Question

Two cubical blocks of side length $a$ and $2a$ are stuck symmetrically as shown in the figure. The combined block is floating in water with the bigger block just submerged completely. The block is pushed down a little and released. Find the time period of its oscillations. Neglect viscosity.

• A. $\pi \sqrt{\frac{2a}{g}}$
• B. $2\pi \sqrt{\frac{2a}{g}}$
• C. $3\pi \sqrt{\frac{2a}{g}}$
• D. $2\pi \sqrt{\frac{a}{g}}$

Answer 1

The correct option is C $3\pi \sqrt{\frac{2a}{g}}$

In equilibrium, $F_B=W$
${\rho }_{\omega }(2a{)}^3=\rho g[(2a{)}^3+a^3]$
$\Rightarrow \frac{\rho }{{\rho }_{\omega }}=\frac{8}{9}$
When the block is depressed by $x$ downwards from is equilibrium position, the excess buoyancy is the restoring force.
Restoring force $=-\left(a^{2}x\right){\rho }_{\omega }g$
i.e $ma=-\left(a^{2}x\right){\rho }_{\omega }g$
or $\left(9a^3\right)\rho .\frac{d^{2}x}{d{t}^2}=-\left(a^2{\rho }_{\omega }g\right)x$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=-\left(\frac{g}{8a}\right)x[\because \frac{\rho }{{\rho }_{\omega }}=\frac{8}{9}]$
$\therefore {\omega }_1=\sqrt{\frac{g}{8a}}$
Time for half oscillation $t_1=\pi \sqrt{\frac{8a}{g}}$

When the block is raised above its mean position by distance $x$, the restoring force is
$F_{rest}=-\left(4a^2{\rho }_{\omega }g\right)x$
$\Rightarrow m\frac{d^{2}x}{d{t}^2}=-\left(4a^2{\rho }_{\omega }g\right)x$
$\Rightarrow 9a^3\rho \frac{d^{2}x}{d{t}^2}=-4a^2{\rho }_{\omega }gx$
$\Rightarrow \frac{d^{2}x}{d{t}^2}=-\left(\frac{g}{2a}\right)x$
$\Rightarrow {\omega }_2=\sqrt{\frac{g}{2a}}$
Time for half oscillation $t_2=\pi \sqrt{\frac{2a}{g}}$
$\therefore T=t_1+t_2=\pi \sqrt{\frac{2a}{g}}[2+1]=3\pi \sqrt{\frac{2a}{g}}$ is the time period of oscillation of block.