Q.A cylindrical piece of cork of density $\rho$ of base area $A$ and height $h$ floats in a liquid of density $\rho_l$. The cork is depressed slightly and then released. Show that the cork oscillates up and down simple harmonically with a period $T = 2\pi\sqrt{\dfrac{h\rho}{\rho_l g}}$ where $\rho$ is the density of cork. (Ignore damping due to viscosity of the liquid).

Concept: Archimedes Principle — the buoyant force equals the weight of the displaced liquid.

Reasoning:

1. At equilibrium, the cork floats with a submerged length $x_0$ such that

   $A x_0 \rho_l g = A h \rho g \quad\Rightarrow\quad x_0 = \frac{h\rho}{\rho_l}$.

2. When depressed by a small distance $y$ below equilibrium, the extra buoyant force is

   $F = - (A y \rho_l) g = -A \rho_l g \, y$.

3. This is a restoring force proportional to displacement: $F = -k y$ with $k = A \rho_l g$. …

Why Archimedes Principle is the key

A floating object is in equilibrium when the weight of the object equals the weight of the liquid it displaces. If you push the cork down a little, it displaces more liquid, so the buoyant force increases. That extra upward force acts like a spring — the more you push, the harder it pushes back. That's the essence of simple harmonic motion: a restoring force proportional to displacement.

The trick here is that the cork is a uniform cylinder, so the displaced volume is simply area times submerged depth. That makes the force-displacement relation perfectly linear.

Step-by-step derivation

1. Equilibrium condition

When the cork floats freely, let the submerged depth be $x_0$. The weight of the cork is $\rho A h g$. The buoyant force equals the weight of displaced liquid: $\rho_l A x_0 g$. At equilibrium:

Cancel $A g$:

This tells us the natural submerged depth. Notice that since $\rho < \rho_l$ (cork floats), $x_0 < h$ — only part of the cork is underwater.

2. Displace the cork and find the net force

Push the cork down by a small distance $y$ (measured from equilibrium, with downward taken as positive). The new submerged depth becomes $x_0 + y$.

The buoyant force now is $\rho_l A (x_0 + y) g$, upward. The weight is still $\rho A h g$, downward. The net force (taking upward as positive) is:

3. Simplify using equilibrium

From step 1, $\rho A h g = \rho_l A x_0 g$. Substitute this in:

So the net force is $F_{\text{net}} = \rho_l A g \, y$, directed upward (since $y$ is downward displacement, a positive $y$ gives a positive upward force — that's a restoring force).

4. Write the equation of motion …

Step 1: At equilibrium, buoyancy balances weight: $\rho_l A x_0 g=\rho A h g \Rightarrow x_0=\dfrac{\rho h}{\rho_l}$.

Step 2: Push the cork down an extra distance $y$ from equilibrium; the buoyant force becomes $\rho_l A(x_0+y)g$ while the weight $\rho Ahg$ is unchanged.

Step 3: Net upward (restoring) force: $F=\rho_l A x_0 g+\rho_l A yg-\rho Ahg=\rho_l Agy$ (the first and third terms cancel by the equilibrium condition of Step 1). …