QJEEAdv09Ph31f0: JEE Advanced 2009 Physics - Class 11 JEEAdv09Ph31f0 Solution: A uniform rod of length L and mass M is pivoted at the centre.…

The restoring torque is

$J = - 2 \times kx\left( {{L \over 2}} \right)\cos \theta = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

Now, $x = {L \over 2}\sin \theta$

Therefore, $J = - k\left( {{{{L^2}} \over 2}} \right)\sin \theta \cos \theta = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

$\Rightarrow \left( {{{ - k{L^2}} \over 4}} \right)\sin 2\theta = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

For small $\theta$ , $\sin 2\theta = 2\theta$ . Therefore,

${{ - k{L^2}\theta } \over 2} = I\left( {{{{d^2}\theta } \over {d{t^2}}}} \right)$

where $I = {{M{L^2}} \over {12}}$ . Therefore,

${{{d^2}\theta } \over {d{t^2}}} = \left( {{{ - 6k} \over M}} \right)\theta = - {\omega ^2}\theta$ (SHM)

$\Rightarrow \omega = \sqrt {{{6k} \over M}}$

Hence, the frequency of oscillation is

${\omega \over {2\pi }} = {1 \over {2\pi }}\sqrt {{{6k} \over M}}$