A uniform rod of length (L) and mass (M) is pivoted at the centre. Its two ends are attached to two springs of equal spring constants (k). The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle (theta) in one direction and released. The frequency of oscillation is. ?
.

Options

A. `(1)/(2pi)sqrt((2k)/(M))`
B. `(1)/(2pi)sqrt((k)/(M))`
C. `(1)/(2pi)sqrt((6k)/(M))`
D. `(1)/(2pi)sqrt((24k)/(M))`

Answer

C

Explanation

The compression of the spring , `x=L/2 theta`

Restoring torque acting on rod , `tau=-(kx)L/2xx2`
`=-(kL/2theta)L=-1/2k L^2 theta`..(i)
If I is the moment of inertie of rod about axis of rotation at O, then `I=(ML)^2/(12)`
From (i), `tau prop theta` and this `tau` brings the rod back to its equilibrium position. If the rod is left free it will excute angular SHM with O as axis of rotation.
Comparing (i), with : `tau=-K theta`
We have spring factor , `K=1/2 kL^2`
Inertia factor=moment of inertia=`(ML)^2/(12)`
`therefore` Frequency of oscillation, `v=1/(2pi)sqrt(("spring factor")/("inertia factor"))=1/(2pi)sqrt((k L^2//12)/(ML^2//12))`
`=1/(2pi)sqrt((6k)/M)`