When $\log$ is pressed downward into the liquid then an upward Buoyant force acting on it which moves the block upward and due to inertia it moves upward from its mean position and then again comedown due to gravity, so net restoring force on block $=$ Buoyant force $-m g$


Let the log be pressed and let the vertical displacement at the equilibrium position be $x_0$.
At equilibrium, when block floats then
$m g=$ buoyant force
$\Rightarrow m g=V \rho g=\left(\rho A x_0\right) g \quad\left[\because m=V \rho=\left(A x_0\right) \rho\right]$
When it is displaced by a further displacement $x$, the buoyant force is $A\left(x_0+x\right) \rho g$.
$\therefore$ Net restoring force $=$ Buoyant force $-$ Weight
$=A\left(x_0+x\right) \rho g-m g$
$=(A \rho g) x$
$$
\left[\because m g=\rho A x_0 g\right]
$$
As displacement $x$ is downward and restoring force is upward,
we can write,
$F_{\text {restoring }}=-(A \rho g) x=-k x$
$F_{\text {restoring }} \propto-x$, so motion is in SHM.
where $k$ (constant) $=A \rho g$
So, the motion is SHM. $\quad(\because F \propto-x)$
Now, Acceleration $a=\frac{F_{\text {restoring }}}{m}=-\frac{k}{m} x$
Comparing with $a=-\omega^2 x$
So, $-\omega^2 x=-\frac{k}{m} x \Rightarrow \omega^2=\frac{k}{m} \Rightarrow \omega=\sqrt{\frac{k}{m}}$ $\frac{2 \pi}{T}=\sqrt{\frac{k}{m}} \Rightarrow T=2 \pi \sqrt{\frac{m}{k}}$
$$
T=2 \pi \sqrt{\frac{m}{A \rho g}}
$$