A particle executes simple harmonic motion along a straight line with an amplitude A . The potential energy is maximum when the displacement is

Options

A. `+-A`
B. zero
C. `+-A/2`
D. `+-A/sqrt2`

Answer

To solve the question, we need to determine the displacement of a particle executing simple harmonic motion (SHM) when its potential energy is at a maximum.

### Step-by-Step Solution:

1. **Understanding Simple Harmonic Motion (SHM)**:
- In SHM, a particle oscillates back and forth around an equilibrium position. The maximum displacement from this equilibrium position is called the amplitude (A).

2. **Energy in SHM**:
- In SHM, the total mechanical energy (E) is conserved and is the sum of kinetic energy (KE) and potential energy (PE).
- The relationship can be expressed as:
\[ E = KE + PE \]
- Kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]
- Potential energy in SHM is given by:
\[ PE = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the displacement from the mean position.

3. **Condition for Maximum Potential Energy**:
- Potential energy is maximum when kinetic energy is minimum. The minimum value of kinetic energy occurs when the particle momentarily stops, which means its velocity (v) is zero.

4. **Finding the Displacement**:
- The velocity is zero at the extreme positions of the motion, which are at \( x = +A \) and \( x = -A \).
- Therefore, the displacement at which the potential energy is maximum is when the particle is at the amplitude positions:
\[ x = +A \quad \text{or} \quad x = -A \]

5. **Conclusion**:
- The potential energy is maximum when the displacement is either at the positive amplitude \( A \) or the negative amplitude \( -A \).

### Final Answer:
The potential energy is maximum when the displacement is \( x = +A \) or \( x = -A \).

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