The magnitude of average accleration in half time period in a simple harmonic motion is

Options

A. `(2Aomega)/(pi)`
B. `(Aomega^(2))/(2pi)`
C. `(Aomega^(2))/(sqrt((2pi))`
D. Zero

Answer

To find the magnitude of the average acceleration in half a time period of simple harmonic motion (SHM), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Average Acceleration**:
Average acceleration is defined as the change in velocity divided by the change in time. Mathematically, it is given by:
\[ \text{Average Acceleration} = \frac{\Delta v}{\Delta t} \]

2. **Identifying the Time Interval**:
In SHM, the time period \( T \) is the time taken to complete one full oscillation. Therefore, half the time period is:
\[ \Delta t = \frac{T}{2} \]

3. **Relating Time Period to Angular Frequency**:
The time period \( T \) can be expressed in terms of angular frequency \( \omega \):
\[ T = \frac{2\pi}{\omega} \]
Thus, half the time period is:
\[ \Delta t = \frac{T}{2} = \frac{\pi}{\omega} \]

4. **Calculating Change in Velocity**:
The maximum velocity \( v_{\text{max}} \) in SHM is given by:
\[ v_{\text{max}} = A \omega \]
where \( A \) is the amplitude. The velocity at \( t = 0 \) is \( v(0) = 0 \) (since the object starts from rest at maximum displacement), and at \( t = \frac{T}{2} \), it reaches maximum velocity:
\[ v\left(\frac{T}{2}\right) = v_{\text{max}} = A \omega \]
Therefore, the change in velocity \( \Delta v \) over half a time period is:
\[ \Delta v = v\left(\frac{T}{2}\right) - v(0) = A \omega - 0 = A \omega \]

5. **Calculating Average Acceleration**:
Now substituting \( \Delta v \) and \( \Delta t \) into the average acceleration formula:
\[ \text{Average Acceleration} = \frac{\Delta v}{\Delta t} = \frac{A \omega}{\frac{\pi}{\omega}} = \frac{A \omega^2}{\pi} \]

6. **Final Expression**:
Thus, the magnitude of the average acceleration in half a time period in simple harmonic motion is:
\[ \text{Average Acceleration} = \frac{A \omega^2}{\pi} \]

### Conclusion:
The correct answer is option B, which corresponds to the derived expression for the average acceleration.