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-

Explanation

Let the rod be pushed downward by a small amount x.
Both the springs are compressed by the same amount x. When the rod is released, the restoring torque is given by
`tau = (k_(1)x) xx (L)/(2) + (k_(2)x) xx (L)/(2) = ((k_(1) + k_(2))xL)/(2)`

now, `sin theta = (x)(L//2) = (2x)/(L)`, in radians, Since `theta` is small,
`sin theta -~ theta`, Thus `theta = 2x//L or x = Ltheta//2`.
Hence, `tau = (k_(1) + k^(2))(thetaL^(2))/(4)`
Let I be the moment of inertia of the rod about P. Then
`I(d^(2)theta)/(dt^(2)) = [((k_(1) + k_(2))L^(2))/(4)]theta `
or `(d^(2)theta)/(dt^(2)) = [((k_(1) + k_(2))L^(2))/(4I)] theta`
SInce `(ed^(2)theta)/(dt^(2)) prop (theta)`, the motion is simple harmonic whose angular frequency is given by
`omega = sqrt(((k_(1) + k_(2))l^(2))/(4I))`
Now, `omega = (2pi)/(T) and I = (ML^(2))/(12)`. Therefore, we have
`(2pi)/(T) = sqrt(((k_(1) + k_(2))L^(2))/(4)) xx (12)/(ML^(2)) = sqrt((3(k_(1) + k_(2)))/(M))` < br> or `T = 2pi sqrt((M)/(3(k_(1) + k_(2)))`