## Cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ1. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period where ρ is the density of cork. | Learn NCERT solution | Education portal Class 11 Physics | Study online Unit-14 Oscillations

### where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Base area of the cork = A
Height of the cork = h
Density of the liquid = ρ1

Density of the cork = ρ

In equilibrium:
Weight of the cork = Weight of the liquid displaced by the floating cork
Let the cork be depressed slightly by x. As a result, some extra water of a certain volume is displaced. Hence, an extra up-thrust acts upward and provides the restoring force to the cork.
Up-thrust = Restoring force, F = Weight of the extra water displaced
F = ­–(Volume × Density × g)
Volume = Area × Distance through which the cork is depressed
Volume = Ax
F = – A ρg …..(i)
Accroding to the force law:
F = kx
k = F/x
where, k is constant
k = F/x = -Aρg ….(ii)
The time period of the oscillations of the cork:
T = 2π √m/     ….(iii)
where,
m = Mass of the cork
= Volume of the cork × Density
= Base area of the cork × Height of the cork × Density of the cork
= Ahρ
Hence, the expression for the time period becomes: