A particle is executing simple harmonic motion (SHM) of amplitude $$A$$, along the $$x$$-axis, about $$x = 0$$. When its potential Energy (PE) equal kinetic energy (KE), the position of the particle will be:

We are told that the particle performs simple harmonic motion (SHM) about the origin, i.e. about $$x = 0$$, with amplitude $$A$$. In SHM the total mechanical energy remains constant and is the sum of the kinetic energy (KE) and the potential energy (PE).

First, we recall the standard results for SHM governed by the restoring force $$F = -kx$$, where $$k$$ is the force (spring) constant:

1. Total energy (always constant): $$E_{\text{total}} = \dfrac12\,kA^2.$$

2. Instantaneous potential energy at displacement $$x$$: $$\text{PE} = \dfrac12\,k x^2.$$

3. Instantaneous kinetic energy at displacement $$x$$: $$\text{KE} = E_{\text{total}} - \text{PE} = \dfrac12\,kA^2 - \dfrac12\,k x^2.$$

The problem asks for the position(s) where the potential energy equals the kinetic energy. Hence we set

$$\text{PE} = \text{KE}.$$

Writing this out explicitly with the above expressions, we have

$$\dfrac12\,k x^2 = \dfrac12\,kA^2 - \dfrac12\,k x^2.$$

Because every term carries the common factor $$\dfrac12\,k$$, we can safely divide both sides of the equation by this factor. Doing so simplifies the equation to

$$x^2 = A^2 - x^2.$$

Now we collect like terms. Adding $$x^2$$ to both sides gives

$$x^2 + x^2 = A^2,$$

or equivalently

$$2x^2 = A^2.$$

To solve for $$x^2$$ we divide by 2:

$$x^2 = \dfrac{A^2}{2}.$$

Taking the square root of both sides, we obtain the magnitude of the displacement:

$$x = \dfrac{A}{\sqrt{2}}.$$

Because the question merely asks for “the position of the particle” when $$\text{PE}=\text{KE}$$ (and provides only positive magnitudes in the options), we choose the positive root. Therefore, the required position from the mean position is $$\dfrac{A}{\sqrt{2}}.$$

Hence, the correct answer is Option D.