A particle is performing linear SHM, along x-axis when the displacement becomes, `(A)/(sqrt2)`, (A is the amplitude), the relation that holds between kinetic energy and potential energy is ……..

Explanation

To solve the problem, we need to analyze the relationship between kinetic energy (KE) and potential energy (PE) of a particle performing simple harmonic motion (SHM) when its displacement is \( x = \frac{A}{\sqrt{2}} \), where \( A \) is the amplitude.

### Step-by-Step Solution:

1. **Understanding 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)
...