Question

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 constants $k$. The springs are fixed to rigid supports as shown in the figure, and 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. $\frac{1}{2\pi }\sqrt{\frac{24k}{M}}$

Answer 1

The correct option is C. $\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$.

Restoring torque $=-2\times K\left(\frac{l}{2}\theta \right)\frac{l}{2}=I\frac{d^2\theta }{d{t}^2}$

Now, $\frac{d^2\theta }{d{t}^2}=\frac{\frac{k{l}^2}{2}(-\theta )}{\frac{M{l}^2}{12}}$

From the relation, $\alpha =-{\omega }^2\theta$ we have

$⟹\omega =\sqrt{\frac{6k}{M}}$

$\therefore$ Frequency, $f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$