A rectangular block of mass m and area of cross-section A floats in a liquid of density `rho`. If it is given a small vertical displacement from equilibrium, it undergoes oscillation with a time period T. Then

Options

A. `T prop sqrt(rho)`
B. `T prop (1)/(sqrt(A))`
C. `T prop (1)/(rho)`
D. `T prop (1)/(sqrt(m))`

Answer

To solve the problem, we will derive the expression for the time period \( T \) of oscillation of a rectangular block floating in a liquid when displaced vertically from its equilibrium position.

### Step-by-Step Solution:

1. **Understanding the Forces**:
When the block is floating, it displaces a volume of liquid equal to the weight of the block. The buoyant force acting on the block is given by:
\[ F_b = \rho g V \]
where \( V \) is the volume of the liquid displaced. For a block of area \( A \) and depth \( x \) submerged, the volume is \( V = A x \). Thus, the buoyant force becomes:
\[ F_b = \rho g A x \]

2. **Weight of the Block**:
The weight of the block is:
\[ F_g = mg \]

3. **Equilibrium Condition**:
At equilibrium, the buoyant force equals the weight of the block:
\[ \rho g A h = mg \]
where \( h \) is the depth of the block submerged at equilibrium.

4. **Displacement from Equilibrium**:
When the block is displaced by a small distance \( x \) downwards, the new submerged depth is \( h + x \). The new buoyant force becomes:
\[ F_b' = \rho g A (h + x) = \rho g A h + \rho g A x \]

5. **Net Force**:
The net force acting on the block when it is displaced is given by:
\[ F_{net} = F_b' - F_g = \rho g A (h + x) - mg \]
Since \( \rho g A h = mg \), we can simplify this to:
\[ F_{net} = \rho g A x \]

6. **Using Newton's Second Law**:
According to Newton's second law, the net force is also equal to mass times acceleration:
\[ F_{net} = -m \frac{d^2x}{dt^2} \]
Setting these two expressions for net force equal gives:
\[ -m \frac{d^2x}{dt^2} = -\rho g A x \]
Rearranging gives:
\[ \frac{d^2x}{dt^2} + \frac{\rho g A}{m} x = 0 \]

7. **Identifying the Angular Frequency**:
This is a standard form of simple harmonic motion, where:
\[ \omega^2 = \frac{\rho g A}{m} \]
Thus, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{\frac{\rho g A}{m}} \]

8. **Finding the Time Period**:
The time period \( T \) of oscillation is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m}{\rho g A}} \]

### Final Expression:
Thus, the time period of oscillation \( T \) is given by:
\[ T = 2\pi \sqrt{\frac{m}{\rho g A}} \]