Derive an expression for the angular frequency of small oscillation of the bob of a simple pendulum when it is immersed in a liquid of density `rho`. Assume the density of the bob as `sigma` and length of the string as `l`.

Explanation

Method 1 :
The bob of the pendulum experiences buoyant due to liquid `= V p g`, in addition to gravitational force. Thus net force on the bob `= (mg - Vpg)`

For small displacement x of the bob, restoring force
`F_(real) = (mg - Vpg) (x)/(t)`
and acceleration `= - (g - ( V pg)/(m)) (x)/(t)`
On comparing with standerd equation of SHM, `a = - omega^(2) x`, we get
`omega = sqrt(((g - (V p g)/(m)))/(l)) = sqrt((g)/(l) (1 - (p)/(sigma)))`
and `T = 2 pi sqrt((l)/(g(1 - (p)/(sigma))`
Method 2
At equilibrium position the pendulum should be vertical. Free body diagram of the bob at equilibrum position
`T + B = mg`
`T = mg - B = V sigma g - Vpg = Vg (sigma - p)`
Equivalent force constant of oscillation
`k_(eq) = (T)/(l) = (Vg (sigma - p))/(l)`
But we know `k_(eq) = m omega^(2)`
`omega = sqrt((k_(eq))/(m)) = sqrt ((Vg(sigma - p))V sigma - l)) = sqrt ((g)/(l) (1 - (p)/(sigma)))`
Hence `T = (2 pi)/(omega) = 2 pi sqrt((l)/(g (1 - (p)/(sigma)))`