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((g)/(l sin alpha))`
B. `sqrt((3g)/(l sin alpha))`
C. `sqrt((3g)/(5l sin alpha))`
D. `sqrt((3g)/(2l sin alpha))`

Answer

D

Explanation

For a small angular displacement `beta` from vertical, the figure is: Restoring torque about 'O' is:

`tau = (T sin beta) l ~~ T betal ………….(1)`.
where `T =` tension in the string `=?` when the rod is in equilibrium position.
`sum tau = 0 = mg ((l)/(2)cos alpha) - T (l cos alpha)`
`rArr T = (mg)/(2) ............(2)`
From figure: `pp' = theta l = betal sin alpha................(3)`
`I = (ml^(2))/(3):` solve for time period.