A pendulum consisting of a rigid rod of negligible mass and length L and bob of mass M is connected to an ideal spring of force constant k at a distance h below its point of suspension. The rod is in equilibrium in vertical position and the spring is at its natural length when it is horizontal. If the pendulum is oscillated in a vertical plane containing spring, frequency of vibration of the system for small values of $θ$ is (Neglect mass of the rod):

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1 2 π L m g L + k L 2 m

1 2 π L g L + k h 2 m

2 π m L 2 m g L + k h

1 2 π L g L + k h m

answer is D.

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Detailed Solution

The extension developed in the string due to small values of $′ θ ′$ is:

$x = h \sin θ ≃ h θ$

Torque about $′ O ′$ :

$τ 0 = (M g sin θ) L + (k x) h$

$τ_{0} = m g θ L + k h^{2} θ = (m g L + k h^{2}) θ$ …..(1)
Also; $τ 0 = I 0 α = m L 2 α$ ...(2)
From (1) and (2):
$\begin{matrix} m L^{2} α = (m g L + k h^{2}) θ \\ \Rightarrow α = \frac{1}{L^{2}} (g L + \frac{k h^{2}}{m}) θ \end{matrix}$