Past exam of the options I course of the University of Oxford / 2013 / K / 2 / Solution

Put $\tau =T/\rho >0$ and take the positive-frequency branch for $k>0$. The phase velocity and group velocity areSince the square root is increasing, minimizing $c_p$ is equivalent to minimizing $g/k+\tau k$. Its derivative is $-g/k^2+\tau$, which changes sign from negative to positive atThe phase speed diverges at both positive endpoints $k\downarrow 0$ and $k\to \infty$, so this stationary point is its global minimum. At $k_c$, $\tau k_c^2=g$, givingThe relative speed is especially transparent:A narrow wave packet moves with $c_g$, while its crests move with $c_p$. For $k<k_c$, $c_p>c_g$: crests move forward through the packet, from its rear towards its front. For $k>k_c$, $c_p<c_g$: crests drift backwards relative to the packet, although both absolute velocities are positive. At $k_c$ the leading-order relative drift vanishes. This is the minimum phase speed of a gravity-capillary wave.