Physics, asked by AaishaFathzz1473, 1 year ago

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

Answers

Answered by prathamesh1855
10
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 x ρ1 g …..(i)
Accroding to the force law:
F = kx 
k = F/x
where, k is constant
k = F/x = -Aρ1 g ….(ii)
The time period of the oscillations of the cork:
T = 2π √m/k      ….(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:


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