A uniform rod of length L and mass M is pivot | vedclass

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Question

(C) When the rod is rotated by a small angle $	heta$,the displacement of each spring end is $x = \frac{L}{2} \sin 	heta \approx \frac{L}{2} 	heta$

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) When the rod is rotated by a small angle $	heta$,the displacement of each spring end is $x = \frac{L}{2} \sin 	heta \approx \frac{L}{2} 	heta$ for small $	heta$.Each spring exerts a restoring force $F = kx = k \frac{L}{2} 	heta$.The restoring torque provided by each spring about the center is $	au = F \cdot \frac{L}{2} = k \left( \frac{L}{2} 	heta ight) \frac{L}{2} = \frac{k L^2}{4} 	heta$.Since there are two springs,the total restoring torque is $	au_{total} = 2 	imes \frac{k L^2}{4} 	heta = \frac{k L^2}{2} 	heta$.The equation of motion for rotational oscillation is $	au = I \alpha$,where $I = \frac{M L^2}{12}$ is the moment of inertia of the rod about its center.So,$\frac{M L^2}{12} \frac{d^2 	heta}{dt^2} = - \frac{k L^2}{2} 	heta$.$\frac{d^2 	heta}{dt^2} = - \left( \frac{6 k}{M} ight) 	heta$.Comparing this with the standard $SHM$ equation $\frac{d^2 	heta}{dt^2} = - \omega^2 	heta$,we get $\omega^2 = \frac{6 k}{M}$,so $\omega = \sqrt{\frac{6 k}{M}}$.The frequency of oscillation is $f = \frac{\omega}{2 \pi} = \frac{1}{2 \pi} \sqrt{\frac{6 k}{M}}$.

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