Question

The figure shows a container having liquid of variable density. The density of liquid varies as $\rho ={\rho }_o(4-\frac{3h}{h_o})$. Here, $h_o\text{and}{\rho }_o$ are constants and $h$ is measured from the bottom of the container. A solid block of small dimensions whose density is $\frac{5}{2}{\rho }_o$ and mass $m$ is released from the bottom of the tank. Then -

• A. The block executes SHM with mean position located at a height $h=\frac{h_o}{2}$ from the bottom of the container.
• B. The block performs oscillatory motion about $h=\frac{h_o}{2}$ but not SHM.
• C. The block oscillates simple harmonically with a frequency $f=\frac{1}{2\pi }\sqrt{\frac{6g}{7h_o}}$
• D. The block oscillates simple harmonically with a frequency $f=\frac{1}{2\pi }\sqrt{\frac{6g}{5h_o}}$

Answer 1

Answer: (A), (D)

The block oscillates simple harmonically with a frequency $f=\frac{1}{2\pi }\sqrt{\frac{6g}{5h_o}}$

Let us identity the mean position of the given system.


Net force on the block at a height $h$ from the bottom is,

$F_{\text{net}}=$ Upthrust $-$ Weight

$\Rightarrow F_{\text{net}}=\left(\frac{m}{\frac{5}{2}{\rho }_o}\right){\rho }_o(4-\frac{3h}{h_o})g-mg$

$[\because$ Upthrust $=\rho gV]$

For mean position, $F_{\text{net}}=0$

$\Rightarrow \left(\frac{m}{\frac{5}{2}{\rho }_o}\right){\rho }_o(4-\frac{3h}{h_o})g=mg...\left(1\right)$

$\Rightarrow h=\frac{h_o}{2}$

So, $h=\frac{h_o}{2}$ is the equilibrium position of the block.

For $h>\frac{h_o}{2}$,

Weight $>$ Upthrust

$i.e.$ net force is downward.

And for $h<\frac{h_o}{2}$,

Weight $<$ Upthrust

$i.e.$ net force is upward.

For upward displacement $x$ from mean position, net downward force is,

$F=-[\left(\frac{m}{\frac{5}{2}{\rho }_o}\right){\rho }_o(4-\frac{3(h+x)}{h_o})g-mg]...\left(2\right)$

Using equation $\left(1\right)$ and $\left(2\right)$, we get,

$F=-\frac{6mg}{5h_o}x$

$\Rightarrow F\propto x$ and negative sign shows restoring nature.

$\Rightarrow$ Oscillations are simple and harmonic in nature.

Also,

$ma=\frac{6mgx}{5h_o}$

$\Rightarrow a=\frac{6gx}{5h_o}$

$\Rightarrow {\omega }^{2}x=\frac{6g}{5h_o}x$

$\Rightarrow \omega =\sqrt{\frac{6g}{5h_o}}$

$\therefore f=\frac{1}{2\pi }\sqrt{\frac{6g}{5h_o}}$

Hence, option $\left(A\right)$ and $\left(D\right)$ are correct.