Average velocity of a particle executing SHM in one complete vibration is :

Options

A. `(omega^(2)A)/(2)`
B. `(omega^(2)A)/(sqrt(2))`
C. zero
D. `Aomega^(2)`

Explanation

To find the average velocity of a particle executing simple harmonic motion (SHM) in one complete vibration, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Motion**:
- A particle in SHM moves back and forth around an equilibrium position. The displacement can be described by the equation:
\[ x(t) = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Find the Velocity**:
- The velocity \( v(t) \) of the particle is the derivative of displacement \( x(t) \) with respect to time \( t \):
\[ v(t) = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]

3. **Average Velocity Over One Complete Cycle**:
- To find the average velocity over one complete cycle, we need to integrate the velocity over one period \( T \) and then divide by the period:
\[ \text{Average Velocity} = \frac{1}{T} \int_0^T v(t) \, dt \]
- Since the velocity function \( v(t) = A \omega \cos(\omega t + \phi) \) oscillates between positive and negative values, we can analyze its behavior over one complete cycle.

4. **Evaluate the Integral**:
- The average value of the cosine function over one complete cycle is zero:
\[ \int_0^T \cos(\omega t + \phi) \, dt = 0 \]
- Therefore, the average velocity becomes:
\[ \text{Average Velocity} = \frac{1}{T} \cdot 0 = 0 \]

5. **Conclusion**:
- The average velocity of a particle executing SHM in one complete vibration is:
\[ \text{Average Velocity} = 0 \]

### Final Answer:
The average velocity of a particle executing SHM in one complete vibration is **0**.

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