Q: A physical pendulum is positioned so that its centre of gravity is above the suspension point. When the pendulum is released it passes the point of stable equilibrium with an angular velocity ω . The period of small oscillations of the pendulum is

(a) $ \displaystyle \frac{4\pi}{\omega} $

(b) $ \displaystyle \frac{2\pi}{\omega} $

(c) $\displaystyle \frac{\pi}{\omega} $

(d) $ \displaystyle \frac{\pi}{2\omega} $

Ans: (a)

$ \displaystyle \frac{1}{2}I\omega^2 $ = mg(2l)

$ \displaystyle I = \frac{4mgl}{\omega^2} $

$ \displaystyle T = 2\pi\sqrt{\frac{I}{mgl}} $

$ \displaystyle T = 2\pi\sqrt{\frac{4mgl}{\omega^2 mgl}} $

$ \displaystyle \frac{4\pi}{\omega} $