A cuboidal piece of wood has dimensions `a, b` and `c`. its relatively density is `d`. it is floating in a large body of water such that side `a` is vertical. It is pushed down a bit and released. The time period of SHM executed by it is

Options

A. `2pisqrt(abc)/g`
B. `2pisqrt(g/(da)`
C. `2pisqrt((bc)/(dg))`
D. `2pisqrt((da)/g)`

Answer

To find the time period of the simple harmonic motion (SHM) executed by the cuboidal piece of wood floating in water, we can follow these steps:

### Step 1: Understand the Forces Acting on the Wood
When the wood is floating, it displaces a volume of water equal to its weight. The downward force (weight of the wood) is equal to the upward buoyant force (weight of the displaced water).

### Step 2: Calculate the Weight of the Wood
The weight of the wood can be expressed as:
\[ F_{\text{down}} = mg = \rho_{\text{wood}} \cdot V \cdot g = d \cdot (abc) \cdot g \]
where \( d \) is the relative density of the wood, \( V \) is the volume of the wood, and \( g \) is the acceleration due to gravity.

### Step 3: Calculate the Buoyant Force
The buoyant force acting on the wood is given by the weight of the displaced water:
\[ F_{\text{up}} = \rho_{\text{water}} \cdot V_{\text{displaced}} \cdot g = 1 \cdot (bcY) \cdot g \]
where \( Y \) is the depth of the submerged part of the wood.

### Step 4: Set Up the Equilibrium Condition
At equilibrium, the downward force equals the upward force:
\[ d \cdot (abc) \cdot g = (bcY) \cdot g \]
Cancelling \( g \) from both sides, we have:
\[ d \cdot abc = bcY \]
From this, we can solve for \( Y \):
\[ Y = \frac{d \cdot a}{1} \]

### Step 5: Determine the Amplitude of Motion
When the wood is pushed down and released, the maximum displacement from the mean position (equilibrium position) is:
\[ A = a - Y = a - d \cdot a = a(1 - d) \]

### Step 6: Calculate the Acceleration
The net force when the wood is displaced by a distance \( x \) from the mean position is given by:
\[ F_{\text{net}} = F_{\text{up}} - F_{\text{down}} \]
This can be expressed as:
\[ F_{\text{net}} = (bc \cdot (Y + x) \cdot g) - (d \cdot abc \cdot g) \]
Substituting \( Y \) into the equation gives:
\[ F_{\text{net}} = (bc \cdot (d \cdot a + x) \cdot g) - (d \cdot abc \cdot g) \]
This simplifies to:
\[ F_{\text{net}} = bcg \cdot x \]

### Step 7: Relate Force to Acceleration
Using Newton's second law, we can relate the net force to acceleration:
\[ F_{\text{net}} = m \cdot a \]
where \( m = d \cdot abc \). Thus:
\[ bcg \cdot x = d \cdot abc \cdot a \]

### Step 8: Find the Angular Frequency
The equation of motion for SHM can be written as:
\[ a = -\omega^2 x \]
From the above, we can derive:
\[ \omega^2 = \frac{bcg}{d \cdot abc} \]
Thus:
\[ \omega = \sqrt{\frac{g}{a(1 - d)}} \]

### Step 9: Calculate the Time Period
The time period \( T \) of SHM is given by:
\[ T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{a(1 - d)}{g}} \]

### Final Answer
The time period of SHM executed by the cuboidal piece of wood is:
\[ T = 2\pi \sqrt{\frac{a(1 - d)}{g}} \]