As illustrated in the figure, a small block undergoes...

Let the block be displaced by a small angle $\theta$ from the lowest point (mean position).

The restoring force acting on the block along the tangential direction is:
$F = -mg \sin\theta$

For small oscillations, $\sin\theta \approx \theta$. Thus, the restoring force becomes:
$F = -mg \theta$

The displacement along the arc is $x = R\theta$, which implies $\theta = \dfrac{x}{R}$. Substituting this into the force equation gives:
$F = -mg \left(\dfrac{x}{R}\right)$

The acceleration of the block is:
$a = \dfrac{F}{m} = -\left(\dfrac{g}{R}\right) x$

Comparing this with the standard equation of simple harmonic motion $a = -\omega^2 x$, we get:
$\omega^2 = \dfrac{g}{R} \Rightarrow \omega = \sqrt{\dfrac{g}{R}}$

The time period of oscillation is:
$T = \dfrac{2\pi}{\omega} = 2\pi \sqrt{\dfrac{R}{g}}$