Question

A Cylindrical piece of cork of density of base area A and height h floats in a liquid of density \(\rho_l\) .The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
\(T\) = \(2π\sqrt{\frac{h\rho}{\rho_lg}}\)
Where \(\rho\) is the density of cork. (Ignore damping due to viscosity of the liquid).

Answer 1

Base area of the cork = \(A\)
Height of the cork = \(h\)
Density of the liquid =\(\rho_l\)
Density of the cork = \(\rho\)
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 = \(Ax\)
∴ \(F\) = \(– A x \rho_lg\) … (i)
According to the force law:
\(F\) = \(kx\)
\(k\) = \(\frac{F}{x}\)
Where, \(k\) is a constant
\(k\) = \(\frac{F}{x}\)=\(-A\rho_lg\)....(ii)
The time period of the oscillations of the cork:
\(T\) = \(2\pi\sqrt{\frac{m}{K}}\)...(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
= \(Ah\rho\)
Hence, the expression for the time period becomes:

\(T\) =\( 2\pi\sqrt{\frac{Ah\rho}{A\rho_lg}}\) 

=\(2\pi\sqrt{\frac{h\rho}{\rho_lg}}\)