A particle performing SHM starts from mean position. The phase of that particle is `pi//s` when it has

Options

A. maximum displacement
B. maximum velocity
C. maximum energy
D. maximum kinetic energy

Explanation

To solve the problem, we need to determine the conditions under which a particle performing Simple Harmonic Motion (SHM) has a phase of \( \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Understanding SHM**:
The equation for the displacement \( y \) of a particle in SHM can be expressed as:
\[ y = A \sin(\omega t) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( t \) is the time.

2. **Finding the Phase**:
The phase of the particle at any time \( t \) is given by \( \omega t \). We want to find out when this phase equals \( \frac{\pi}{2} \):
\[ \omega t = \frac{\pi}{2} \]

3. **Solving for Time**:
Rearranging the equation gives:
\[ t = \frac{\frac{\pi}{2}}{\omega} = \frac{\pi}{2\omega} \]

4. **Velocity in SHM**:
The velocity \( v \) of the particle in SHM is given by the derivative of the displacement:
\[ v = \frac{dy}{dt} = A \omega \cos(\omega t) \]
At \( \omega t = \frac{\pi}{2} \), we have:
\[ v = A \omega \cos\left(\frac{\pi}{2}\right) = A \omega \cdot 0 = 0 \]
This indicates that the particle is at the maximum displacement when the phase is \( \frac{\pi}{2} \).

5. **Conclusion**:
Therefore, the particle will have a phase of \( \frac{\pi}{2} \) when it is at its maximum displacement, which corresponds to the mean position of the motion.

### Final Answer:
The phase of the particle is \( \frac{\pi}{2} \) when it is at its maximum displacement.

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