The `KE` and `PE` , at is a particle executing `SHM` with amplitude `A` will be equal when its displacement is

Options

A. `A sqrt(2)`
B. `(A)/(sqrt(2))`
C. ` (A)/(2)`
D. `A`

Answer

To find the displacement at which the kinetic energy (KE) and potential energy (PE) of a particle executing simple harmonic motion (SHM) are equal, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formulas**:
- The kinetic energy (KE) of a particle in SHM is given by:
\[ KE = \frac{1}{2} m \omega^2 (A^2 - x^2) \]
- The potential energy (PE) of a particle in SHM is given by:
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

2. **Set KE Equal to PE**:
- We want to find the displacement \( x \) where \( KE = PE \):
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2 \]

3. **Cancel Common Terms**:
- Since \( \frac{1}{2} m \omega^2 \) is common on both sides, we can cancel it out:
\[ A^2 - x^2 = x^2 \]

4. **Rearrange the Equation**:
- Rearranging gives:
\[ A^2 = 2x^2 \]

5. **Solve for \( x^2 \)**:
- Divide both sides by 2:
\[ x^2 = \frac{A^2}{2} \]

6. **Take the Square Root**:
- Taking the square root of both sides gives:
\[ x = \frac{A}{\sqrt{2}} \]

### Conclusion:
The displacement \( x \) at which the kinetic energy and potential energy are equal is:
\[ x = \frac{A}{\sqrt{2}} \]

### Final Answer:
The correct option is \( \frac{A}{\sqrt{2}} \).

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