= Solution
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:
$$
\psi=\operatorname{Re}\{A\sin(k_yy)e^{i(\omega t-k_xx)}\},\qquad
k_y=\frac{n\pi}{b},\quad n=1,2,\ldots.
$$
The $n=0$ sine gives no displacement. <separation of variables> gives
$$
\boxed{\omega^2=v_m^2\left[k_x^2+\left(\frac{n\pi}{b}\right)^2\right]},
\qquad
f_{c,n}=\frac{nv_m}{2b}.
$$
For a propagating branch with $k_x>0$,
$$
\boxed{v_p=\frac{\omega}{k_x}
=\frac{v_m}{\sqrt{1-(\omega_{c,n}/\omega)^2}},\qquad
v_g=\frac{d\omega}{dk_x}
=v_m\sqrt{1-(\omega_{c,n}/\omega)^2}}.
$$
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 \b[$k_x=175.6\ \mathrm{m^{-1}}$] for travel toward positive $x$.
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