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. `(4A)/(T)`
D. `(4sqrt2A)/(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 \( \frac{T}{4} \), we can follow these steps:

### Step 1: Understand the Definition of Average Velocity
The average velocity \( V_{\text{avg}} \) is defined as the total distance traveled divided by the total time taken. Mathematically, it can be expressed as:
\[ V_{\text{avg}} = \frac{d}{t} \]
where \( d \) is the distance traveled and \( t \) is the time.

### Step 2: Identify the Time Interval
In this case, the time interval given is \( t = \frac{T}{4} \).

### Step 3: Determine the Maximum Distance Traveled
To find the maximum average velocity, we need to maximize the distance \( d \) traveled in the time interval \( \frac{T}{4} \).

In SHM, the particle moves in a circular path, and after a time \( \frac{T}{4} \), it covers an angle of \( \frac{\pi}{2} \) radians (90 degrees).

### Step 4: Calculate the Distance Traveled
The maximum distance traveled in this time can be thought of as the vertical distance covered in the circular motion. The maximum vertical distance (which corresponds to the amplitude) can be expressed as:
\[ d = 2A \sin(\theta) \]
where \( \theta \) is the angle covered.

For \( \theta = \frac{\pi}{4} \):
\[ d = 2A \sin\left(\frac{\pi}{4}\right) = 2A \cdot \frac{1}{\sqrt{2}} = \sqrt{2}A \]

### Step 5: Substitute into the Average Velocity Formula
Now, substituting the maximum distance \( d \) into the average velocity formula:
\[ V_{\text{avg}} = \frac{d}{t} = \frac{\sqrt{2}A}{\frac{T}{4}} = \frac{4\sqrt{2}A}{T} \]

### Conclusion
Thus, the maximum possible average velocity in time \( \frac{T}{4} \) is:
\[ \boxed{\frac{4\sqrt{2}A}{T}} \]

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