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A horizontal rod of mass m and length L is pivoted at one end. The rod's other end is supported by a spring of force constant k. The rod is displaced by a small angle q from its horizontal equilibrium position and released. The angular frequency of the subsequent simple harmonic motion is

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$\sqrt{\frac{3 k}{m}}$

$\sqrt{\frac{k}{3 m}}$

$\sqrt{\frac{3 k}{m} + \frac{3 g}{2 L}}$

$\sqrt{\frac{k}{m}}$

answer is A.

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Detailed Solution

Restoring torque $τ = k y L = \frac{m L^{2}}{3} α$

$\Rightarrow K L^{2} θ = \frac{m L^{2}}{3} . α$

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