Where, $l=$ length of part immersed in liquid.
When it is given in downward displacement $y$, restoring force (upward direction) on block is
$F=-[A(l+y) \rho g-m g]$
$=-[A(l+y) \rho g-A l \rho g]$
$=-A \rho g y$
i.e. $F \propto-y$ or $a \propto-\gamma,$ so it execute SHM (inertia factor). Mass of block $=m$
Spring factor $=A \rho g$
Time period $=2 \pi \sqrt{\frac{\text { Inertia factor }}{\text { spring factor }}}$
$T=2 \pi \sqrt{\frac{m}{A \rho g}}$
$\Rightarrow T^{2} \propto \frac{m}{A \rho}$
$(A)\;y= sin\omega t-cos\omega t$
$(B)\;y=sin^3\omega t$
$(C)\;y=5cos\left( {\frac{{3\pi }}{4} - 3\omega t} \right)$
$(D)\;y=1+\omega t+{\omega ^2}{t^2}$