi}e^{i(kx-\omega t)}=e^{i(kx-\omega t + \pi)}$

Answer 2

The fact that the interface is a node follows from physical considerations rather from some equations.If you imagine sound to be travelling in air ,you can visualize air molecules moving as a result of wave propagation .But the molecules near the interface are stuck up and can hardly move . On the other hand if you consider a water wave travelling through a closed container then the container walls represent a anti node as water molecules can easily move .Because this is a case of transverse wave .