A cylindrical block of wood of mass m and area cross-section A is floating in water (density = `rho`) when its axis vertical. When pressed a little and the released the block starts oscillating. The period oscillations is

Options

A. `(2pisqrt(m/(rhobidwedgeg))`
B. `2pisqrt((mg)/(rhoA))`
C. `2pisqrt((rhoAg)/(m))`
D. `2pisqrt(rhoA)/(mg))`

Answer

To find the period of oscillation of a cylindrical block of wood floating in water, we can follow these steps:

### Step 1: Understand the Forces Acting on the Block
When the block is floating, it experiences two main forces:
- The weight of the block (W = mg, acting downwards)
- The buoyant force (F_b, acting upwards)

At equilibrium, the buoyant force equals the weight of the block:
\[ F_b = W \]
\[ F_b = \rho_{\text{water}} \cdot V_{\text{submerged}} \cdot g \]
Where \( V_{\text{submerged}} = A \cdot h \) (A is the area of cross-section and h is the height submerged).

### Step 2: Write the Equilibrium Condition
At equilibrium, we can express:
\[ \rho_{\text{water}} \cdot A \cdot h \cdot g = mg \]
This gives us the relationship between the submerged height and the mass of the block.

### Step 3: Consider the Displacement
When the block is pressed down slightly and then released, it will oscillate. Let \( x \) be the additional displacement from the equilibrium position. The new submerged height becomes \( h + x \).

### Step 4: Calculate the New Buoyant Force
The new buoyant force when the block is displaced by \( x \) is:
\[ F_b' = \rho_{\text{water}} \cdot A \cdot (h + x) \cdot g \]
\[ F_b' = \rho_{\text{water}} \cdot A \cdot h \cdot g + \rho_{\text{water}} \cdot A \cdot x \cdot g \]

### Step 5: Find the Net Force
The net force \( F \) acting on the block when displaced is:
\[ F = W - F_b' \]
Substituting the expressions we have:
\[ F = mg - \left( \rho_{\text{water}} \cdot A \cdot h \cdot g + \rho_{\text{water}} \cdot A \cdot x \cdot g \right) \]
Using the equilibrium condition \( F_b = mg \), we can simplify:
\[ F = - \rho_{\text{water}} \cdot A \cdot g \cdot x \]

### Step 6: Relate to Simple Harmonic Motion
This net force can be expressed in the form of Hooke's Law:
\[ F = -kx \]
Where \( k = \rho_{\text{water}} \cdot A \cdot g \).

### Step 7: Find the Time Period
The time period \( T \) of oscillation for a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting for \( k \):
\[ T = 2\pi \sqrt{\frac{m}{\rho_{\text{water}} \cdot A \cdot g}} \]

### Final Answer
The period of oscillation of the cylindrical block of wood floating in water is:
\[ T = 2\pi \sqrt{\frac{m}{\rho_{\text{water}} \cdot A \cdot g}} \]

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