Introduce
$$
\tau=f_0t,
\qquad
X=\frac{f_0x}{V},
\qquad
Y=\frac{f_0y}{V},
\qquad
\epsilon=\widetilde\beta=\frac{\beta V}{f_0^2}.
$$
Because the speed is $V$, write
$$
\dot x=V\sin\phi,
\qquad
\dot y=V\cos\phi.
$$
The momentum equations give
$$
\frac{d\phi}{d\tau}=1+\epsilon Y,
\qquad
\frac{dX}{d\tau}=\sin\phi,
\qquad
\frac{dY}{d\tau}=\cos\phi.
$$
Use an <asymptotic expansion>
$$
\phi=\tau+\epsilon\phi_1+O(\epsilon^2),
\quad
X=X_0+\epsilon X_1+O(\epsilon^2),
\quad
Y=Y_0+\epsilon Y_1+O(\epsilon^2).
$$
The zeroth-order <inertial oscillation> is
$$
X_0=1-\cos\tau,
\qquad
Y_0=\sin\tau.
$$
At first order,
$$
\phi_1'=\sin\tau,
\qquad
\phi_1=1-\cos\tau,
$$
and integration with the initial conditions gives
$$
X_1=\sin\tau-\frac14\sin2\tau-\frac12\tau,
$$
$$
Y_1=\cos\tau-1+\frac12\sin^2\tau.
$$
Therefore
$$
\boxed{
x(t)=\frac V{f_0}
\left[
1-\cos\tau
+\widetilde\beta
\left(\sin\tau-\frac14\sin2\tau-\frac12\tau\right)
\right]
+O(\widetilde\beta^2)},
$$
$$
\boxed{
y(t)=\frac V{f_0}
\left[
\sin\tau
+\widetilde\beta
\left(\cos\tau-1+\frac12\sin^2\tau\right)
\right]
+O(\widetilde\beta^2)}.
$$
The periodic terms describe a slightly distorted clockwise circle. The secular term in $x$ is a westward drift with velocity
$$
\boxed{
U_{\rm drift}=-\frac{\beta V^2}{2f_0^2}}.
$$
This <beta drift of an inertial oscillation> occurs because the <Coriolis parameter>, and hence the turning rate, is larger on the poleward half of the orbit than on its equatorward half. A trajectory sketch therefore consists of clockwise loops whose centers move steadily westward; the parcel starts at the westernmost point of its first loop and initially travels northward.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 333 / 1 / c / Solution

On a <beta plane>, the turning rate of an <inertial oscillation> is larger on the poleward side of its orbit than on the equatorward side. For initial speed $V$ and $\beta V/f_0^2\ll1$, this asymmetry produces westward mean drift of magnitude $\beta V^2/(2f_0^2)$.

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