Q.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 S.H.M. with a time period.
$T = 2\pi\sqrt{\dfrac{m}{A\rho g}}$
where $m$ is mass of the body and $\rho$ is density of the liquid.

When a body floats, the buoyant force balances its weight. When displaced, the change in buoyant force provides the restoring force, leading to Simple Harmonic Motion (S.H.M.).

1. Equilibrium: Let the log be submerged to a depth $h_0$ at equilibrium. The buoyant force $F_B = A h_0 \rho g$ balances the weight of the log $mg$. Thus, $mg = A h_0 \rho g$.
2. Displacement and Restoring Force: When the log is pressed down by a small distance $x$, the additional volume submerged is $Ax$. The increase in buoyant force is $\Delta F_B = (Ax)\rho g$. This increased buoyant force acts upwards, opposing the downward displacement $x$. …

When an object floats in a liquid, it is in equilibrium. This means the upward buoyant force exerted by the liquid perfectly balances the downward gravitational force (weight) of the object. If we disturb this equilibrium by pushing the object down slightly, the buoyant force increases, creating an upward net force. When released, this net force pushes the object back towards its equilibrium position. As it passes the equilibrium, its inertia carries it upwards, reducing the buoyant force and creating a downward net force. This continuous oscillation around the equilibrium position is the hallmark of Simple Harmonic Motion, provided the restoring force is directly proportional to the displacement and acts opposite to it.

Let's derive the time period for this motion step-by-step.

1. Equilibrium Position: Consider the log floating in water. Let $h_0$ be the height of the log submerged in the water at its equilibrium position. The volume of water displaced by the log at equilibrium is $V_0 = A h_0$. According to Archimedes' Principle, the buoyant force $F_B$ is equal to the weight of the displaced liquid.

At equilibrium, this buoyant force balances the weight of the log, $W = mg$.

This equation defines the equilibrium submerged height $h_0$.

2. Displacement from Equilibrium:

Now, imagine the log is pressed down by a small additional distance $x$ from its equilibrium position.

The new submerged height of the log becomes $h_0 + x$.

The new volume of water displaced is $V' = A(h_0 + x)$.

3. Net Restoring Force: The new buoyant force $F_B'$ acting on the log is:

Since the log is pushed downwards, the buoyant force $F_B'$ is now greater than its weight $mg$. The net force acting on the log will be an upward restoring force, trying to bring it back to equilibrium.
The net force $F_{net}$ is the difference between the buoyant force and the weight of the log:

Substituting the expressions for $F_B'$ and using Equation 1 ($mg = A h_0 \rho g$):

This net force is directed upwards, opposite to the downward displacement $x$. If we consider the downward direction as positive for displacement, then the restoring force acting upwards must be negative.
Therefore, the restoring force $F$ is: …

Step 1: At equilibrium, the buoyant force balances the log's weight: $Ah_0\rho g = mg$, where $h_0$ is the equilibrium submerged depth.

Step 2: Push the log down by a small extra depth $x$. The additional submerged volume is $Ax$, producing an additional upward buoyant force $\Delta F = A x\rho g$.

Step 3: This extra force opposes the displacement, so the net restoring force is $F = -(A\rho g)x$ — linear in $x$ and directed opposite to it, which is exactly the condition for SHM, with effective spring constant $k = A\rho g$. …