particle is executing SHM of amplitude A and angular frequency omega.The average acceleration of particle for half the time period is : (Starting from mean position)

Options

A. (a)`(2Aomega^(2))/(pi)`
B. (b)`(Aomega^(2))/(pi)`
C. (c)`(3Aomega^(2))/(2pi)`
D. (d)`(Aomega^(2))/(2)`

Explanation

To solve the problem of finding the average acceleration of a particle executing Simple Harmonic Motion (SHM) for half the time period, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding SHM**:
The particle is executing SHM with amplitude \( A \) and angular frequency \( \omega \). The motion starts from the mean position (which we can denote as \( x = 0 \)).

2. **Position Function**:
The position of the particle as a function of time \( t \) can be expressed as:
\[ x(t) = A \sin(\omega t) \]

3. **Velocity Function**:
The velocity \( v(t) \) is the derivative of the position function:
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t) \]

4. **Acceleration Function**:
The acceleration \( a(t) \) is the derivative of the velocity function:
\[ a(t) = \frac{dv}{dt} = -A \omega^2 \sin(\omega t) \]

5. **Average Acceleration Calculation**:
The average acceleration \( \bar{a} \) over a time interval \( T \) is given by:
\[ \bar{a} = \frac{1}{T} \int_0^T a(t) \, dt \]
For half the time period \( T = \frac{T}{2} = \frac{\pi}{\omega} \).

6. **Setting Up the Integral**:
We need to calculate the integral from \( 0 \) to \( \frac{\pi}{\omega} \):
\[ \bar{a} = \frac{1}{\frac{\pi}{\omega}} \int_0^{\frac{\pi}{\omega}} -A \omega^2 \sin(\omega t) \, dt \]

7. **Evaluating the Integral**:
The integral of \( \sin(\omega t) \) is:
\[ \int \sin(\omega t) \, dt = -\frac{1}{\omega} \cos(\omega t) \]
Therefore, we evaluate:
\[ \int_0^{\frac{\pi}{\omega}} \sin(\omega t) \, dt = -\frac{1}{\omega} \left[ \cos(\omega t) \right]_0^{\frac{\pi}{\omega}} = -\frac{1}{\omega} \left( \cos(\pi) - \cos(0) \right) = -\frac{1}{\omega} \left( -1 - 1 \right) = \frac{2}{\omega} \]

8. **Substituting Back**:
Now substitute back into the average acceleration formula:
\[ \bar{a} = \frac{\omega}{\pi} \left( -A \omega^2 \cdot \frac{2}{\omega} \right) = -\frac{2A \omega^2}{\pi} \]

9. **Final Result**:
The average acceleration of the particle for half the time period is:
\[ \bar{a} = -\frac{2A \omega^2}{\pi} \]