The density $\rho$ of a liquid varies with depth h from the free surface as $\rho = kh.$ A small body of density ${\rho }_1$ is released from the surface of liquid. The body will

From the $F.B.D$ of the body $a=\frac{V{\rho }_{1}g-V\rho g}{V{\rho }_1}=\left(1-\frac{kh}{{\rho }_1}\right)g$ = $-\frac{k}{{\rho }_1}(h-\frac{{\rho }_1}{k})$

Since a $\propto -(h-\frac{{\rho }_1}{k}) , the motion is SHM.$

Therefore $V\frac{dv}{dh}=\left(1-\frac{kh}{{\rho }_1}\right)g \Rightarrow \frac{V^2}{2}=\left(h-\frac{k}{2{\rho }_2}{h}^2\right)g$

Therefore $V=0 at h=2{\rho }_1/k$

also  $a=0 h={\rho }_1/k$ is the mean position.

Therefore  Amplitude $={\rho }_1/k$