A uniform metre scale of length 1m is balanced on a fixed semi–circular cylinder of radius 30 cm as shown in figure. One end of the scale is slightly depressed and released. The time period (in seconds) of the resulting simple harmonic motion is (Take $g=10m{s}^{-2}$ )

 

The magnitude of the restoring torque = force $\times$ perpendicular distance 

$=mg\times AB=mg\times R\sin \theta$

Since $\theta$ is small, $\sin \theta \simeq \theta .$ Here $\theta$ is expressed is radian. The equation of motion of the scale is $I\frac{d^2\theta }{d{t}^2}=-mgR\theta$ Or $I\frac{d^2\theta }{d{t}^2}=\left(\frac{-mgR}{I}\right)\theta$

$\therefore \omega =\sqrt{\frac{I}{mgR}}.NowI=\frac{m{L}^2}{12}$ Hence $T=\frac{\pi L}{\sqrt{3gR}}$

Using the values $L=1m,g=10m{s}^{-2}$ and R = 0.3m, we get $T=\pi /3$ second.