Question

A thin uniform vertical rod of mass $m$ and length $l$ pivoted at point $O$ is shown in the given figure. The combined stiffness of the springs is equal to $k$. The mass of the spring is negligible. Find the frequency of small oscillation.

Answer 1

As $\theta$ is small we consider $l\theta$ as linear displacement. So, each spring has change in length is equal to $l\theta$.

The restoring force $=-2kl\theta$

The moment of inertia of rod about one of its end is $I=\frac{m{l}^2}{3}$

Where, $l$=length of rod

and, $m$=mass of rod

Moment of restoring force about $o$ is $=-2kl\theta .OA=-2kl\theta .l\cos \theta$

Angular acceleration of rod

$\frac{d^2\theta }{{dt}^2}=\frac{Momentofrestoringforceabout0}{I}=2kl\theta .l\cos \theta$

$\frac{d^2\theta }{{dt}^2}=\frac{-2k{l}^2\theta }{\frac{m{l}^2}{3}}$ (as $\theta$ is small, $\cos \theta \approx 1$)

$\Rightarrow \frac{d^2\theta }{{dt}^2}+\left(\frac{6k}{m}\right)\theta =0$

Angular frequency $\omega =\sqrt{\frac{6k}{m}}$

frequency, $f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{6k}{m}}$