 Solution

For a narrow packet, phase velocity is $u_p=\omega /k$ and group velocity is $u_g=d\omega /dk$. Set $b=\sigma /{\rho }_1$. The uncovered surface hasEquality occurs at $k_{\ast }^2=g/b$, also the minimum of $u_p$. Numerically,The long-wave gravity limit has $u_g/u_p\to 1/2$; the short-wave capillary limit has $u_g/u_p\to 3/2$.

A compact packet subsequently spreads because different wavenumbers have different group speeds. In the gravity regime $\omega \simeq \sqrt{gk}$, so smaller $k$ and longer waves move ahead. Individual crests move twice as fast as the envelope, appearing at its rear and disappearing at its front. In the capillary regime $\omega \simeq \sqrt{b}{k}^{3/2}$, shorter waves move ahead; the envelope overtakes its slower individual crests, which drift backward relative to it. Thus both the wavelength ordering within the spread packet and the crest motion relative to its envelope reverse.

For the sheet define $c=a{\rho }_2/{\rho }_1=8.0\times 10^{-5}\mathrm{m}$. The elastic-cover frequency ratio for water waves isAt small $k$, $F=1-ck/2+O(k^2)$: sheet inertia lowers the frequency, while bending first enters at higher order. At sufficiently large $k$, bending dominates the restoring force and sheet inertia dominates the extra mass, giving $F\sim \sqrt{B/(\sigma c)}{k}^{1/2}$. The curve initially dips below one, reaches a minimum, then rises and crosses one. This asymptote describes the supplied ideal dispersion law; arbitrarily large $k$ eventually violates a thin-sheet continuum model.

For a pinned-square surface displacement mode, $k_{nm}=\pi \sqrt{n^2+m^2}/L$ with $n,m\ge 1$. The fundamental has $k_{11}=74.05{\mathrm{m}}^{-1}$. The uncovered frequency is $4.3729\mathrm{Hz}$ and $F(k_{11})=0.997051$, soThis uses a square two-dimensional surface mode, not the one-dimensional $\pi /L$ estimate. A clamped plate or additional hydraulic boundary conditions would require a different mode calculation.