For small angular displacements $\theta$, the spring stretches by a distance $x \approx r\theta$. The spring exerts a restoring force $F_s = -kx = -kr\theta$. This force creates a restoring torque $\tau$ about the axle.

$$\tau = F_s \cdot r = (-kr\theta)r = -kr^2\theta$$

The equation for rotational motion is $\tau = I\alpha$, where $I$ is the moment of inertia and $\alpha = d^2\theta/dt^2$ is the angular acceleration. For a hoop rotating about its center, $I = mR^2$.

$$I\alpha = -kr^2\theta$$ $$mR^2\alpha = -kr^2\theta$$ $$\alpha = -\frac{kr^2}{mR^2}\theta$$

This is the equation for simple harmonic motion, $\alpha = -\omega^2\theta$, with the angular frequency $\omega$ given by:

$$\omega = \sqrt{\frac{kr^2}{mR^2}} = \frac{r}{R}\sqrt{\frac{k}{m}}$$

For the case $r=R$, the expression simplifies. For the case $r=0$, the spring is attached at the axle, creates no torque, and thus no oscillation.