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 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. $\frac{1}{2\pi }\sqrt{\frac{24k}{M}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{2k}{M}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{k}{M}}$

Answer 1

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

$\text{ANSWER:}$

Spring force acting at the end gives restoring torque

$\tau =rF\sin \theta$

$\Rightarrow {\tau }_0=-\left(kx\frac{L}{2}\cos \theta \right)\times 2$

Value of displacement of spring,

$x=\frac{L}{2}\sin \theta$

For the value of angular displacement tends to zero,

$\theta \to 0$

$\Rightarrow \cos \theta \cong 1,\sin \theta \cong \theta$

Then torque,

$\Rightarrow \tau =-k\frac{L}{2}\times \frac{L}{2}\theta =-\frac{k{L}^2}{2}\theta \dots \left(i\right)$

From equation $\left(i\right),$
$\Rightarrow \tau =I\alpha =\frac{k{L}^2}{2}\theta$

$\Rightarrow \frac{M{L}^2}{12}\alpha =-\frac{k{L}^2}{2}\theta$

$\Rightarrow \alpha =-\frac{6k}{M}\theta =-{\omega }^2\theta$

$\Rightarrow \omega =\sqrt{\frac{6k}{M}}$

Time period,

$T=\frac{2\pi }{\omega }$

$\Rightarrow T=2\pi \sqrt{\frac{M}{6k}}$

Frequency,

$f=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$

Final answer: $\left(c\right)$