Question

Figure shows a container having liquid of variable density.The density of liquid varies as $\rho ={\rho }_0(4-\frac{3h}{h_0})$ Here $h_0$ and ${\rho }_0$ are constants and h is measured from bottom of the container. A solid block of small dimensions whose density is $\frac{5}{2}{\rho }_0$ and mass m is released from bottom of the tank. Prove that the block will execute simple harmonic motion. Find the frequency of oscillation ?

Answer 1

Net force on the block at a height h from the bottom is
$F_{net}=$upthrust-weight (upwards)
$=\left(\frac{m}{\frac{5}{2}{\rho }_0}\right){\rho }_0(4-\frac{3h}{h_0})g-mg$
$F_{net}=0$ at $h=\frac{h_0}{2}$
So, $h=\frac{h_0}{2}$ is the equilibrium position of the block.
For $h>\frac{h_0}{2}$, weight> upthrust
i.e., net force is downwards and for $h<\frac{h_0}{2}$ weight< upthrust
i.e., net force is upwards.
For upward displacement x from mean position, net downward force is
$F=-[\left(\frac{m}{\frac{5}{2}{\rho }_0}\right){\rho }_0\{4-\frac{3(h+x)}{h_0}\}g-mg](h=\frac{h_0}{2})$
$\therefore$ $F=-\frac{6mg}{5h_0}x$ (i)
(because at $h=\frac{h_0}{2}$ upthrust and weight are equal)
Since $F\propto -x$
Oscillations are simple harmonic in nature. Rewriting Eq. (i)
$ma=\ast \frac{6mgx}{5h_0}$
or $a=-\frac{6g}{5h_0}x$
$\therefore$ $f=\frac{1}{2\pi }\sqrt{\left|\frac{a}{x}\right|}$
$f=\frac{1}{2\pi }\sqrt{\frac{6g}{5h_0}}$