Two particles are executing simple harmonic motion of same amplitude motion of same amplitude A and frequency along the x-axis. Their mean position is separated by distance${{\text{X}}_{0}}\left( {{\text{X}}_{0}}>\text{A} \right)$ . If the maximum separation between them is $\left( {{\text{X}}_{0}}+\text{A} \right)$ , the phase difference between their motion is:
A.$\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$
B.$\dfrac{\text{ }\!\!\pi\!\!\text{ }}{3}$
C.$\dfrac{\text{ }\!\!\pi\!\!\text{ }}{4}$
D.$\dfrac{\text{ }\!\!\pi\!\!\text{ }}{6}$

Hint: If the separation between the particles is maximum this means that if one particle is at mean position (phase 0), the other particle is at extreme position (phase$\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$). So phase difference is calculated by subtracting the phases of two particles.

Complete answer:
It is given that the maximum separation between two oscillating particles in simple harmonic motion is$\left( {{\text{X}}_{0}}+\text{A} \right)$
It is the maximum distance which means that if one particle is at a mean position, then the other will be at an extreme position and vice versa. So, if the phase of one particle is 0, then that of another particle is $\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$.
Hence, the phase difference is given as$\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}-0=\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}$

So the correct option is (A).

Note:
Simple harmonic motion is a particle motion where restoring force on the moving object is directly proportional to the object’s displacement magnitude and acts towards the object’s equilibrium position.
When the body is at mean position, its displacement x=0. So, f minimum =0. Acceleration is minimum (zero).
When the body is at extreme position, the magnitude of its displacement x=0 is $\text{f}={{\text{w}}^{2}}\text{a}$ . Thus at extremes position, the magnitude of acceleration is maximum.