= Solution
For the initial radius $r_0=l/2$ to be circular, the radial equation requires
$$
\frac{mv_0^2}{r_0}=Mg,\qquad
v_0^2=\frac{Mgl}{2m}.
$$
Consider a small radial perturbation at fixed conserved <angular momentum> $J$. Using $J^2/(mr_0^3)=Mg$ and $r=r_0+\epsilon$ in the radial equation gives
$$
(M+m)\ddot\epsilon
=Mg\left[\left(1+\frac\epsilon{r_0}\right)^{-3}-1\right]
=-\frac{3Mg}{r_0}\epsilon+O(\epsilon^2).
$$
Consequently the linearized radial motion is <simple harmonic motion>:
$$
\boxed{\ddot\epsilon+\omega_r^2\epsilon=0,\qquad
\omega_r=\sqrt{\frac{3Mg}{(M+m)r_0}}=\sqrt{\frac{6Mg}{(M+m)l}}.}
$$
The positive curvature of the <effective potential> establishes radial stability. This is the small-amplitude limit, not an exact finite-amplitude sinusoid. A perturbation that changes $J$ shifts the circular radius; its radial oscillations are about the new radius to the same order. A radial displacement or radial impulse preserving $J$ gives the stated oscillations about $l/2$.
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