 Solution

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 $7.35\mathrm{mm}$.