Question 26: Oscillations

Question

A uniform rod of length L and mass M is pivoted at the centre. Its two ends are attached to two springs of equal spring constants $$k$$. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle $$	heta$$ in one direction and released. The frequency of oscillation is

Accepted Answer

The restoring torque is
$$J =  - 2 	imes kx\left( {{L \over 2}} ight)\cos 	heta  = I\left( {{{{d^2}	heta } \over {d{t^2}}}} ight)$$
Now, $$x = {L \over 2}\sin 	heta $$
Therefore, $$J =  - k\left( {{{{L^2}} \over 2}} ight)\sin 	heta \cos 	heta  = I\left( {{{{d^2}	heta } \over {d{t^2}}}} ight)$$
$$ \Rightarrow \left( {{{ - k{L^2}} \over 4}} ight)\sin 2	heta  = I\left( {{{{d^2}	heta } \over {d{t^2}}}} ight)$$
For small $$	heta$$, $$\sin 2	heta  = 2	heta $$. Therefore,
$${{ - k{L^2}	heta } \over 2} = I\left( {{{{d^2}	heta } \over {d{t^2}}}} ight)$$
where $$I = {{M{L^2}} \over {12}}$$. Therefore,
$${{{d^2}	heta } \over {d{t^2}}} = \left( {{{ - 6k} \over M}} ight)	heta  =  - {\omega ^2}	heta $$ (SHM)
$$ \Rightarrow \omega  = \sqrt {{{6k} \over M}} $$
Hence, the frequency of oscillation is
$${\omega  \over {2\pi }} = {1 \over {2\pi }}\sqrt {{{6k} \over M}} $$

Answer

$${1 \over {2\pi }}\sqrt {{{2k} \over M}} $$

Answer

$${1 \over {2\pi }}\sqrt {{k \over M}} $$

Answer

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

Answer

$${1 \over {2\pi }}\sqrt {{{24k} \over M}} $$

Oscillations

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