A massless rod is pivoted at point $O$ . A string carrying a mass $m$ at one end is attached to the point $A$ on the rod.
The period of small vertical oscillation of mass $m$ around its equilibrium position is

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$T = 2 π \frac{b}{a} \sqrt{\frac{m}{k}}$

$T = 2 π \frac{a}{b} \sqrt{\frac{m}{k}}$

$T = π \frac{a}{b} \sqrt{\frac{m}{k}}$

$T = π \frac{b}{a} \sqrt{\frac{m}{k}}$

answer is A.

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Detailed Solution

In this device, the restoring torque is constituted by spring force $(F = k b θ)$ . So for small $θ$

$τ_{r e s t} = - (k b θ) \times b$

and $α = \frac{τ_{r e s t}}{I} = \frac{k b^{2}}{m a^{2}} (- θ)$

On comparing with, $α = - ω^{2} θ$ , we get

$ω = \sqrt{\frac{k b^{2}}{m a^{2}}}$ or $T = 2 π \sqrt{\frac{{ma}^{2}}{{kb}^{2}}}$

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