Q.
A uniform rod of length $L$ and mass $M$ is pivoted at the centre. Its two ends are attached to two springs of equal spring constant $k.$ The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $\theta$ in one direction and released. The frequency of oscillation is
A) $\frac{1}{2\pi}\sqrt{\frac{2k}{M}}$
B) $\frac{1}{2\pi}\sqrt{\frac{k}{M}}$
C) $\frac{1}{2\pi}\sqrt{\frac{6k}{M}}$
D) $\frac{1}{2\pi}\sqrt{\frac{24k}{M}}$
Explanation
(c) Torque about$\ P = - (kx)\frac{L}{2} + \left( - kx\frac{L}{2} \right) = - kxL = - k\frac{L^{2}}{2}\theta$
For small angle $\theta,\ x = \frac{L}{2}\theta;\ \tau = - I\alpha$
[image:14_image64.emf]
$\Rightarrow - \frac{KL^{2}}{2}\theta = \frac{ML^{2}}{12}\alpha$
$\Rightarrow \frac{- 6K\theta}{M} = \alpha$
$\Rightarrow \omega = \sqrt{\frac{6K}{M}}$ and $f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\frac{6K}{M}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.