Two particles executing SHM of the same amplitude and frequency on parallel lines side by side. They cross one another when moving in opposite directions each time their displacement is $\dfrac{\sqrt{3}}{2}$ times their amplitude. The phase difference between them is:

since $x=A\sin { \left( \omega t+\phi \right) }$

$\\ \dfrac { \sqrt{3}A }{ 2 } =A\sin { \left( \omega t+\phi \right) } \\ \Rightarrow \sin { \left( \omega t+\phi \right) } =\dfrac{\sqrt{3}}{2} \\ \Rightarrow \omega t+\phi =60^{ \circ }or\quad 120^{ \circ }$

so the one particle has phase of $60^{ \circ }$ and another has $120^{ \circ }$

So the phase difference is $120^{ \circ }-60^{ \circ }=60^{ \circ }$