Question

Two light identical springs of spring constant $k$ are attached horizontally at the two ends of a uniform horizontal rod $\text{AB}$ of length $l$ and mass $m.$ The rod is pivoted at its centre $\text{O}$ and can rotate freely in horizontal plane. The other ends of two springs are fixed to rigid supports as shown in the figure. The rod is gently pushed through a small angle and released, the linear frequency of the resulting oscillation is :

• A. $\frac{1}{2\pi }\sqrt{\frac{3k}{m}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{2k}{m}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{6k}{m}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

Answer 1

The correct option is C $\frac{1}{2\pi }\sqrt{\frac{6k}{m}}$

Let the rod is displaced through a small angle $\theta$.



The linear displacement of the ends of the rod is given by,

$x=\frac{l}{2}\sin \theta$

Net torque due to spring force,

$\tau =-2kx{r}_{\perp }=-2k\times \frac{l}{2}\sin \theta \times \frac{l}{2}\cos \theta$

$\Rightarrow \tau =-\frac{k{l}^2\sin 2\theta }{4}$

$\Rightarrow \tau =-\frac{k{l}^2}{4}\times 2\theta$, as angle is small

$\Rightarrow \tau =-\frac{k{l}^2\theta }{2}$

$\Rightarrow -\frac{k{l}^2\theta }{2}=\frac{m{l}^2}{12}\alpha$

$\Rightarrow \alpha =-\frac{6k}{m}\theta =-{\omega }^2\theta$

$\Rightarrow \omega =\sqrt{\frac{6k}{m}}$

So, the linear frequency,

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

Hence, option $\left(C\right)$ is the correct answer.