 Solution

A wave packet is a localized superposition of nearby wavenumbers. Its rapid carrier crests have phase velocity $v_p=\omega /k$; the slowly varying envelope translates at the group velocity. Dispersion means that $\omega (k)$ is not simply a constant times $k$, so different carrier components have different phase velocities. Expand around $k_0$:The integral is an envelope depending on $x-{\omega }^{\prime }(k_0)t$, so $v_g=d\omega /dk$ at the central frequency. Neglected curvature causes packet spreading. For a deep-water gravity wave, $\omega =\sqrt{gk}$, hence $v_p=\sqrt{g/k}$ andFor crest crossings through a dispersive wave packet, an instantaneous group containing six crests has length approximately $L=6\lambda$. A stationary ship takes $L/v_g$ to encounter the group; its crest-arrival frequency is $v_p/\lambda$. ConsequentlyIn the envelope frame the ship moves backwards at $v_g$ while crests move forwards at $v_p-v_g=v_g$. Their relative speed is $2v_g$, explaining why counting a snapshot and counting crossings differ.

To estimate deep-water swell distance from period-dependent arrival times, assume a localized storm emits both periods at the same time and place, propagation is in deep water, and currents and refraction can be neglected. If its distance is $D$, the arrival lag obeysWith $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $D=15g(86400)/\pi \simeq 4.05\times 10^6\mathrm{m}$, about 4050 km. The corresponding travel times are three and four days. A moving or long-lived storm with frequency-dependent emission times would invalidate the unique distance inference.