Displacement between maximum potential energy position and maximum kinetic energy position for a particle executing `S.H.M` is

Options

A. `+- (a)/(2)`
B. `+-a`
C. `+- 2a`
D. `+-1`

Answer

To solve the problem of finding the displacement between the maximum potential energy position and the maximum kinetic energy position for a particle executing Simple Harmonic Motion (S.H.M), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding S.H.M**:
- In Simple Harmonic Motion, a particle oscillates about a mean position. The motion can be visualized as moving between two extreme positions (maximum displacement) on either side of the mean position.

2. **Identifying Key Positions**:
- The **mean position** is where the particle has maximum kinetic energy (KE) and zero potential energy (PE).
- The **extreme positions** (maximum displacement) are where the particle has maximum potential energy and zero kinetic energy. These positions are at a distance equal to the amplitude (A) from the mean position.

3. **Defining Positions**:
- Let’s denote the mean position as 0.
- The right extreme position (maximum potential energy) can be denoted as +A.
- The left extreme position (also maximum potential energy) can be denoted as -A.

4. **Calculating Displacement**:
- The displacement between the maximum potential energy position (which is at +A) and the maximum kinetic energy position (which is at 0) is calculated as:
\[ \text{Displacement} = \text{Position of max PE} - \text{Position of max KE} = (+A) - (0) = +A \]
- Similarly, if we consider the left extreme position (-A), the displacement would be:
\[ \text{Displacement} = (-A) - (0) = -A \]

5. **Conclusion**:
- Therefore, the displacement between the maximum potential energy position and the maximum kinetic energy position is given as:
\[ \text{Displacement} = \pm A \]

### Final Answer:
The displacement between the maximum potential energy position and the maximum kinetic energy position for a particle executing S.H.M is \( \pm A \).

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