A particle execute SHM and its position varies with time as `x = A sin omega t`. Its average speed during its motion from mean position to mid-point of mean and extreme position is

Options

A. zero
B. `(3 A omega)/(pi)`
C. `(A omega)/(2pi)`
D. `(2A omega)/(pi)`

Answer

To find the average speed of a particle executing simple harmonic motion (SHM) from the mean position to the midpoint of the mean and extreme position, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**:
The position of the particle in SHM is given by the equation:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude, and \( \omega \) is the angular frequency.

2. **Identify Key Positions**:
- The mean position is at \( x = 0 \).
- The extreme position is at \( x = A \).
- The midpoint between the mean position and the extreme position is at \( x = \frac{A}{2} \).

3. **Calculate Total Distance Travelled**:
The total distance travelled by the particle from the mean position to the midpoint is:
\[ \text{Distance} = \frac{A}{2} \]

4. **Determine the Time Taken to Reach Midpoint**:
To find the time taken to reach \( x = \frac{A}{2} \), we set up the equation:
\[ \frac{A}{2} = A \sin(\omega t) \]
Dividing both sides by \( A \) gives:
\[ \frac{1}{2} = \sin(\omega t) \]
The angle whose sine is \( \frac{1}{2} \) is \( \frac{\pi}{6} \) radians. Thus:
\[ \omega t = \frac{\pi}{6} \]
Therefore, the time \( t \) taken to reach the midpoint is:
\[ t = \frac{\pi}{6\omega} \]

5. **Calculate Average Speed**:
The average speed \( v_{\text{avg}} \) is given by the formula:
\[ v_{\text{avg}} = \frac{\text{Total Distance}}{\text{Total Time}} \]
Substituting the values we found:
\[ v_{\text{avg}} = \frac{\frac{A}{2}}{\frac{\pi}{6\omega}} = \frac{A}{2} \cdot \frac{6\omega}{\pi} = \frac{3A\omega}{\pi} \]

### Final Answer:
Thus, the average speed of the particle during its motion from the mean position to the midpoint of the mean and extreme position is:
\[ v_{\text{avg}} = \frac{3A\omega}{\pi} \]