Question

A horizontal rod of mass m and length L is pivoted smoothly at one end. The rod’s other end is supported by a spring of force constant k. The rod is rotated (in the vertical plane) by a small angle from its horizontal equilibrium position and released. The angular frequency of the subsequent simple harmonic motion is

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Restoring torque:
$\tau =kyL=K{L}^2\theta$ (Since $y\cong L\theta$ from figure)
$\Rightarrow K{L}^2\theta =\frac{m{L}^2}{3}\alpha \Rightarrow \alpha =\frac{3k}{m}\theta$
$\Rightarrow \omega =\sqrt{\frac{3k}{m}}$ (torque due to mg was already balanced so it is not taken in calculation)