Question

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?

Answer 1

`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)`