The periodic time of a particle executing a linear S.H.M. is `(2pi)/(omega)`. If its velocity at a dis"tan"ce b from the mean position is `sqrt(3)bomega`, then the path length of the particle is

Options

A. b
B. 2b
C. 3b
D. 4b

Explanation

To solve the problem, we need to find the path length of a particle executing linear Simple Harmonic Motion (S.H.M.) given its periodic time and velocity at a certain distance from the mean position.

### Step-by-Step Solution:

1. **Understand the given information**:
- The periodic time \( T \) is given by the formula:
\[ T = \frac{2\pi}{\omega} \]
- The velocity \( v \) at a distance \( b \) from the mean position is given as:
\[ v = \sqrt{3} b \omega \]

2. **Relate velocity in S.H.M.**:
- The velocity of a particle in S.H.M. at a distance \( x \) from the mean position can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( A \) is the amplitude of the motion.

3. **Substitute the given distance \( b \)**:
- Substitute \( x = b \) into the velocity equation:
\[ \sqrt{3} b \omega = \omega \sqrt{A^2 - b^2} \]

4. **Simplify the equation**:
- Cancel \( \omega \) from both sides (assuming \( \omega \neq 0 \)):
\[ \sqrt{3} b = \sqrt{A^2 - b^2} \]

5. **Square both sides**:
- Squaring both sides gives:
\[ 3b^2 = A^2 - b^2 \]

6. **Rearranging the equation**:
- Rearranging the equation results in:
\[ A^2 = 4b^2 \]
- Therefore, we find the amplitude \( A \):
\[ A = 2b \]

7. **Calculate the path length**:
- The path length (the distance traveled in one complete oscillation) is equal to the total distance covered in one complete cycle, which is:
\[ \text{Path Length} = 4A = 4 \times 2b = 8b \]

### Final Answer:
The path length of the particle is \( 8b \).

---