Maximum velocity ini SHM is `v_(m)`. The average velocity during motion from one extreme point to the other extreme point will be

Options

A. `(pi)/(2)v_(m)`
B. `(2/pi)v_(m)`
C. `(4/pi)v_(m)`
D. `(pi/4)v_(m)`

Answer

To find the average velocity during motion from one extreme point to the other extreme point in Simple Harmonic Motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Maximum Velocity in SHM**:
The maximum velocity \( v_m \) in SHM is given by the formula:
\[ v_m = \omega A \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

2. **Identify the Motion from One Extreme to Another**:
When the particle moves from one extreme position to the other, it covers a total displacement of \( 2A \) (from \( -A \) to \( +A \)).

3. **Determine the Time Taken**:
The time taken to move from one extreme to the other is half of the time period \( T \) of the motion:
\[ \text{Time taken} = \frac{T}{2} \]

4. **Calculate the Average Velocity**:
The average velocity \( v_{avg} \) can be calculated using the formula:
\[ v_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} \]
Substituting the values:
\[ v_{avg} = \frac{2A}{\frac{T}{2}} = \frac{2A \cdot 2}{T} = \frac{4A}{T} \]

5. **Relate Average Velocity to Maximum Velocity**:
We know that \( \omega = \frac{2\pi}{T} \). Therefore, we can express \( T \) in terms of \( \omega \):
\[ T = \frac{2\pi}{\omega} \]
Substituting this into the average velocity equation:
\[ v_{avg} = \frac{4A}{\frac{2\pi}{\omega}} = \frac{4A \cdot \omega}{2\pi} = \frac{2A \omega}{\pi} \]

6. **Substitute for Maximum Velocity**:
Now, substituting \( \omega A \) for \( v_m \):
\[ v_{avg} = \frac{2}{\pi} v_m \]

### Final Result:
Thus, the average velocity during motion from one extreme point to the other extreme point is:
\[ v_{avg} = \frac{2}{\pi} v_m \]