Question

A rod of mass M and length L is pivoted about its end O as shown in the figure. Find the period of SHM.

• (A) $2\pi \sqrt{\frac{3L}{2g}}$
• (B) $2\pi \sqrt{\frac{2L}{3g}}$
• (C) $2\pi \sqrt{\frac{E}{9}}$
• (D) $2\pi \sqrt{\frac{L}{3g}}$

Answer 1

The Correct Option is B

Explanation:
$\tau =\frac{MgL}{2}\sin \theta$For the small angle, $\sin \theta \approx \theta$And we have $\tau =I\alpha =I\frac{d^2\theta }{d{t}^2}$So, $\left(\frac{M{L}^2}{3}\right)\frac{d^2\theta }{d{t}^2}=-\frac{MgL}{2}\theta$$\frac{d^2\theta }{d{t}^2}=-\frac{3g}{2L}\theta$So, the angular frequency is $\omega =\sqrt{\frac{3g}{2L}}$$\therefore T=2\pi \sqrt{\frac{2L}{3g}}$Hence, the correct option is (B).