Question

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

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

Answer 1

Answer: (C)

Let the rod be depressed by a small amount $x$ (in the figure). Both the springs are compressed by $x$ . When the rod is released, the restoring torque is given by
$\tau =\left(kx\right)\times \frac{1}{2}+\left(kx\right)\times \frac{1}{2}=\left(kx\right)1$
Now $tan\theta =\frac{x}{1/2}=\frac{2x}{l}$ .
Since $\theta$ is small, $tan\theta \simeq \theta$ , where $\theta$ is expressed in radian.
Thus $\theta =\frac{2x}{l}$ or $x=\frac{\theta l}{2}$
$\tau =k\left(\frac{\theta 1}{2}\right)\times 1=\frac{{\left(k\theta 1\right)}^2}{2}$

If $I$ is the moment of inertia of the rod about $O$ , then
$I\frac{d^2\theta }{{\left(dt\right)}^2}=-\left(\frac{{\left(kl\right)}^2}{2}\right)\theta$
Or $\frac{d^2\theta }{{\left(dt\right)}^2}=-\left(\frac{{\left(kl\right)}^2}{2I}\right)\theta$
Since $\frac{d^2\theta }{{\left(dt\right)}^2}\propto \left(-\theta \right)$ , the motion is simple harmonic whose angular frequency is given by
$\omega =\sqrt{\frac{k{l}^2}{2I}}$
Now, $\omega =\frac{2\pi }{T}$ and $I=\frac{m{l}^2}{12}$ . Therefore, we have
$\frac{2\pi }{T}=\sqrt{\frac{k{l}^2}{2}\times \frac{12}{m{l}^2}}=\sqrt{\frac{6k}{m}}$
Or $T=\pi \sqrt{\frac{2m}{3k}},$ which is choice $\pi \sqrt{\frac{2m}{3k}}$