A particle performs SHM with a period `T` and amplitude a. The mean velocity of particle over the time interval during which it travels `a//2` from the extreme position is

Options

A. `6a//T`
B. `2 a//T`
C. `3 a//T`
D. `a//2T`

Explanation

To solve the problem of finding the mean velocity of a particle performing Simple Harmonic Motion (SHM) as it travels a distance of \( \frac{a}{2} \) from the extreme position, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle is performing SHM with an amplitude \( a \) and a period \( T \). The maximum displacement from the mean position is \( a \).

2. **Determine the Initial and Final Positions**:
- The particle starts from the extreme position at \( x = a \) and travels to \( x = \frac{a}{2} \).

3. **Calculate the Distance Traveled**:
- The distance traveled by the particle from the extreme position to the point \( \frac{a}{2} \) is:
\[ \text{Distance} = a - \frac{a}{2} = \frac{a}{2} \]

4. **Finding the Time Taken**:
- The displacement of the particle can be described by the equation:
\[ x(t) = a \cos(\omega t) \]
- Here, \( \omega = \frac{2\pi}{T} \).
- To find the time taken to reach \( x = \frac{a}{2} \):
\[ \frac{a}{2} = a \cos(\omega t) \]
- Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ \frac{1}{2} = \cos(\omega t) \]
- The angle \( \omega t \) for which \( \cos(\omega t) = \frac{1}{2} \) is:
\[ \omega t = \frac{\pi}{3} \]
- Therefore, the time \( t \) taken is:
\[ t = \frac{\pi/3}{\omega} = \frac{\pi/3}{\frac{2\pi}{T}} = \frac{T}{6} \]

5. **Calculating the Mean Velocity**:
- The mean velocity \( v_{mean} \) is defined as the total distance traveled divided by the total time taken:
\[ v_{mean} = \frac{\text{Distance}}{\text{Time}} = \frac{\frac{a}{2}}{\frac{T}{6}} = \frac{a}{2} \cdot \frac{6}{T} = \frac{3a}{T} \]

### Final Answer:
The mean velocity of the particle over the time interval during which it travels \( \frac{a}{2} \) from the extreme position is:
\[ v_{mean} = \frac{3a}{T} \]