The <isopycnal displacement> is
$$
\xi=-\frac{f_0}{N^2}\psi'_z.
$$
For the $n$th cosine mode,
$$
\xi=
\frac{f_0A}{N^2}\frac{n\pi}{H}
\sin\frac{n\pi z}{H}.
$$
Its maximum magnitude is therefore
$$
\boxed{
|\xi|_{\max}
=\frac{f_0|A|}{N^2}\frac{n\pi}{H}
=\frac{gn\pi}{N^2H}|\eta|_{\max}}.
$$
For $H=3000\ {\rm m}$, $N=10^{-3}\ {\rm s}^{-1}$, and $|\eta|_{\max}=10^{-2}\ {\rm m}$,
$$
|\xi|_{\max}
\simeq
\frac{9.81\,n\pi}{(10^{-3})^2(3000)}
(10^{-2}\ {\rm m})
\simeq103n\ {\rm m}.
$$
The first baroclinic mode therefore displaces interior density surfaces by about $10^2\ {\rm m}$ even though the surface moves only one centimetre.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 333 / 3 / b / iii / Solution

{title2=$\xi$}
{wiki}

An isopycnal displacement is the vertical displacement $\xi$ of a constant-density surface. In linear stratified quasi-geostrophic flow,
$$
\xi=-\frac{f_0}{N^2}\psi_z.
$$


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