The arrangement shown of a uniform rod of length l and mass m, pivoted at the centre and its two ends attached to two springs of spring constant `K_(1)` and `K_(2)` . The other end of springs are fixed to two rigid supports the rod which is free to rotate in the horizontal plane , is pushed by a small angle in one direction and released . Teh frequency of oscillation of the rod is

Options

A. `(1)/(2pi)sqrt(K_(1)+K_(2))/(m)`
B. `(1)/(2pi)sqrt(K_(1)+K_(2))/(3m)`
C. `(1)/(2sqrt3pi)sqrt(K_(1)+K_(2))/(m)`
D. `(sqrt3)/(2pi)sqrt(K_(1)+K_(2))/(m)`

Answer

D

Explanation

Restoring torque about O is
`tau = [K_(1) (l)/(2)theta xx (l)/(2) + K_(2)(l)/(2)theta xx (l)/(2)]`
`= (K_(1) + K_(2))(l^(2)theta)/(4)`
Moment of inertia of rod about its centre is
`I = (ml^(2))/(12)`
Angular acceleration
`((K_(1) + K_(2))l^(2)theta)/((4)/((ml^(2))/12)) = 3(K_(1) + K_(2))/(m)theta`
`alpha = omega^(2)theta`
`therefore omega^(2) = (3(K_(1) + K_(2)))/(m)`
`omega = sqrt((3(K_(1) + K_(2)))/(m))`
Frequency of oscillation,
`f = (omega)/(2pi) = (sqrt(3))/(2pi) sqrt((K_(1) + K_(2)))(,m)`