= Solution
Linearize about water at rest with uniform surface elevation $h_0$. Write $h=h_0+\eta$, $u=v$, $S_0=S(x,h_0)$ and $T_0=S_h(x,h_0)$. To first order the <shallow water equations> become
$$
T_0\eta_t+(S_0v)_x=0,\qquad v_t+g\eta_x=0.
$$
Differentiate continuity in time and eliminate $v_t$:
$$
T_0\eta_{tt}-g(S_0\eta_x)_x=0.
$$
For a time-harmonic standing disturbance, use $\eta(x,t)=\operatorname{Re}[\eta(x)e^{-i\omega t}]$. Then
$$
(S_0\eta')'+\frac{\omega^2T_0}{g}\eta=0.
$$
Dividing by $S_0$ proves
$$
\boxed{\eta''+\frac{S_0'}{S_0}\eta'+k^2\eta=0,\qquad
k^2(x)=\frac{\omega^2}{c_0^2(x)},\quad c_0^2=\frac{gS_0}{T_0}.}
$$
This identifies the coefficient as the local long-wave wavenumber. It is constant under the fixed-shape geometry used in part (a), when $c_0$ is independent of $x$; that assumption is needed for the constant-coefficient mode calculation in part (c).
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