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

Answer

To solve the problem, we need to find the value of \( x \) in the expression for the mean velocity of a particle performing simple harmonic motion (SHM). The mean velocity is given as \( \frac{xa}{2T} \), where \( T \) is the period of the motion and \( a \) is the amplitude.

### Step-by-Step Solution:

1. **Understanding the Motion**:
- The particle starts at the extreme position (maximum displacement) at \( t = 0 \).
- The equation of motion for SHM can be expressed as:
\[ x(t) = a \cos(\omega t) \]
- Here, \( \omega = \frac{2\pi}{T} \).

2. **Determine the Displacement**:
- The particle travels a distance of \( \frac{a}{2} \) from the extreme position.
- If the particle starts at \( a \) (extreme position), after traveling \( \frac{a}{2} \), its new position will be:
\[ x = a - \frac{a}{2} = \frac{a}{2} \]

3. **Finding the Time Taken**:
- Set the displacement equal to \( \frac{a}{2} \):
\[ \frac{a}{2} = a \cos(\omega t) \]
- Dividing both sides by \( a \):
\[ \frac{1}{2} = \cos(\omega t) \]
- From the cosine function, we know:
\[ \omega t = \frac{\pi}{3} \quad \text{(since } \cos(\frac{\pi}{3}) = \frac{1}{2}\text{)} \]
- Therefore, the time \( t \) can be expressed as:
\[ t = \frac{\pi}{3\omega} \]

4. **Calculating the Mean Velocity**:
- The mean velocity \( v_{avg} \) is defined as the total distance traveled divided by the total time taken:
\[ v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{\frac{a}{2}}{t} \]
- Substituting \( t \):
\[ v_{avg} = \frac{\frac{a}{2}}{\frac{\pi}{3\omega}} = \frac{a}{2} \cdot \frac{3\omega}{\pi} = \frac{3a\omega}{2\pi} \]

5. **Relating \( \omega \) to the Period \( T \)**:
- Recall that \( \omega = \frac{2\pi}{T} \):
\[ v_{avg} = \frac{3a \cdot \frac{2\pi}{T}}{2\pi} = \frac{3a}{T} \]

6. **Equating to the Given Mean Velocity**:
- We are given that the mean velocity is also expressed as:
\[ v_{avg} = \frac{xa}{2T} \]
- Setting the two expressions for mean velocity equal to each other:
\[ \frac{3a}{T} = \frac{xa}{2T} \]
- Canceling \( a \) and \( T \) (assuming \( a \neq 0 \) and \( T \neq 0 \)):
\[ 3 = \frac{x}{2} \]
- Solving for \( x \):
\[ x = 6 \]

### Final Answer:
The value of \( x \) is \( 6 \).