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 springs is negligible. The angular frequency of small oscillation is

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

Answer 1

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

Let the following be the diagram in a slightly disturbed position
Restoring torque about $O$ is
$\tau =mg\left(\frac{l}{2}\right)sin\theta +k_{1}x\left(lcos\theta \right)+k_{2}x\left(lcos\theta \right)$
$=\frac{mgl}{2}sin\theta +kxlcos\theta$

$(k_1+k_2=k)$
$(\theta \to small,sin\theta \to \theta ,cos\theta \to 1,x=l\theta )$

$\tau =[\frac{mgl}{2}+k{l}^2]\theta$,
$\frac{m{l}^2}{3}\alpha =[\frac{mgl}{2}+k{l}^2]\theta$
$\frac{m{l}^2}{3}{\omega }^2\theta =[\frac{mgl}{2}+k{l}^2]\theta$

From the above equation, the final value of $\omega$ can be found out as:
$\omega =\sqrt{\frac{3g}{2l}+\frac{3k}{m}}$