The value of phase at maximum distance from the mean position of a particle in S.H.M. is :

Options

A. `pi//2`
B. `pi`
C. Zero
D. `2pi`

Answer

To solve the question regarding the value of phase at maximum distance from the mean position of a particle in Simple Harmonic Motion (SHM), we will follow these steps:

### Step-by-Step Solution:

1. **Understanding the SHM Equation**:
The general equation for the displacement \( X \) of a particle in SHM is given by:
\[ X = A \sin(\omega t + \phi) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, \( t \) is time, and \( \phi \) is the phase constant.

2. **Identifying Maximum Distance**:
The maximum distance from the mean position in SHM is equal to the amplitude \( A \). Therefore, at maximum displacement:
\[ X = A \]

3. **Setting Up the Equation**:
To find the phase when the displacement is at its maximum, we set \( X = A \):
\[ A = A \sin(\omega t + \phi) \]

4. **Simplifying the Equation**:
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ 1 = \sin(\omega t + \phi) \]

5. **Finding the Phase**:
The sine function reaches its maximum value of 1 at specific angles. The general solution for when \( \sin(\theta) = 1 \) is:
\[ \theta = \frac{\pi}{2} + 2n\pi \]
where \( n \) is any integer.

6. **Determining the Phase Value**:
Therefore, substituting back for \( \theta \):
\[ \omega t + \phi = \frac{\pi}{2} + 2n\pi \]
If we take \( n = 0 \) for the first instance, we have:
\[ \omega t + \phi = \frac{\pi}{2} \]
Thus, the phase at maximum displacement is:
\[ \phi = \frac{\pi}{2} - \omega t \]
For the maximum distance, we are primarily interested in the phase value itself, which is \( \frac{\pi}{2} \).

### Conclusion:
The value of the phase at maximum distance from the mean position of a particle in SHM is:
\[ \frac{\pi}{2} \]