a
Displacement form mean position
$=\mathrm{A}-\left(\frac{2-\sqrt{3}}{2}\right) \mathrm{A}=\frac{\sqrt{3}}{2} \mathrm{A}$
So phase of the particle which is going
towards $+x$ is $\phi_{1}=\frac{\pi}{3}$
and for other particle $\phi_{2}=2 \pi-\frac{\pi}{3}=\frac{5 \pi}{3}$
phase difference $=\frac{5 \pi}{3}-\frac{\pi}{3}$
$=\frac{4 \pi}{3}$ or $\left(2 \pi-\frac{4 \pi}{3}\right)$
$=\frac{4 \pi}{3}$ or $\frac{2 \pi}{3}$