Two bodies performing $SHM$ have same amplitude and frequency. Their phases at a certain instant are as shown in the figure. The phase difference between them is
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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 $\frac{2 \pi}{3}$

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