A 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 T = \( \dfrac{ 2π \sqrt{hρ}}{ρ_1g}\) where ρ is the density of cork. (Ignore damping due to viscosity of the liquid)

Asked by Abhisek | 1 year ago |  114

1 Answer

Solution :-

Given

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 excess water of a certain volume is displaced.

Thus, 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 x Density x g)

Volume = Area x Distance through which the cork is depressed

Volume = Ax

Therefore,

F = – Ax x ρ1g …. (1)

According to the force law:

F = kx

k = \(\dfrac{F}{X}\)

where,

k is constant

k = \( \dfrac{F}{X}\) = – Aρ1g ……. (2)

The time period of the oscillations of the cork:

T = 2π (\( \sqrt{\dfrac{m}{k} }\)) ……. (3)

Where,

m = Mass of the cork

= Volume of the cork x Density

= Base area of the cork x Height of the cork x Density of the cork

= Ahρ

Therefore, the expression for the time period becomes:

T = 2π \( \sqrt{\dfrac{Ahρ}{Aρ_1g}}\)

T = 2π\( \sqrt{\dfrac{hρ}{ρ_1g}}\)

Answered by Pragya Singh | 1 year ago

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