= Solution
In a central potential, $J=mr^2\dot\theta$ is conserved. Substituting it into $U=m(\dot r^2+r^2\dot\theta^2)/2+V(r)$ gives
$$
U=\frac m2\dot r^2+V_{\rm eff}(r),\qquad V_{\rm eff}=\frac{J^2}{2mr^2}+V(r).
$$
While the string stays taut, $V=k(r-L)^2/2$. Circular <force> balance gives $mr_0\omega_0^2=k(r_0-L)$, hence $\omega_0^2=k(r_0-L)/(mr_0)$. A radial impulse changes no <angular momentum>. The <radial oscillation of an elastic-string orbit> follows by expanding the same fixed-$J$ effective potential about its minimum:
$$
m\ddot{\delta r}=-V_{\rm eff}''(r_0)\delta r,\quad
\Omega^2=\frac{k}{m}+3\omega_0^2=\frac{k}{m}\left(4-\frac{3L}{r_0}\right),
$$
so
$$
\boxed{\frac\Omega{\omega_0}=\sqrt{\frac{4r_0-3L}{r_0-L}}.}
$$
For an outward impulse, one convenient linear approximation is $r=r_0+\epsilon\sin\Omega t$ and $\theta=\omega_0t+[2\omega_0\epsilon/(r_0\Omega)](\cos\Omega t-1)$. The angular modulation follows from $\dot\theta=J/(mr^2)$ and is needed for a consistent trajectory sketch.
\Image[/past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-elastic-string-orbits.png]
{title=Small radial-impulse trajectories: threefold modulation and the harmonic-orbit limit}
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