Answer
The hoop oscillates with the same frequency as a simple pendulum of length equal to the diameter of the hoop.
Work Step by Step
We can find the period of the hoop's oscillations:
$T = 2\pi~\sqrt{\frac{I}{mgL_{cm}}}$
$T = 2\pi~\sqrt{\frac{2mr^2}{mgr}}$
$T = 2\pi~\sqrt{\frac{2r}{g}}$
Since the diameter is twice the radius, the diameter is $2r$. We can write the period of a simple pendulum with the length $2r$:
$T = 2\pi~\sqrt{\frac{2r}{g}}$
Since the period $T$ of each system is equal and $f = \frac{1}{T}$, then the frequency of each system is equal.
The hoop oscillates with the same frequency as a simple pendulum of length equal to the diameter of the hoop.
