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

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Question

If stick is rotated through a small angle $	heta,$ the

spring is stretched by a distance $\frac{\mathrm{L}}{2} 	heta .$ Therefore

restoring fo

Vedclass · Accepted Answer

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

Vedclass · Answer

If stick is rotated through a small angle $	heta,$ the

spring is stretched by a distance $\frac{\mathrm{L}}{2} 	heta .$ Therefore

restoring force is spring $=\mathrm{K}\left(\frac{\mathrm{L}}{2} 	hetaight) .$ Each spring

causes a torque $	au(=\mathrm{Fd})=\left(\frac{\mathrm{KL} 	heta}{2}ight) \frac{\mathrm{L}}{2}$ in the

same direction.

Therefore equation of ratational motion is $	au=I \alpha$

$-2\left(\frac{\mathrm{KL} 	heta}{2}ight) \frac{\mathrm{L}}{2}=\mathrm{I} \alpha$ with $\mathrm{I}=\frac{\mathrm{ML}^{2}}{12}$

$\alpha=-\left(\frac{6 K}{M}ight) 	heta$

Standard equation of angular $S.H.M.$ is

$\alpha=-\omega^{2} 	heta$

$\omega^{2}=\frac{6 \mathrm{K}}{\mathrm{M}}$

frequency $\quad \mathrm{f}=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{6 \mathrm{K}}{\mathrm{M}}}$

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 figure, the strings are in their natural length. When the stick is displaced through a small angle $\theta $ and released. The stick

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