Velocity at mean position of a particle executing SHM is v. Velocity of the particle at a distance equal to half of the amplitude will be

Options

A. `(v)/(2)`
B. `(v)/(sqrt(2))`
C. `(sqrt(3))/(2)v`
D. `(sqrt(3))/(4)` v

Explanation

To solve the problem, we need to find the velocity of a particle executing Simple Harmonic Motion (SHM) when it is at a distance equal to half of the amplitude (x = A/2), given that the velocity at the mean position (x = 0) is v.

### Step-by-Step Solution:

1. **Understand the Formula for Velocity in SHM**:
The velocity \( v \) of a particle in SHM at a position \( x \) is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

2. **Velocity at the Mean Position**:
At the mean position, \( x = 0 \):
\[ v = \omega \sqrt{A^2 - 0^2} = \omega A \]
We are given that this velocity is \( v \):
\[ v = \omega A \]

3. **Calculate Velocity at \( x = A/2 \)**:
Now, we need to find the velocity when \( x = A/2 \):
\[ v_{A/2} = \omega \sqrt{A^2 - (A/2)^2} \]

4. **Simplify the Expression**:
Substitute \( (A/2)^2 \):
\[ v_{A/2} = \omega \sqrt{A^2 - \frac{A^2}{4}} = \omega \sqrt{\frac{4A^2}{4} - \frac{A^2}{4}} = \omega \sqrt{\frac{3A^2}{4}} \]
This simplifies to:
\[ v_{A/2} = \omega \cdot \frac{\sqrt{3}A}{2} \]

5. **Relate to the Given Velocity \( v \)**:
We know from step 2 that \( \omega A = v \). Thus, we can substitute \( \omega A \) with \( v \):
\[ v_{A/2} = \frac{\sqrt{3}}{2} \cdot v \]

6. **Conclusion**:
Therefore, the velocity of the particle at a distance equal to half of the amplitude is:
\[ v_{A/2} = \frac{\sqrt{3}}{2} v \]

### Final Answer:
The velocity of the particle at a distance equal to half of the amplitude is \( \frac{\sqrt{3}}{2} v \).