Hint: In a periodic motion, the body repeats its motion after a certain time interval.

Step: Find the acceleration acting on the ball.
When the ball bearing is released from a point slightly above the lower point the only restoring force is the force due to gravity which tries to take it at the mean position.
Thus, when the displacement is \(\theta\),
\(\begin{aligned} & F=-m g \sin \theta \\ &\Rightarrow m a=-m g \sin \theta \\ & \Rightarrow a=-g \sin \theta \\ & \Rightarrow \frac{d^2 x}{d t^2}=-g \sin \theta \end{aligned}\)
\(\Rightarrow a = \frac{d^2x}{dt^2}= -g\times \frac{x}{R}~~~[\sin\theta\approx\theta= \frac{x}{R}]\)
\(\Rightarrow a =-\omega^2 x~~~\left[\omega = \sqrt{\frac{g}{R}}\right] \Rightarrow a\propto-x\)
Therefore, the motion is periodic and SHM.
Hence, option (1) is the correct answer.