A uniform rod of length `L` and mass `M` is pivotedat the centre. Its two ends are attached to two springs of equal spring constants. `k`. The springs as shown in the figure, and the rod is free to oscillate in hte horizontal plane. the rod is gently pushed through a small angle `theta` in one direction and released. the frequency of oscilllation 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

`tau_0=-(kxL/2costheta)xx2`
`x=L/2sintheta`
For small `theta`, `costheta=1`, `sintheta=theta`
`T=-kL/2L/2theta=-(kL^2)/(2)theta`
`Ialpha-(kL^2)/(2)theta`
`(ML^2)/(12)alpha=-(kL^2)/(2)theta`
`alpha=-(6k)/(M)theta=-omega^2theta`
`T=2pisqrt((M)/(6k))`
`f=1/T=(1)/(2pi)sqrt((6k)/(M))`