In the figure is shown a vessel filled with a liquid whose density varies with height h from the bottom as $\rho ={\rho }_0(4-\frac{3h}{h_0})$, where ${\rho }_0$ is a constant and $h_0$ is the height of the liquid in the container. A solid block of small dimensions of density $\frac{5}{2}{\rho }_0$ and mass m is released from the bottom of the tank. Then,

Density of block is ${\rho }_{block}=\frac{5}{2}{\rho }_0$. Resultant force acting on the block at the height h from bottom is  $F=\left[\frac{m}{\frac{5}{2}{\rho }_0}\right]{\rho }_0[4-\frac{3h}{h_0}]g-mg$
When the block is in equilibrium position, $F=0$
$$\begin{gathered} \therefore \frac{2}{5}(4-\frac{3h}{h_0})=1 \\ \Rightarrow h=\frac{h_0}{2} \end{gathered}$$

Let the block has moved up by a distance x from mean position.
Now, the resultant force is

$F_R=\left[\frac{m}{\frac{5}{2}{\rho }_0}\right]{\rho }_0\{4-\frac{3(\frac{h_0}{2}+x)}{h_0}\}g-mg=-\left[\frac{6mg}{5h_0}\right]x$

Acceleration of the block is $a=\frac{F_R}{m}=-\left(\frac{6g}{5h_0}\right)x$. So, the block performs SHM and  $\omega =\sqrt{\frac{6g}{5h_0}}$

$f=\frac{1}{2\pi }\sqrt{\frac{6g}{5h_0}}$