\textbf{1. Equilibrium Condition:} In equilibrium, the weight of the cylinder is balanced by the buoyant force. Let \(h_0\) be the depth of immersion in equilibrium. Weight of the cylinder: \(W = mg\) Buoyant force: \(F_B = V_{submerged} \rho g = (A h_0) \rho g\) In equilibrium, \(W = F_B\): \[mg = A h_0 \rho g\] \[h_0 = \frac{m}{A \rho}\] \textbf{2. Displacement and Restoring Force:} Let the cylinder be displaced vertically downwards by a small distance \(y\) from equilibrium. New depth of immersion: \(h_0 + y\) New buoyant force: \(F_B' = A (h_0 + y) \rho g\) Restoring force is the net upward force: \[F_{restoring} = F_B' - W = A (h_0 + y) \rho g - mg\] Substitute \(mg = A h_0 \rho g\): \[F_{restoring} = A h_0 \rho g + A y \rho g - mg = mg + A y \rho g - mg = A y \rho g\] Since the restoring force is in opposite direction to displacement, we can write: \[F_{restoring} = - (A \rho g) y\] \textbf{3. Simple Harmonic Motion (SHM):} Since \(F_{restoring} \propto -y\), the motion is Simple Harmonic Motion. The spring constant \(k\) is given by: \[k = A \rho g\] \textbf{4. Time Period of Oscillation:} The time period of SHM is given by: \[T = 2\pi \sqrt{\frac{m}{k}}\] Substitute \(k = A \rho g\): \[T = 2\pi \sqrt{\frac{m}{A \rho g}}\] \textbf{5. Final Answer:} The time period of small vertical oscillation is \(2\pi \sqrt{\frac{m}{A \rho g}}\). \[\boxed{2\pi \sqrt{\frac{m}{A \rho g}}}\]