Assertion : Average kinetic energy in one oscillation during SHM of a body is `(1)/(4)momega^(2)A^(2)`.
Reason : Maximum kinetic energy is `(1)/(2)momega^(2)A^(2)`.

Options

A. If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
B. If both Assertion and Reason are correct but Reason is not the correct explanation of Assertion.
C. If Assertion is true but Reason is false.
D. If Assertion is false but Reason is true.

Explanation

To solve the given question, we need to analyze both the assertion and the reason provided regarding the average kinetic energy and maximum kinetic energy in Simple Harmonic Motion (SHM).

### Step-by-step Solution:

1. **Understanding Maximum Kinetic Energy**:
- The maximum kinetic energy (KE_max) of a particle in SHM is given by the formula:
\[ KE_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2 \]
- The maximum velocity \( v_{\text{max}} \) of a particle in SHM is \( \omega A \), where \( \omega \) is the angular frequency and \( A \) is the amplitude.
- Substituting \( v_{\text{max}} \) into the kinetic energy formula:
\[ KE_{\text{max}} = \frac{1}{2} m (\omega A)^2 = \frac{1}{2} m \omega^2 A^2 \]
- This confirms that the reason given in the question is correct.

2. **Calculating Average Kinetic Energy**:
- The instantaneous kinetic energy (KE) at any point in time can be expressed as:
\[ KE = \frac{1}{2} m v^2 \]
- The instantaneous velocity \( v \) in SHM is given by:
\[ v = \frac{dx}{dt} = \omega A \cos(\omega t) \]
- Therefore, the instantaneous kinetic energy becomes:
\[ KE = \frac{1}{2} m (\omega A \cos(\omega t))^2 = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) \]

3. **Finding the Average Kinetic Energy**:
- To find the average kinetic energy over one complete cycle, we need to calculate the average value of \( KE \):
\[ \text{Average } KE = \frac{1}{T} \int_0^T KE \, dt \]
- Since \( KE = \frac{1}{2} m \omega^2 A^2 \cos^2(\omega t) \), we can factor out the constant:
\[ \text{Average } KE = \frac{1}{2} m \omega^2 A^2 \cdot \frac{1}{T} \int_0^T \cos^2(\omega t) \, dt \]
- The average value of \( \cos^2(\omega t) \) over one complete cycle is \( \frac{1}{2} \):
\[ \text{Average } KE = \frac{1}{2} m \omega^2 A^2 \cdot \frac{1}{2} = \frac{1}{4} m \omega^2 A^2 \]
- This confirms the assertion that the average kinetic energy in one oscillation during SHM is \( \frac{1}{4} m \omega^2 A^2 \).

4. **Conclusion**:
- Both the assertion and the reason are true:
- Assertion: Average kinetic energy is \( \frac{1}{4} m \omega^2 A^2 \) (True)
- Reason: Maximum kinetic energy is \( \frac{1}{2} m \omega^2 A^2 \) (True)
- However, the reason does not explain the assertion directly.

### Final Answer:
The correct answer is option 2: Both assertion and reason are correct, but the reason is not the correct explanation of the assertion.