A solid cylinder of radius r and mass M rests on a curved path  of radius R as  shown figure. When displaced through a small angle $\theta$  as shown and left to itself, it  executes simple harmonic motion. Then the time period of oscillation is (Assume  that the cylinder rolls without slipping)

Restoring torque acting on a cylinder after a small displacement $\theta$ , about an axis passing through point of contact O with the curved path, $\tau =-Mgr\sin \theta$, for small angles   $\sin \theta \approx \theta$
Therefore,  $\tau =-Mgr\theta$………….(1)
Angular acceleration of the cylinder is  $\alpha =\frac{d^2\varphi }{d{t}^2}$ 
 From the diagram, $r\varphi =(R-r)\theta$ 
   $\varphi =\frac{(R-r)\theta }{r}$ 
   Consequently,  $\alpha =\frac{d^2}{d{t}^2}\left(\frac{(R-r)\theta }{r}\right)$
   $\alpha =\left(\frac{(R-r)}{r}\right)\frac{d^2\theta }{d{t}^2}$……………………. (2)
   Moment of inertia of the cylinder about a point of contact is $I=\frac{3}{2}M{r}^2$   
   Using the definition of torque  $\tau =I\alpha$ and substituting equation (2) in it,
  $\tau =\left(\frac{3}{2}M{r}^2\right)\left(\frac{(R-r)}{r}\right)\frac{d^2\theta }{d{t}^2}$   
   Substituting equation (1) in the above relation,
 $-Mgr\theta =\left(\frac{3}{2}M{r}^2\right)\left(\frac{(R-r)}{r}\right)\frac{d^2\theta }{d{t}^2}$  
   $-Mg\theta =\left(\frac{3}{2}M\right)(R-r)\frac{d^2\theta }{d{t}^2}$
   $-g\theta =\left(\frac{3}{2}\right)(R-r)\frac{d^2\theta }{d{t}^2}$
  This equation can be written as  $0=\frac{d^2\theta }{d{t}^2}+\left(\frac{2}{3(R-r)}\right)g\theta$
   Where  $\frac{d^2\theta }{d{t}^2}=\alpha$, Therefore  $-\left(\frac{2}{3(R-r)}\right)g\theta =\alpha$
   Comparing this with the equation of motion for angular simple harmonic motion,  $\alpha =-{\omega }^2\theta$ ,
  ${\omega }^2=\left(\frac{2g}{3(R-r)}\right)$
 Hence angular frequency is  $\omega =\sqrt{\left(\frac{2g}{3(R-r)}\right)}$
$\frac{2\pi }{T}=\sqrt{\left(\frac{2g}{3(R-r)}\right)}$, where   $\omega =\frac{2\pi }{T}$
 
$T=2\pi \sqrt{\left(\frac{3(R-r)}{2g}\right)}$