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(C) Let the rod be rotated by a small angle $	heta$. The displacement of end $A$ is $x = \frac{l}{2} 	heta$.The restoring force in each spring is $F

Vedclass · Accepted Answer

$\pi \sqrt {\frac{2m}{{3k}}} $

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(C) Let the rod be rotated by a small angle $	heta$. The displacement of end $A$ is $x = \frac{l}{2} 	heta$.The restoring force in each spring is $F = kx = k \left( \frac{l}{2} 	heta ight)$.The restoring torque 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{kl^2}{2} 	heta$.The equation of motion for rotational oscillation is $	au = I \alpha$,where $I = \frac{ml^2}{12}$ is the moment of inertia of the rod about its center.So,$\frac{kl^2}{2} 	heta = I \alpha = \frac{ml^2}{12} \alpha$.This gives $\alpha = \frac{6k}{m} 	heta$.Comparing this with the standard $SHM$ equation $\alpha = \omega^2 	heta$,we get $\omega^2 = \frac{6k}{m}$,so $\omega = \sqrt{\frac{6k}{m}}$.The time period $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{6k}} = \pi \sqrt{\frac{4m}{6k}} = \pi \sqrt{\frac{2m}{3k}}$.

The ends of a rod of length $l$ and mass $m$ are attached to two identical springs. The rod is free to rotate about its centre $O$. The rod is depressed slightly at end $A$ and released. The time period of the resulting oscillation is

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