Question

A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and rod is free to oscillate in the horizontal plane.The rod is gently pushed through a small angle $\theta$ in one direction and released.The frequency of oscillation is

• A. $\frac{1}{2\pi }\sqrt{\frac{2k}{M}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{k}{M}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{24k}{M}}$

Answer 1

Answer: (C)

$\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$

Restoring torque $=2\times F_{spring}\frac{l}{2}$
$F_{spring}=k\delta x$
For small angular displacement $\theta$,
$\delta x=\frac{l}{2}\theta$
Restoring torque $=-2\times k\left(\frac{l}{2}\theta \right)\frac{l}{2}=I\alpha$
where $\alpha$ is angular acceleration of the rod
$\alpha =\frac{d^2\theta }{d{t}^2}$
Restoring torque $=-k\left(\frac{l^2}{2}\theta \right)=\frac{I{d}^2\theta }{d{t}^2}$
$\frac{d^2\theta }{d{t}^2}=\frac{\frac{k{l}^2}{2}(-\theta )}{\frac{M{l}^2}{12}}$
this is equation of simple harmonic motion where $\alpha =C\theta$
angular frequency is given by $\Rightarrow \omega =\sqrt{C}=\sqrt{\frac{6k}{M}}$
frequency is given by $\frac{\omega }{2\pi }$ $=\frac{1}{2\pi }\sqrt{\frac{6k}{M}}$