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(C) Let the rod be rotated by a small angle $	heta$. The displacement of each end of the rod is $x = \frac{l}{2} 	heta$.Each spring exerts a restori

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the rod be rotated by a small angle $	heta$. The displacement of each end of the rod is $x = \frac{l}{2} 	heta$.Each spring exerts a restoring force $F = kx = k \frac{l}{2} 	heta$.The restoring torque $	au$ about the center $O$ is $	au = 2 	imes (F 	imes \frac{l}{2}) = 2 	imes (k \frac{l}{2} 	heta 	imes \frac{l}{2}) = \frac{k l^2}{2} 	heta$.The moment of inertia of the rod about its center is $I = \frac{ml^2}{12}$.Using the rotational form of Newton's second law,$	au = -I \alpha$,where $\alpha$ is the angular acceleration:$\frac{k l^2}{2} 	heta = -(\frac{ml^2}{12}) \alpha \implies \alpha = -(\frac{6k}{m}) 	heta$.Comparing this with the standard equation for angular simple harmonic motion $\alpha = -\omega^2 	heta$,we get $\omega^2 = \frac{6k}{m}$,so $\omega = \sqrt{\frac{6k}{m}}$.The frequency $f$ is given by $f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{6k}{m}}$.

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