A particle executes `SHM` with time period `T` and amplitude `A`. The maximum possible average velocity in time `T//4` is-

Options

A. `(2A)/(T)`
B. `(4A)/(T)`
C. `(8A)/(T)`
D. `(4sqrt(2)A)/(T)`

Explanation

To solve the problem of finding the maximum possible average velocity of a particle executing Simple Harmonic Motion (SHM) over a time period of \( T/4 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**:
- A particle in SHM can be modeled as moving in a circular motion, where its projection on a diameter represents the SHM. The amplitude \( A \) is the maximum displacement from the mean position.

2. **Identify the Time Period**:
- The time period of the SHM is given as \( T \). Thus, the time for \( T/4 \) is \( T/4 \).

3. **Maximum Velocity in SHM**:
- The maximum velocity \( V_{\text{max}} \) of a particle in SHM is given by the formula:
\[ V_{\text{max}} = \omega A \]
- Here, \( \omega \) (angular frequency) is related to the time period by:
\[ \omega = \frac{2\pi}{T} \]
- Therefore, substituting for \( \omega \):
\[ V_{\text{max}} = \frac{2\pi A}{T} \]

4. **Calculate the Average Velocity**:
- The average velocity \( V_{\text{avg}} \) over a time interval is defined as:
\[ V_{\text{avg}} = \frac{\text{Total Displacement}}{\text{Total Time}} \]
- In the first \( T/4 \) of the motion, the particle moves from the mean position (0) to the maximum displacement \( A \) and then back to the mean position. The maximum displacement during this time can be calculated.

5. **Displacement in \( T/4 \)**:
- During \( T/4 \), the particle will cover a distance corresponding to \( \frac{\pi}{2} \) radians in circular motion, which translates to a displacement of:
\[ \text{Displacement} = A \sin\left(\frac{\pi}{2}\right) = A \]
- However, the average velocity is calculated as the total distance traveled divided by the time taken. The particle moves from 0 to \( A \) and back to 0, thus covering a distance of \( A \) in \( T/4 \).

6. **Final Calculation of Average Velocity**:
- The average velocity over the time \( T/4 \) can be expressed as:
\[ V_{\text{avg}} = \frac{A}{T/4} = \frac{4A}{T} \]

7. **Maximum Possible Average Velocity**:
- To find the maximum possible average velocity, we consider the maximum displacement covered in \( T/4 \), which is:
\[ V_{\text{avg}} = \frac{2A \sin(\frac{\pi}{4})}{T/4} = \frac{2A \cdot \frac{1}{\sqrt{2}}}{T/4} = \frac{4A}{T\sqrt{2}} = \frac{4\sqrt{2}A}{T} \]

### Conclusion:
The maximum possible average velocity in time \( T/4 \) is:
\[ \frac{4\sqrt{2}A}{T} \]