Past exam of the physics A course of the University of Cambridge / 2014 / Paper 2 / Section B / B7

Reflection at the fixed edges combines waves with opposite $k_y$ into a transverse standing wave while retaining their common longitudinal propagation factor. The membrane waveguide scalar transverse displacement satisfies the wave equation and vanishes at both edges:The $n=0$ sine gives no displacement. separation of variables givesFor a propagating branch with $k_x>0$,Their product is $v_m^2$. The phase velocity can exceed the material wave speed; energy transport follows the group velocity.

Here the cutoff sequence is $10n\mathrm{kHz}$. At $15\mathrm{kHz}$ only $n=1$ propagates, with $k_x=175.6{\mathrm{m}}^{-1}$ for travel toward positive $x$.

At $5\mathrm{kHz}$ every transverse mode is below cutoff. There is no propagating guided mode in an infinite uniform guide; excitation produces only evanescent penetration. For the slowest-decaying mode write $\psi \propto e^{-\kappa x}$, so $k_x=-i\kappa$ under the preceding convention, with $\kappa =\sqrt{(\pi /b{)}^2-(2\pi f/v_m{)}^2}=136.0{\mathrm{m}}^{-1}$, giving a penetration length about $7.35\mathrm{mm}$.

At $50\mathrm{kHz}$ modes $n=1,2,3,4$ can propagate. Their positive $k_x$ values are approximately $769.5,719.8,628.3,471.2{\mathrm{m}}^{-1}$. Mode $n=5$ is a nonpropagating mode at waveguide cutoff: $k_x=0$ and $v_g=0$, so it does not carry longitudinal power as a traveling mode. Higher modes are evanescent. The injected transverse profile determines which permitted modes are actually excited.