A rectangular block of mass m and area of cross-section A floats in a liquid of density $\mathrm{\rho }$. If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T then:

Force applied on the body will be equal to upthrust for vertical oscillations.
Let block is displaced through x m, then the weight of displaced water or upthrust (upwards)
= _-Ax$\mathrm{\rho }$g
Where A is area of cross-section of the block and $\mathrm{\rho }$ is its density. This must be equal to force (= ma) applied, where m is mass of the block and a is acceleration.
$\therefore$ ma = _ Ax$\mathrm{\rho }$g
or $\mathrm{a} = -\frac{\mathrm{A\rho g}}{\mathrm{m}}\mathrm{x} = -{\mathrm{\omega }}^2\mathrm{x}$

This is the equation of simple harmonic motion.

Time period of oscillation

$T = \frac{2\mathrm{\pi }}{\omega }= 2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{A\rho g}}} \Rightarrow \mathrm{T} \propto \frac{1}{\sqrt{\mathrm{A}}}$