$A$ cylindrical log of wood of height $h$ and area of cross-section $A$ floats in water. It is pressed and then released. Show that the log would execute $SHM$ with a time period $T = 2\pi \sqrt{\frac{m}{A\rho g}}$,where $m$ is the mass of the body and $\rho$ is the density of the liquid.

(N/A) Let the log be pressed downward by a small displacement $y$. The additional volume of liquid displaced by the log is $V_{disp} = A \times y$.
The additional buoyant force $F_b$ acting on the log due to this extra displaced volume is equal to the weight of the displaced liquid:
$F_b = V_{disp} \times \rho \times g = (A \times y) \times \rho \times g = (A \rho g) y$.
Since this buoyant force acts in the direction opposite to the displacement $y$,the restoring force $F$ is:
$F = - (A \rho g) y$.
This equation is of the form $F = -ky$,where $k = A \rho g$ is the effective force constant.
Since the restoring force is directly proportional to the displacement and directed towards the equilibrium position,the motion of the log is Simple Harmonic Motion $(SHM)$.
The time period $T$ of $SHM$ is given by:
$T = 2\pi \sqrt{\frac{m}{k}}$.
Substituting $k = A \rho g$ into the formula:
$T = 2\pi \sqrt{\frac{m}{A \rho g}}$.
Thus,it is proved that the log executes $SHM$ with the given time period.