Question

A cylindrical piece of cork of density of base area $A$ and height $h$ floats in a liquid of density ${\rho }_1$. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period
$T=2\pi \sqrt{\frac{h\rho }{{\rho }_{1}g}}$
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 $={\rho }_1$

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
$F=-(Volume\times Density\times g)$
Volume $=$ Area $\times$ Distance through which the cork is depressed
Volume $=Ax$

$\therefore F=-Ax\times {\rho }_{1}g$ ...(i)
Accroding to the force law:

$F=kx$

$k=F/x$
where, $k$ is constant

$k=F/x=-A{\rho }_{1}g$ ...(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 $\times$ Density
$=$ Base area of the cork $\times$ Height of the cork $\times$ Density of the cork
$=Ah\rho$
Hence, the expression for the time period becomes:

$T=2\pi \sqrt{\frac{Ah\rho }{A{\rho }_{1}g}}=2\pi \sqrt{\frac{h\rho }{{\rho }_{1}g}}$