Please help?

A uniform rod of length L and mass M is pivoted at its center.It's 2 ends are attached to 2 springs of equal spring constants k.The springs are fixed to rigid supports as shown and the rod is free to oscillate in the horizontal plane.It's gently pushed through a small angle theta in one direction and released.What is the frequen cy of oscillation?enter image source here

1 Answer
Apr 15, 2018

omega = sqrt(1/12 M/k)

Explanation:

Assuming a small theta such that sintheta approx theta

I ddot theta =- L ^2 k theta

here

I = 1/12 M L^2 then

1/12 M L^2 ddot theta = L^2 k theta

or

1/12 M/k ddot theta + theta = 0 rArr omega = sqrt(1/12 M/k)