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## Homework Statement

A pendulum of mass ##m## and length ##L## is connected to a spring as shown in figure. If the bob is displaced slightly from its mean position and released, it performs simple harmonic motion. What is the angular frequency of the bob?

## Homework Equations

Angular frequency for angular simple harmonic motion is given by $$\omega=\sqrt{\frac{\alpha}{\theta}}=\sqrt{\frac{K}{I}}$$

Where

##\theta=## Displaced angle

##\alpha=## Angular acceleration

##I=## Moment of inertia

##K=## Equivalent force constant

## The Attempt at a Solution

I have done the problem in the following way:

Consider a small angular displacement of ##\theta##. The torque about the the hinge at any instant is then given by $$\Gamma = -mgL\sin\theta-kxh$$ where ##x## is the displacement of the spring. But ##x=h\sin\theta## and therefore our equation becomes $$\Gamma = -mgL\sin\theta -mh^2\sin\theta$$

But ##\sin\theta \approx \theta## for small ##theta##. Therefore the torque is given by $$\Gamma = I\alpha=-mgL\theta -mh^2\theta$$ Thus we have ##\Gamma \propto -\theta##, which is the condition for angular SHM. Recalling that ##I=mL^2## we finally obtain $$\omega=\sqrt{\frac{\alpha}{\theta}} = \sqrt{\frac{K}{I}} = \sqrt{\frac{kh^2+mgL}{mL^2}}$$

Please look and tell me if there's anything wrong with the solution. Thank you.