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Question: A cylindrical piece of cork of density of base area A and height h floats in a liquid of density \({...

A cylindrical piece of cork of density of base area A and height h floats in a liquid of density ρ1{{\rho }_{1}} . The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
T=2πhρρ1gT=2\pi \sqrt{\dfrac{h\rho }{{{\rho }_{_{1}}}g}}
Where ρ is the density of cork. (Ignore damping due to viscosity of the liquid).

Explanation

Solution

Archimedes principle is the basic for solving this question. The equation of force including volume, density and acceleration due to gravity will also help in doing this. According to Archimedes’ principle, in equilibrium, the weight of the cork will be the same as the weight of the fluid displaced by the cork.

Complete step by step answer:
Archimedes' principle states that ,’the upward buoyant force exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces.’
According to Archimedes’ principle, in equilibrium, the weight of the cork will be the same as the weight of the fluid displaced by the cork.

Let the base area of the cork = A.
Height of the cork =h
Density of liquid displaced=ρ1{{\rho }_{1}}
Density of the floating cork=ρ

Cork has been dipped in the liquid more by a distance x and thus some extra volume of liquid got displaced from it. Hence, some restoring force is gained by the cork due to the upthrust force acting upwards.

Restoring force
F=(ρvg)F=-\left( \rho vg \right)
Where volume v= area× distance
Therefore force
F=Ax×ρ1gF=-Ax\times {{\rho }_{1}}g
We know that according to force law,
k=FXk=\dfrac{F}{X}
k is the spring constant.
Now substituting this in above equation,

k=Fx=Aρ1gk=\dfrac{F}{x}=-A{{\rho }_{1}}g……………. (1)
Time period of oscillation of a cork
T=2πmkT=2\pi \sqrt{\dfrac{m}{k}}…………… (2)
m= Mass of the cork
= Volume of the cork × Density
= Base area × Height × Density of the cork
=Ahρ=Ah\rho …………….. (3)
Substituting (2) and (3) in equation (1)
T=2πAhρAρ1g=2πhρρ1gT=2\pi \sqrt{\dfrac{Ah\rho }{A{{\rho }_{1}}g}}=2\pi \sqrt{\dfrac{h\rho }{{{\rho }_{1}}g}}
Hence proved.

Note:
Archimedes principle should be known and the equation of time period of oscillation should also be taken care of.
The equation for solving this question is
T=2πAhρAρ1g=2πhρρ1gT=2\pi \sqrt{\dfrac{Ah\rho }{A{{\rho }_{1}}g}}=2\pi \sqrt{\dfrac{h\rho }{{{\rho }_{1}}g}}
According to Archimedes’ principle, in equilibrium, the weight of the cork will be the same as the weight of the fluid displaced by the cork.