Question

A thin uniform vertical rod of mass $m$ and length $l$ is hinged about the point $O$ as shown in the figure. The combined stiffness of springs is equal to $k$. Considering the springs to be light, find the angular frequency of small oscillations.

• A. $\omega =\sqrt{\frac{3k}{m}+\frac{3g}{l}}$
• B. $\omega =\sqrt{\frac{3k}{m}+\frac{g}{l}}$
• C. $\omega =\sqrt{\frac{k}{m}+\frac{3g}{l}}$
• D. $\omega =\sqrt{\frac{3k}{m}+\frac{3g}{2l}}$

Answer 1

The correct option is D $\omega =\sqrt{\frac{3k}{m}+\frac{3g}{2l}}$

Let the individual spring constants be $k_1$ and $k_2$ respectively. Let us rotate the bar by small angle $\theta$.


From the FBD, restoring torque about point ${}^{\prime }{O}^{\prime }$
$\tau =-[mg\frac{l}{2}\sin \theta +k_{1}x(l\cos \theta )+k_{2}x(l\cos \theta \left)\right]$
$\tau =-\left[\right(k_1+k_2)lx\cos \theta +mg\frac{l}{2}\sin \theta ].......1)$
Since, the elongation in one spring equals the compression in the other spring, we can say that, Both springs are in parallel combinations i.e $k_1+k_2=k$
Since $\theta$ is small, $\sin \theta \approx \theta$ and $\cos \theta \approx 1$
From the diagram we can say that , $x=l\theta$
From the above given data, $\left(1\right)$ can be written as,
$\tau =-[k{l}^2+mg\frac{l}{2}]\theta$
But we know that, $\tau =I\alpha$
Using this we get,
$I\alpha =-[k{l}^2+mg\frac{l}{2}]\theta$
Moment of inertia of rod about an axis passing through the end point ${}^{\prime }{O}^{\prime }$ is given by $I=\frac{m{l}^2}{3}$
$\Rightarrow \alpha =\frac{-[k{l}^2+mg\frac{l}{2}]\theta }{\frac{m{l}^2}{3}}$
Comparing this with $\alpha =-{\omega }^2\theta$, we get
$\omega =\sqrt{\frac{3k}{m}+\frac{3g}{2l}}$
Thus, option (d) is the correct answer.