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

Answer:

#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)#