A thin rod of length L and area of cross section S is pivoted at its lowest point P inside a stationary, homogeneous and non-viscous liquid. The rod is free to rotate in a vertical plane about a horizontal axis passing through P. The density `d_(1)`of the rod is smaller than the density `d_(2)` of the liquid. The rod is displaced by a small angle theta from its equilibrium position and then released. Shown that the motion of the rod is simple harmonic and determine its angular frequency in terms of the given parameters ___________ .

Explanation

If the rod is displaced through an angle `alpha` from its equilibrium position i.e., vertical position then it will be under the thrust U and its own weight mg. There will be a net torque, which will try to restore the rod to its equilibrium position.
Its equilibrium position.
Weight of the rod acting downward
`mg = SLd_(1)g`
Buoyant force acting upwards
`U = SLd_(2)g`
Net force acting on the rod in the upward direction
`= SL (d_(2) - d_(1)) g`
where the area of cross-section of the rod is S
Net restoring torque `tau = SL (d_(2) - d_(1)) g xx (L)/(2) sin alpha`
`tau = SL (d_(2) - d_(1)) g xx (L)/(2) alpha ( :. sin alpha = alpha)`
If `alpha` is in clockwise direction, then `tau` will be in anticlockwise direction. Thus
`tau = -(1)/(2) SL^(2) (d_(2) - d_(1)) g alpha`
`tau = 1 alpha = -(3)/(SL^(3) d_(1)) xx (1)/(2) SL^(2) (d_(2) - d_(1)) g alpha or (d^(2) alpha)/(dt^(2)) = - (3g)/(2L) ((d_(2) - d_(1))/(d_(1))) alpha`
This is the equation of angular S.H.M
For which the angular frequency `omega` is given by
`omega^(2) = (3g)/(2L) ((d_(2) - d_(1))/(d_(1)))`
The angular frequency `omega = sqrt((3g)/(2) ((d_(2) - d_(1))/(d_(1))))`
Therefore time period `T = (2pi)/(omega) = 2pi sqrt((2L)/(3g) ((d_(1))/(d_(2) - d_(1))))`