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 `theta` from its horizontal equilibrium position and released. The angular frequency of the subsequent simple harmonic motion is

Options

A. `sqrt((3k)/(m))`
B. `sqrt((k)/(3m))`
C. `sqrt((3k)/(m)+(3g)/(2L))`
D. `sqrt((k)/(m))`

Answer

A

Explanation

Restoring torque
`tau=kyL=(mL^2)/(3)KL^2theta`
`implieskL^2theta=(mL^2)/(3)` `alphaimpliesalpha=(3k)/(m)theta`
`impliesT=2pisqrt((theta)/(alpha))=2pisqrt((mtheta)/(3ktheta))=2pisqrt((m)/(3k))`
`impliesomega=(2pi)/(T)=(2pi)/(2pi)sqrt((3k)/(m))`
`sqrt((3k)/(m))`