A uniform stick of mass M and length L is piv | vedclass

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(C) When the stick is rotated through a small angle $	heta$,each spring is stretched (or compressed) by a distance $x = \frac{L}{2} 	heta$.The resto

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executes $S.H.M.$ of frequency $\frac{1}{{2\pi }}\sqrt {\frac{{6K}}{M}}$

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(C) When the stick is rotated through a small angle $	heta$,each spring is stretched (or compressed) by a distance $x = \frac{L}{2} 	heta$.The restoring force in each spring is $F = Kx = K \left( \frac{L}{2} 	heta ight)$.Each spring exerts a restoring torque $	au$ about the pivot point given by $	au = F \cdot d = \left( K \frac{L}{2} 	heta ight) \frac{L}{2} = \frac{KL^2}{4} 	heta$.Since both springs exert torques in the same direction to oppose the displacement,the total restoring torque is $	au_{total} = 2 	imes \left( \frac{KL^2}{4} 	heta ight) = \frac{KL^2}{2} 	heta$.The equation of rotational motion is $	au_{total} = -I \alpha$,where $I = \frac{ML^2}{12}$ is the moment of inertia of the stick about its centre.Substituting the values: $-\frac{KL^2}{2} 	heta = \left( \frac{ML^2}{12} ight) \alpha$.Solving for angular acceleration $\alpha$: $\alpha = -\left( \frac{KL^2}{2} 	imes \frac{12}{ML^2} ight) 	heta = -\left( \frac{6K}{M} ight) 	heta$.Comparing this with the standard equation for angular $S.H.M.$,$\alpha = -\omega^2 	heta$,we get $\omega^2 = \frac{6K}{M}$,so $\omega = \sqrt{\frac{6K}{M}}$.The frequency of oscillation is $f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{6K}{M}}$.

$A$ uniform stick of mass $M$ and length $L$ is pivoted at its centre. Its ends are tied to two springs each of force constant $K$. In the position shown in the figure,the springs are in their natural length. When the stick is displaced through a small angle $\theta$ and released,the stick:

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