Question

For the arrangement shown, a solid block of mass $M$ is connected to another block of mass $m$ by string. The block of mass $m$ is connected to the spring. The cross-sectional area of the submerged block is $A$. Initially, the submerged block is in equilibrium position and is displaced slightly inside the liquid of density $\rho$. The motion of mass $M$ is a simple harmonic motion. What is its time period?

• A. $2\pi \sqrt{\frac{2M}{(2K+\rho Ag)}}$
• B. $2\pi \sqrt{\frac{2M}{(K+\rho Ag)}}$
• C. $2\pi \sqrt{\frac{M}{(2K+\rho Ag)}}$
• D. $2\pi \sqrt{\frac{M}{(K+\rho Ag)}}$

Answer 1

The correct option is B $2\pi \sqrt{\frac{2M}{(K+\rho Ag)}}$

Let the block of mass $m$ be at equilibrium under the tension $T$ at position $x_o$. Also, let $V$ be the volume of the block of mass $M$ inside the liquid.


We have, $T=K{x}_0......\left(i\right)$ and
$T+V\rho g=Mg......\left(ii\right)$
From (i) - (ii), we get
$K{x}_0=Mg-V\rho g\dots \left(iii\right)$
When a block of mass $M$ is displaced by $\Delta x$ downwards, then


We have, $ma=K(x_0+\Delta x)-T^{\prime }....\left(iv\right)$ and,
$Ma=T^{\prime }+(V+A\Delta x)\rho g-Mg....\left(v\right)$
Adding equation $\left(iv\right)$ and $\left(v\right)$,
$(M+m)a=K(x_0+\Delta x)+(V+A\Delta x)\rho g-Mg$
[using (iii)]
$\Rightarrow (M+m)a=-(k+A\rho g)\Delta x$
$\Rightarrow a=-\frac{(K+A\rho g)}{M+m}dx$
$\therefore \omega =\sqrt{\frac{K+A\rho g}{M+m}}$
Thus, $T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{M+m}{K+A\rho g}}$