A vertical pole of length `l` , density `rho` , area of cross section A floats in two immiscible liquids of densities `rho_(1)`and `rho_(2)`. In equuilibrium possition the bottom end is at the interface of the liquids. When the cylinder is displaced vertically, find the time period of oscillation.

Answer

`pi [sqrt((rho l)/(rho_(1) g)) + sqrt((rho l)/(rho_(2) g)) ]`

Explanation

In equation position , net force on pole is zero.
`rho Alg = rho_(1) A l_(1) g` (i)
In displaced possition, net force.
`F_(1) = rho A l g - rho_(1) A l_(1) g - rho_(2) A y g = - A g y`
Acceleration of the pole. `a_(1) = - (rho_(2) Ag)/(rho A l) y`
Hence in the liquid of density `r_(2)`.
`omega_(1) = sqrt(( rho_(2) g)/(rho l)) implies T_(1) = 2 pi sqrt((rho l)/(rho_(2) g))`
The pole system is in the liquid of density `rho_(2)` for half time period only. When it moves in the liquid of density `rho_(1)`, the net force is `F_(2) = - rho A l g + (l_(1) - y') A rho_(1) g = - rho_(1) A g y'`
`a_(2) = - (rho_(2) A g)/(rho A l) y' implies omega_(2) = sqrt((rho_(1) g)/(rho l))`
`T_(2) = 2 pi sqrt((rho l)/(rho_(1) g))`
Total time period of oscillation of the cylinder is
`T = (T_(1))/(2) + (T_(2))/(2) = pi [sqrt((rho l)/(rho_(1) g)) + sqrt((rho l)/(rho_(2) g)) ]`