Question
A rectangular block of mass $$m$$ and area of cross-section $$A$$ floats in a liquid of density $$\rho .$$ If it is given a small vertical displacement from equilibrium, it undergoes oscillation with a time period $$T.$$ Then
A.
$$T \propto \sqrt \rho $$
B.
$$T \propto \frac{1}{{\sqrt A }}$$
C.
$$T \propto \frac{1}{\rho }$$
D.
$$T \propto \frac{1}{{\sqrt m }}$$
Answer :
$$T \propto \frac{1}{{\sqrt A }}$$
Solution :
Let block be displaced through $$x$$ $$m,$$ then weight of displaced water or upthrust, (upwards) is given by Archimedes principle
$${F_b} = - Ax\rho g$$
where, $$A$$ is the area of cross-section of the block and $$\rho $$ is its density. This must be equal to force $$\left( { = ma} \right)$$ applied, where, $$m$$ is the mass of the block and $$a$$ is the acceleration.
$$\therefore ma = - Ax\rho g\,\,{\text{or}}\,\,a = - \frac{{A\rho g}}{m}x = - {\omega ^2}x$$
This is the equation of simple harmonic motion. Time period of oscillation,
$$\eqalign{
& T = \frac{{2\pi }}{\omega } = 2\pi \sqrt {\frac{m}{{A\rho g}}} \cr
& \Rightarrow T \propto \frac{1}{{\sqrt A }} \cr} $$