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

Explanation

`I = (1)/(3) M (2L)^(2) = (4)/(3) ML^(2)`
Force applied by the spring is `F = - ks`
`implies F = - k (2L theta)`
(`theta` is the angular displacement from the equalibrium possition). Further
`tau = |bar tau| = |bar r xx bar F| = - 4L^(2) k sin theta = - 4L^(2) k theta`
Also, `tau = I alpha = I ddot theta = - 4L^(2) k theta`
` implies ddot theta + (3k)/(M) theta = 0` `implies omega_(0) = sqrt((3k)/(M))`