A particle starts performing SHM on a smooth horizontal plane and it is released from `x = A/2` and it is moving in -ve x-direction then `phi =?`

Options

A. `pi/6`
B. `5pi/6`
C. `2pi/3`
D. `pi/3`

Answer

To solve the problem, we need to determine the phase angle \( \phi \) of a particle performing simple harmonic motion (SHM) given that it is released from \( x = \frac{A}{2} \) and is moving in the negative x-direction.

### Step-by-Step Solution:

1. **Understanding the SHM Equation**:
The displacement \( x \) of a particle in SHM can be expressed as:
\[ x = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( t \) is the time,
- \( \phi \) is the phase angle.

2. **Initial Conditions**:
At the moment of release, we set \( t = 0 \) and \( x = \frac{A}{2} \). Substituting these values into the SHM equation gives:
\[ \frac{A}{2} = A \sin(\phi) \]

3. **Simplifying the Equation**:
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\phi) \]

4. **Finding Possible Values of \( \phi \)**:
The values of \( \phi \) that satisfy \( \sin(\phi) = \frac{1}{2} \) are:
\[ \phi = \frac{\pi}{6} \quad \text{and} \quad \phi = \frac{5\pi}{6} \]

5. **Determining the Direction of Motion**:
We know that the particle is moving in the negative x-direction. To find out which \( \phi \) corresponds to this direction, we need to analyze the velocity.

6. **Velocity in SHM**:
The velocity \( v \) of the particle is given by the derivative of the displacement:
\[ v = \frac{dx}{dt} = A \omega \cos(\omega t + \phi) \]
At \( t = 0 \):
\[ v = A \omega \cos(\phi) \]
Since the particle is moving in the negative x-direction, we require:
\[ v < 0 \quad \Rightarrow \quad \cos(\phi) < 0 \]

7. **Analyzing the Cosine Values**:
- For \( \phi = \frac{\pi}{6} \), \( \cos\left(\frac{\pi}{6}\right) > 0 \) (positive).
- For \( \phi = \frac{5\pi}{6} \), \( \cos\left(\frac{5\pi}{6}\right) < 0 \) (negative).

8. **Conclusion**:
Since we need \( \cos(\phi) < 0 \) for the velocity to be negative, we conclude that:
\[ \phi = \frac{5\pi}{6} \]

### Final Answer:
\[ \phi = \frac{5\pi}{6} \]