= Solution
Comparing with the standard <wave equation> gives the shallow-water <wave speed>
$$
\boxed{c=\sqrt{gd}.}
$$
Twice differentiating the proposed <standing wave> multiplies its spatial factors by $-k_x^2$ and $-k_y^2$, and its temporal factor by $-\omega^2$. It solves the <wave equation> precisely when
$$
\boxed{\omega^2=gd(k_x^2+k_y^2).}
$$
To obtain the local <gravitational potential energy>, integrate through a water column of height $d+h$ and unit horizontal area:
$$
V(t)=\int_0^{d+h}\rho gz\,dz=\frac{\rho g}{2}(d+h)^2.
$$
The time averages are $\langle h\rangle=0$ and $\langle h^2\rangle=A^2\cos^2(k_xx)\cos^2(k_yy)/2$. Thus
$$
\boxed{\langle V\rangle=\frac12\rho gd^2+\frac14\rho gA^2\cos^2(k_xx)\cos^2(k_yy),\quad c_1=\frac12\rho gd^2,\quad c_2=\frac14\rho gA^2.}
$$
The constant is retained because the specified zero of <gravitational potential> is the tank bottom. This is local time-averaged <potential energy>, not the sum of <potential energy> and <kinetic energy>.
The no-normal-flow wall conditions are <Neumann boundary conditions>. Differentiating the cosine factors at the opposite walls gives
$$
\boxed{k_x=\frac{m\pi}{a},\qquad k_y=\frac{n\pi}{b},\qquad m,n=0,1,2,\ldots.}
$$
Negative integers give the same cosine modes. The formal $m=n=0$ case has $\omega=0$: it is a uniform change of the mean level, not a finite-<frequency> oscillation. In a closed tank with fixed water volume even that mean-level change is excluded from the oscillatory disturbance.
Since $a<b$, the lowest finite <frequency> is the mode $(m,n)=(0,1)$:
$$
\boxed{h=A\cos(\pi y/b)\cos(\omega t),\qquad \omega=\frac\pi b\sqrt{gd}.}
$$
It is uniform across the shorter direction, has opposite-sign displacement <antinodes> at the two ends of the longer direction, and a displacement <node> midway along it.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-water-modes.png]
{title=Original mode shapes for a closed rectangular tank and an ideal basin with one open end}
For the open-ended <Bay of Fundy> model, approximate the ocean mouth as a displacement <node> and the closed head as an <antinode>. The fundamental is a <quarter-wave basin resonance>, so its <wavelength> is $4L$ and
$$
T=\frac{4L}{\sqrt{gd}},\qquad \boxed{d=\frac{16L^2}{gT^2}.}
$$
Taking a roughly $12$-hour <semidiurnal tide> gives $d\simeq79\ \mathrm m$. Using the principal lunar period of approximately $12\ \mathrm h\ 25\ \mathrm{min}$ gives $d\simeq73\ \mathrm m$. \b[The idealized estimate is therefore about $7\text{–}8\times10^1\ \mathrm m$.] The lunar period is half of a roughly $24$-hour $50$-minute lunar day.<About Harmonic Constituents by National Oceanic and Atmospheric Administration>[Examples M2 and S2: two lunar peaks per 24 hours 50 minutes, compared with two solar peaks per 24 hours]{ref}
This estimate treats the mouth as an ideal open boundary and replaces the real basin by a uniform depth. The supplied large tidal height motivates <resonance>; it is not enough by itself to determine damping or an amplification factor in this model.
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