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:

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

=$2\pi\sqrt{\frac{h\rho}{\rho_lg}}$