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

Options

A. `Asqrt2`
B. `(A)/(2)`
C. `(A)/(sqrt2)`
D. `Asqrt((2)/(3))`

Explanation

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. **Understanding the Energy Expressions**:
- The kinetic energy (KE) of a particle in SHM is given by:
\[ KE = \frac{1}{2} mv^2 \]
- The velocity \( v \) can be expressed in terms of displacement \( x \) as:
\[ v = \omega \sqrt{A^2 - x^2} \]
- Therefore, substituting for \( v \):
\[ KE = \frac{1}{2} m (\omega \sqrt{A^2 - x^2})^2 = \frac{1}{2} m \omega^2 (A^2 - x^2) \]

2. **Potential Energy Expression**:
- The potential energy (PE) in SHM is given by:
\[ PE = \frac{1}{2} kx^2 \]
- The spring constant \( k \) can be related to \( m \) and \( \omega \) as:
\[ k = m \omega^2 \]
- Thus, substituting for \( k \):
\[ PE = \frac{1}{2} m \omega^2 x^2 \]

3. **Setting KE Equal to PE**:
- According to the problem, we need to find the displacement \( x \) where \( KE = PE \):
\[ \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 x^2 \]
- We can cancel \( \frac{1}{2} m \omega^2 \) from both sides (assuming it is not zero):
\[ A^2 - x^2 = x^2 \]

4. **Solving for \( x^2 \)**:
- Rearranging the equation gives:
\[ A^2 = 2x^2 \]
- Dividing both sides by 2:
\[ x^2 = \frac{A^2}{2} \]

5. **Finding \( x \)**:
- Taking the square root of both sides:
\[ x = \frac{A}{\sqrt{2}} \]

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