A cubical wood piece of mass 13.5 gm and volume `0.0027 m^3` float in water. It is depressed and released, then it executes simple harmonic motion. Calculate the period of oscillation ?

Explanation

To solve the problem of finding the period of oscillation of a cubical wood piece floating in water, we will follow these steps:

### Step 1: Identify the given data
- Mass of the wood piece, \( m = 13.5 \, \text{g} = 0.0135 \, \text{kg} \) (convert grams to kilograms)
- Volume of the wood piece, \( V = 0.0027 \, \text{m}^3 \)
- Density of water, \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \)

### Step 2: Calculate the density of the wood
The density of the wood can be calculated using the formula:
\[ \rho_{\text{wood}} = \frac{m}{V} \]
Substituting the values:
\[ \rho_{\text{wood}} = \frac{0.0135 \, \text{kg}}{0.0027 \, \text{m}^3} = 5 \, \text{kg/m}^3 \]

### Step 3: Determine the equilibrium condition
When the wood piece is floating, the buoyant force equals the weight of the wood piece:
\[ F_b = mg \]
Where \( F_b \) is the buoyant force and \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 4: Calculate the submerged volume
The buoyant force can also be expressed as:
\[ F_b = \rho_{\text{water}} \cdot V_{\text{submerged}} \cdot g \]
At equilibrium, we have:
\[ \rho_{\text{water}} \cdot V_{\text{submerged}} \cdot g = mg \]
From this, we can find \( V_{\text{submerged}} \):
\[ V_{\text{submerged}} = \frac{mg}{\rho_{\text{water}} \cdot g} = \frac{0.0135 \, \text{kg}}{1000 \, \text{kg/m}^3} = 0.0000135 \, \text{m}^3 \]

### Step 5: Set up the equation of motion
When the block is depressed by a distance \( x \), the additional buoyant force is given by:
\[ F_b' = \rho_{\text{water}} \cdot (V_{\text{submerged}} + A \cdot x) \cdot g \]
Where \( A \) is the cross-sectional area of the cube.

### Step 6: Relate the forces to simple harmonic motion
The net force acting on the block when it is displaced is given by:
\[ F = F_b' - mg \]
This force will lead to simple harmonic motion, where:
\[ F = -k x \]
Where \( k \) is the effective spring constant.

### Step 7: Find the angular frequency \( \omega \)
The angular frequency \( \omega \) for simple harmonic motion is given by:
\[ \omega = \sqrt{\frac{k}{m}} \]
From the buoyant force relationship, we can derive:
\[ k = \rho_{\text{water}} \cdot A \cdot g \]
Thus,
\[ \omega = \sqrt{\frac{\rho_{\text{water}} \cdot A \cdot g}{m}} \]

### Step 8: Calculate the period \( T \)
The period of oscillation \( T \) is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{m}{\rho_{\text{water}} \cdot A \cdot g}} \]

### Step 9: Substitute the values
Assuming the side length \( a \) of the cube is \( 0.3 \, \text{m} \):
\[ A = a^2 = (0.3)^2 = 0.09 \, \text{m}^2 \]
Substituting the values:
\[ T = 2\pi \sqrt{\frac{0.0135}{1000 \cdot 0.09 \cdot 9.81}} \]
Calculating this gives:
\[ T \approx 2.43 \, \text{s} \]

### Final Answer
The period of oscillation of the cubical wood piece is approximately \( T \approx 2.43 \, \text{s} \).

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