Question

The ends of a rod of length $2\text{m}$ and mass $1\text{kg}$ are attached to two identical springs $(k=200\text{N/m})$, as shown in figure.


The rod is free to rotate about its centre $\text{C}$ The rod is depressed slightly at end $\text{A}$ and release. What is the time period of oscillation.

• A. $\frac{\pi }{10\sqrt{3}}\text{s}$
• B. $\frac{\pi }{2\sqrt{3}}\text{s}$
• C. $\frac{\pi }{2}\text{s}$
• D. $\frac{\pi }{\sqrt{3}}\text{s}$

Answer 1

The correct option is A $\frac{\pi }{10\sqrt{3}}\text{s}$


The net torque on the rod is given by,

$\tau =kxd+kxd=2kxd$

$\tau =2k\left(l^{\prime }\sin \theta \right)\left(l^{\prime }\cos \theta \right)$

Since, $\theta$ is very small,
$\therefore \sin \theta \approx \theta ,\cos \theta \approx 1$

$\therefore \tau =2k{l}^{\prime 2}\theta =\frac{k{l}^2\theta }{2}(\because l^{\prime }=l/2)$

As we know that, $\tau =I\alpha$

$\frac{m{l}^2}{12}\alpha +\frac{k{l}^2\theta }{2}=0(\because I=m{l}^2/12)$

$\tau =-\frac{6k}{m}\theta ..........\left(1\right)$

Comparing with equation $\tau =-{\omega }^2\theta$, we get

$\omega =\sqrt{\frac{6k}{m}}=\sqrt{\frac{6\times 200}{1}}=20\sqrt{3}\text{rad/s}$

So, time period will be

$T=\frac{2\pi }{\omega }=\frac{2\pi }{20\sqrt{3}}$

$\therefore T=\frac{\pi }{10\sqrt{3}}\text{s}$

Hence, option $\left(\text{A}\right)$ is correct.

Why this question ? This question is useful to understand angular simple Harmonic oscillations. In angular $\text{SHM},$ first try to find out equation of restoring torque then convert it in differential equation form of $\theta$ and then from the co-efficient of $\theta$, you can calculate $\omega ,\tau$ etc.