= Solution
<Irrotational> motion in this <simply connected> fluid region has a <velocity potential>, $u=\nabla\phi$. <Incompressibility> gives $\nabla\cdot u=0$, hence $\Delta\phi=0$.
The moving <surface> is a <material surface>: differentiating $y-\eta(x,t)=0$ along a particle gives $\eta_t+\phi_x\eta_x=\phi_y$ at $y=\eta$. Linearizing about a flat stationary <surface> drops the product term and evaluates at zero, giving $\eta_t=\phi_y$ there. The <Unsteady Bernoulli equation>, after dropping quadratic <velocity> terms and absorbing the time-only constant into $\phi$, is $\phi_t+p/\rho+gy=0$. The linear <capillary pressure> at the <surface> is $p=-T\eta_{xx}$ relative to atmosphere, in the sign convention of the source. Therefore
$$
\boxed{\phi_y=\eta_t,\qquad\phi_t+g\eta-\frac T\rho\eta_{xx}=0\quad(y=0).}
$$
For $\eta=A\cos(kx-\omega t)$ choose the <harmonic potential> decaying at infinite depth,
$$
\phi=\frac{A\omega}{k}e^{ky}\sin(kx-\omega t),\qquad y<0.
$$
Its vertical <derivative> at zero equals $A\omega\sin(kx-\omega t)=\eta_t$. Substitution into the dynamic condition gives $-\omega^2/k+g+Tk^2/\rho=0$. Thus
$$
\boxed{\omega^2=gk+\frac T\rho k^3,\qquad c=\frac\omega k=\sqrt{\frac gk+\frac T\rho k}.}
$$
The two positive terms are the gravitational and capillary contributions to the <restoring force>.
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