= 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 has
$$
u_p^2=\frac gk+bk,\qquad
\frac{u_g}{u_p}=\frac{g+3bk^2}{2(g+bk^2)}.
$$
Equality occurs at $k_*^2=g/b$, also the minimum of $u_p$. Numerically,
$$
\boxed{k_*=368.9\ \mathrm{m^{-1}},\quad\lambda_*=1.703\ \mathrm{cm},\quad
u_p=u_g=(4gb)^{1/4}=0.2305\ \mathrm{m\,s^{-1}}}.
$$
The long-wave gravity limit has $u_g/u_p\to1/2$; the short-wave capillary limit has $u_g/u_p\to3/2$.
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 $c=a\rho_2/\rho_1=8.0\times10^{-5}\ \mathrm m$. The <elastic-cover frequency ratio for water waves> is
$$
\boxed{F(k)=\frac{\omega_s}{\omega_0}
=\left[\frac{1+Bk^4/(\rho_1g+\sigma k^2)}{1+ck}\right]^{1/2}}.
$$
At 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 $n,m\geq1$. The fundamental has $k_{11}=74.05\ \mathrm{m^{-1}}$. The uncovered <frequency> is $4.3729\ \mathrm{Hz}$ and $F(k_{11})=0.997051$, so
$$
\boxed{f_s=4.3600\ \mathrm{Hz},\qquad\Delta f=-0.0129\ \mathrm{Hz}}.
$$
This 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.
\Image[/past-exam-of-the-physics-a-course-of-the-university-of-cambridge/2012/paper-2-covered-water-dispersion.png]
{title=Original sheet-to-uncovered frequency ratio, with the pinned-square fundamental and the ideal large-wavenumber asymptote}
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