Interference — Question 5

Question 5

A rigid reflection and a pressure-release reflection exchange extrema

A plane wave reflects normally from a boundary at $x=0$. The medium occupies $x<0$, the wavelength is $0.68\,\mathrm m$, and the incident pressure amplitude is $P$. Compare a rigid boundary with pressure reflection coefficient $+1$ and a pressure-release boundary with coefficient $-1$.

1. Find the pressure-amplitude envelope for each boundary.

2. Locate the nearest interior pressure minimum in the rigid case and nearest pressure maximum in the pressure-release case.

Question 5 - Solution

A rigid reflection and a pressure-release reflection exchange extrema

(a) Method and calculation

For rigid reflection, the two equal pressure waves add to$$p_{\mathrm{rigid}}=2P\cos kx\cos\omega t,\qquad A_p=2P|\cos kx|.$$For pressure-release reflection, subtraction instead gives$$p_{\mathrm{release}}=2P\sin kx\sin\omega t,\qquad A_p=2P|\sin kx|.$$

(b) Interpretation and further calculation

The nearest interior rigid pressure node and release-boundary pressure antinode both lie at$$x=-\lambda/4=-0.170\,\mathrm m.$$At the boundary itself the roles reverse: rigid reflection gives amplitude $2P$, while pressure-release reflection gives zero pressure perturbation. Equal path lengths do not ensure reinforcement when the reflection adds a phase inversion.

Physical check

The release boundary constrains pressure; the rigid boundary constrains normal velocity. Interference must satisfy the appropriate boundary condition.