= Solution
{c}
Seek the travelling forced component of the <surface elevation> in the form $\eta=A\cos\chi$, where $\chi=x-ct$ and $A$ is a signed coefficient. The forcing has <wavenumber> one. <Separation of variables> in the <Laplace equation>, followed by the bed <no-penetration boundary condition>, gives
$$
\phi=B\cosh(y+h)\sin\chi.
$$
Since $\eta_t=Ac\sin\chi$, the surface <kinematic boundary condition> requires $B\sinh h=Ac$, so
$$
\boxed{\phi=\frac{Ac}{\sinh h}\cosh(y+h)\sin(x-ct)}.
$$
At $y=0$, $\phi_t=-Ac^2\coth h\cos\chi$. The dynamic member of the <linearized free-surface boundary conditions> therefore becomes
$$
A(g-c^2\coth h)+\frac{p_1}{\rho}=0.
$$
Consequently the signed response and the nonnegative <amplitude> are
$$
\boxed{A=-\frac{p_1}{\rho(g-c^2\coth h)},\qquad
a(c)=|A|=\frac{p_1}{\rho\,|g-c^2\coth h|}}.
$$
The <travelling-pressure response of finite-depth surface gravity waves> is opposite in phase to the imposed <pressure> below the resonant speed and in phase above it. These expressions describe the travelling forced component; freely propagating <surface gravity waves> can be superposed to meet additional <initial conditions>.
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