= Solution
Let $\chi=kx-\omega t$, with $k>0$. The <surface elevation> is $\eta=\alpha\cos\chi$, so $\eta_t=\alpha\omega\sin\chi$. Use <separation of variables> in the interior <Laplace equation>:
$$
\phi=A\cosh(k(z+h))\sin\chi.
$$
The $\sinh(k(z+h))$ term is excluded by the bed <no-penetration boundary condition>. The linear surface <kinematic boundary condition> determines
$$
Ak\sinh(kh)=\alpha\omega,
\qquad
\boxed{\phi(x,z,t)=\frac{\alpha\omega}{k\sinh(kh)}
\cosh(k(z+h))\sin(kx-\omega t)}.
$$
An irrelevant additive constant is omitted. At $z=0$,
$$
\phi_t=-\frac{\alpha\omega^2}{k}\coth(kh)\cos\chi,
\qquad \eta_{xxxx}=\alpha k^4\cos\chi.
$$
The linear dynamic boundary condition then reduces to
$$
-\frac{\omega^2}{k}\coth(kh)+g+\frac B\rho k^4=0.
$$
Consequently the <dispersion relation of a floating elastic plate> with zero plate inertia is
$$
\boxed{\omega^2=\left(gk+\frac B\rho k^5\right)\tanh(kh),\qquad
\omega(k)=\sqrt{\left(gk+\frac B\rho k^5\right)\tanh(kh)}}.
$$
The negative-frequency branch gives propagation in the opposite direction. For signed <wavenumber>, replace $k$ by $|k|$ in the squared-frequency formula. Setting $B=0$ recovers the finite-depth <surface gravity wave> relation, a useful sign check.
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