Question

A uniform cubical block of density ${\rho }_b\left(\text{in g/cc}\right)$ and length a is released from rest from the position where one of its horizontal surface (lower) just touches the water (density ${\rho }_w=1\text{g/cc}$). The time period of oscillation of the block is $({\rho }_b=\frac{2}{3}\text{g/cc},\text{neglect viscosity and surface tension})$

• A. $2\pi \sqrt{\frac{p_{b}a}{p_{w}g}}$
• B. $2\pi \sqrt{\frac{p_{w}a}{p_{b}g}}$
• C. $\frac{4\pi }{3}\sqrt{\frac{p_{b}a}{p_{w}g}}+\frac{p_b}{p_w-p_b}\sqrt{\frac{4p_{w}a}{3p_{b}g}}$
• D. $\frac{2\pi }{3}\sqrt{\frac{p_{b}a}{p_{w}g}}+\frac{p_b}{p_w-p_b}\sqrt{\frac{2p_{w}a}{3p_{b}g}}$

Answer 1

The correct option is C $\frac{4\pi }{3}\sqrt{\frac{p_{b}a}{p_{w}g}}+\frac{p_b}{p_w-p_b}\sqrt{\frac{4p_{w}a}{3p_{b}g}}$

In case of equilibrium,

$B=mg$
$\rho \left(y_{)}A\right)g={\rho }_{b}aAg$
$\Rightarrow y_0=\frac{{\rho }_{b}a}{{\rho }_w}$

If we displace the body in downdwards direction, then
$\Rightarrow F_{restoring}=-(B-mg)$
$\Rightarrow F_{restoring}=-\left({\rho }_w\right(y_0+y)Ag-{\rho }_{b}Aag)$
$\Rightarrow F_{restoring}=-{\rho }_w{y}_{0}Ag-{\rho }_{w}yAg+{\rho }_{b}Aag$

Substituting value of $y_0$ we get,
$F_{restoring}=-{\rho }_{b}Aag-{\rho }_{w}yAg+{\rho }_{b}Aag$
$\therefore F_{restoring}=-\left({\rho }_{w}Ag\right)y$

Since restoring force is proportional to displacement from mean position, hence body is performing SHM motion.

So we can write,
$F_{restoring}=-m{\omega }^{2}y=-\left({\rho }_{w}Ag\right)y$
$\Rightarrow {\rho }_{b}Aa{\omega }^{2}y={\rho }_{w}Agy$
$\Rightarrow \omega =\sqrt{\frac{{\rho }_{w}g}{{\rho }_{b}a}}$

Now since $y_0=\frac{2a}{3}$ ,
This situation can be better understood from the diagram below.

So amplitude can be calculated as
$A=\frac{a}{2}+\frac{a}{6}=\frac{2a}{3}$

But by the time body comes down $\frac{a}{3}$, it will be completely submerged, and from there on buoyancy force does not increase more. So the motion ceases to be SHM.

So phasor of the motion of body will be,
$\cos \varphi =\frac{y}{A}=\frac{\frac{a}{3}}{\frac{2a}{3}}=\frac{1}{2}$
$\Rightarrow \varphi =\frac{\pi }{3}$

So body is performing SHM only for the phase of $2\pi -2\varphi =\frac{4\pi }{3}$.
Hence time duration when body is performing SHM will be
$T_1=\frac{4\pi }{3}\sqrt{\frac{{\rho }_{b}a}{{\rho }_{w}g}}$

Now for time taken for uniformly accelerated motion:

$a=\frac{B-mg}{m}$
$a=\frac{{\rho }_w-{\rho }_b}{{\rho }_b}g$

Velocity when body has just reached its completely submerged position,
$v=\omega \sqrt{A^2-y^2}$
$v=\sqrt{\frac{{\rho }_{w}g}{{\rho }_{b}a}}\sqrt{\frac{4a^2}{9}-\frac{a^2}{9}}$
$v=\sqrt{\frac{{\rho }_{w}ga}{3{\rho }_b}}$

So time taken during uniformly accelerated motion,
$T_2=\frac{2v}{a}=2\sqrt{\frac{{\rho }_{w}a}{3{\rho }_{b}g}}\frac{{\rho }_b}{{\rho }_w-{\rho }_b}$

So total time $=T_1+T_2=\frac{4\pi }{3}\sqrt{\frac{p_{b}a}{p_{w}g}}+\frac{p_b}{p_w-p_b}\sqrt{\frac{4p_{w}a}{3p_{b}g}}$