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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
executes non-periodic motion
executes periodic motion which is not simple harmonic
executes $S.H.M.$ of frequency $\frac{1}{{2\pi }}\sqrt {\frac{{6K}}{M}}$
executes $S.H.M.$ of frequency $\frac{1}{{2\pi }}\sqrt {\frac{{K}}{2M}}$
Solution
If stick is rotated through a small angle $\theta,$ the
spring is stretched by a distance $\frac{\mathrm{L}}{2} \theta .$ Therefore
restoring force is spring $=\mathrm{K}\left(\frac{\mathrm{L}}{2} \theta\right) .$ Each spring
causes a torque $\tau(=\mathrm{Fd})=\left(\frac{\mathrm{KL} \theta}{2}\right) \frac{\mathrm{L}}{2}$ in the
same direction.
Therefore equation of ratational motion is $\tau=I \alpha$
$-2\left(\frac{\mathrm{KL} \theta}{2}\right) \frac{\mathrm{L}}{2}=\mathrm{I} \alpha$ with $\mathrm{I}=\frac{\mathrm{ML}^{2}}{12}$
$\alpha=-\left(\frac{6 K}{M}\right) \theta$
Standard equation of angular $S.H.M.$ is
$\alpha=-\omega^{2} \theta$
$\omega^{2}=\frac{6 \mathrm{K}}{\mathrm{M}}$
frequency $\quad \mathrm{f}=\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \sqrt{\frac{6 \mathrm{K}}{\mathrm{M}}}$
