= Solution
For the trial displacement $z=\sin(qy)e^{i(kx-\omega t)}$, direct differentiation gives
$$
\frac{\partial^2z}{\partial x^2}=-k^2z,\qquad
\frac{\partial^2z}{\partial y^2}=-q^2z,\qquad
\frac{\partial^2z}{\partial t^2}=-\omega^2z.
$$
The two-dimensional <wave equation> is therefore satisfied if
$$
\boxed{\omega^2=\frac T\sigma(k^2+q^2)=c_m^2(k^2+q^2),\qquad c_m=\sqrt{T/\sigma}.}
$$
Here $k$ is the longitudinal <wavenumber>, $q$ the transverse <wavenumber>, and $c_m$ the speed of an unconstrained membrane wave. The physical displacement is the real part, multiplied by an arbitrary displacement amplitude. $T$ has units $\mathrm{N\,m^{-1}}$ and $\sigma$ is <surface mass density> in $\mathrm{kg\,m^{-2}}$, so $T/\sigma$ has units $\mathrm{m^2\,s^{-2}}$ as required.
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