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

Answer

C

Explanation

Figure shows that rod at an angle (theta) with respect to its equilibrium position. Both the springs are streched by length `(l theta)/(2)`
(##JMA_CHMO_C10_009_S01##).
The restoring torque due to the springs.
`tau = -2` (Restoring force) xx perpendicular distance
`tau = - 2 k((l theta)/(2)) xx (l)/(2) = -k(l^2)/(2) theta`....(i) ltbRgt If (I) is the moment of interia of the rod about (M) then
`tau = I prop = I (d^2 theta)/(d t^2)` ....(ii)
From (i) & (ii) we get
`I(d^2 theta)/(d t^2) = -(l^2 theta)/(2) rArr (d^2 theta)/(d t^2)= - (k)/(I) (l^2)/(2) theta = (- k)/(M l^2//12)(l^2)/(2) theta`
Comparing it with the standard equation of rotational (SHM) we get
`(d^2 theta)/(d t^2) =-omega^2 theta rArr omega^2 = (6 k)/(M) rArr omega = sqrt(6 k)/(M)`
`rArr 2 pi v = sqrt(6 k)/(M) rArr v = (1)/(2 pi)sqrt( 6k)/ (M)`.