Past exam of the fluids and waves course of the University of Oxford / 2021 / 3 / b / i / Solution

The denominator of the travelling response vanishes atFor the positive speeds in the paper, a finite travelling response exists whenever $c\ne c^{\ast }$. At $c=c^{\ast }$, the dynamic condition would reduce to $p_1/\rho =0$, which is impossible. The finite-depth dispersion relation of a surface gravity wave with wavenumber $k$ is ${\omega }^2=gk\tanh (kh)$. At $k=1$, its phase speed $\omega /k$ equals $c^{\ast }$: the moving pressure pattern drives a natural surface gravity wave at its own speed, producing resonance.

The restriction concerns a bounded travelling response. To distinguish this from existence of a solution of an initial value problem, set $T=\tanh h$, $P=p_1/\rho$, and ${\omega }_{\ast }^2=gT$. Ifthe linearized free-surface boundary conditions reduce to the forced harmonic oscillatorFor example, if the forcing is switched on at $t=0$ from a flat surface with zero initial velocity, then, for $c\ne {\omega }_{\ast }$,This satisfies $H(0)=\dot{H}(0)=0$ and contains both free oscillations and the forced component. At $c={\omega }_{\ast }$ a solution of the initial value problem still exists, but its particular solution containswhich grows linearly with time. Thus the ideal inviscid flow model predicts resonant growth rather than a finite constant amplitude. Additional free oscillations can set the initial conditions but cannot remove that secular growth.