= Solution
{c}
The exact free-surface kinematic condition is $\eta_t+\phi_x\eta_x=\phi_y$ evaluated at $y=\eta$. Linearizing about a flat <surface> at rest drops products of small quantities and evaluates at $y=0$, giving $\phi_y=\eta_t$. <Bernoulli's equation> with constant atmospheric pressure gives $\phi_t+g\eta=0$ to the same order. The rigid bed has no normal flow, hence $\phi_y=0$ at $y=-h$.
In the fluid, <incompressibility> and the velocity-potential representation give $\Delta\phi=0$. For $k>0$, take
$$
\phi(x,y,t)=\frac{Ag}{\omega}
\frac{\cosh(k(y+h))}{\cosh(kh)}\sin(kx-\omega t).
$$
It is harmonic, satisfies the bed condition and the linear dynamic condition. At the <surface> its vertical derivative is $Agk\tanh(kh)\sin(kx-\omega t)/\omega$, while $\eta_t=A\omega\sin(kx-\omega t)$. Therefore the kinematic condition is equivalent to
$$
\boxed{\omega^2=gk\tanh(kh),\qquad
c=\frac\omega k=\sqrt{\frac gk\tanh(kh)}.}
$$
Finally $\tanh q=q-q^3/3+O(q^5)$. Taking the square root gives
$$
\boxed{c=\sqrt{gh}\left(1-\frac16k^2h^2+O((kh)^4)\right).}
$$
The first term is nondispersive shallow-water speed; the next term is the leading finite-depth dispersion correction.
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