It is a physical pendulum, the time period of which is,

$\text{T}=2\text{π}\sqrt{\frac{\text{I}}{\text{mgl}}}$

Here, $I$ = moment of inertia of the ring about the point of suspension

          $= m{r}^2 + m{r}^2$

          $= 2m{r}^2$

​and   l = distance of the point of suspension from the centre of gravity = r

$\therefore \text{T}=2\text{π}\sqrt{\frac{2{\text{mr}}^2}{\text{mgr}}}=2\text{π}\sqrt{\frac{\text{2r}}{\text{g}}}$

$\therefore \text{Angular frequency}\text{ω}=\frac{2\text{π}}{\text{T}}$

$\text{or}\text{ω}=\sqrt{\frac{\text{g}}{\text{2r}}}$