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 1

Answer: (B)

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)
$=-A x \rho g$
Where $A$ is area of cross-section of the block and $\rho$ is its density. This must be equal to force $(=m a)$ applied, where $m$ is mass of the block and $a$ is acceleration.
$\therefore \quad m a=-A x \rho g$
or $a=-\frac{A \rho g}{m} x=-\omega^{2} x$
This is the cquation of simple harmonic motion.
Time period of oscillation
$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m}{A \rho g}} $
$\Rightarrow T \propto \frac{1}{\sqrt{A}}$