Question

The ends of a rod of length $l$ and mass $m$ are attached to two identical springs as shown in figure. The rod is free to rotate about its centre ${}^{\prime }{\text{O}}^{\prime }$. The rod is depressed slightly at end $\text{A}$ and released. The time period of the resulting oscillation is

• A. $T=2\pi \sqrt{\frac{2m}{k}}$
• B. $T=2\pi \sqrt{\frac{m}{2k}}$
• C. $T=\pi \sqrt{\frac{2m}{3k}}$
• D. $T=2\pi \sqrt{\frac{3m}{k}}$

Answer 1

The correct option is C $T=\pi \sqrt{\frac{2m}{3k}}$


Let the rod be depressed by a small distance $x$. Both the springs are compressed by $x$. When the rod is released, the restoring torque about ${}^{\prime }{\text{O}}^{\prime }$ is given by
$\tau =-\left(kx\right)\times \frac{l}{2}+-\left(kx\right)\times \frac{l}{2}=-\left(kx\right)l.....\left(1\right)$
[positive denotes anticlockwise]
From the diagram, $\sin \theta =\frac{x}{\frac{l}{2}}=\frac{2x}{l}$
$\theta$ is very small $\therefore \sin \theta \approx \theta$.
where $\theta$ is expressed in radian.
Thus $\theta =\frac{2x}{l}$ or $x=\frac{\theta }{2}l$.

From $\left(1\right)$, we can write that
$\tau =-k\left(\frac{\theta l}{2}\right)\times l=-\frac{k\theta l^2}{2}\circlearrowleft$
But we know that $\tau =I\alpha$
Moment of inertia of the rod about ${}^{\prime }{\text{O}}^{\prime }$ $I=\frac{m{l}^2}{12}$
Thus, we get
$\frac{m{l}^2}{12}\alpha =-\frac{k{l}^2}{2}\theta$
$\Rightarrow \alpha =-\frac{6k}{m}\theta$
Comparing with $\alpha =-{\omega }^2\theta$ we get,
${\omega }^2=\frac{6k}{m}$
Time period of oscillations
$T=\frac{2\pi }{\omega }=\pi \sqrt{\frac{2m}{3k}}$
Thus, option (c) is the correct answer.