= Solution
{c}
Choose the orientation $J>0$. To leading order in the radial disturbance, the azimuthal <angular frequency> is $\omega_\theta=J/(mr_0^2)$, so
$$
\boxed{\frac{\omega_r}{\omega_\theta}=\sqrt{k+2}.}
$$
For either orientation the positive <frequency> ratio is $\omega_r/|\omega_\theta|$. At $k=-1$, the ratio is $1$: there is one radial oscillation per revolution. The bound <Kepler orbit> is an <ellipse> with the force centre at a focus, with one closest and one furthest approach per revolution. Its exact <polar coordinates> equation is
$$
r(\theta)=\frac{p}{1+e\cos\theta},\qquad
p=\frac{J^2}{m|\beta|},\qquad 0<e\ll1.
$$
At $k=2$, the ratio is $2$: there are two radial oscillations per revolution. The <isotropic harmonic oscillator> trajectory is an <ellipse> centred on the force centre. An exact parametrization is
$$
x(t)=a\cos\Omega t,\qquad y(t)=b\sin\Omega t,\qquad
\Omega=\sqrt{\frac{2\beta}{m}},\qquad a\simeq b.
$$
Here $r^2=a^2\cos^2\Omega t+b^2\sin^2\Omega t$ repeats twice during one full circuit. The two different placements of the force centre distinguish these <ellipses>. These are the two all-bound-orbit cases of <Bertrand's theorem>.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2022-central-orbit-sketches.png]
{title=Nearly circular ellipses: one radial cycle about a focus for the inverse-distance potential, and two about the centre for the quadratic potential}
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