$x_1=A \sin \left(\omega t+\phi_1\right) \text {...1 }$
Equation of motion of particle 2,
$x_2=A \sin \left(\omega t+\phi_2\right) \text {...2 }$
From (1) and (2)
$x_2-x_1=A \sin \left(\omega t+\phi_2\right)-A \sin \left(\omega t+\phi_1\right)$
$=A\left[\sin \left(\omega t+\phi_2\right)-\sin \left(\omega t+\phi_1\right)\right]$
Using, $\sin C-\sin D=2 \cos \left(\frac{C+D}{2}\right) \cdot \sin \left(\frac{C-D}{2}\right)$
$x_2-x_1=2 A \cos \left(\omega t+\frac{\phi_1+\phi_2}{2}\right) \cdot \sin \left(\frac{\phi_2-\phi_1}{2}\right)$
Given that, $\left(x_0+x_2-x_1\right)_{\max }=x_0+A$
$\Rightarrow\left(x_2-x_1\right)_{\max }=A$
$\text { To get }\left(x_2-x_1\right)_{\max } \text {, we }$
$\Rightarrow 2 A \sin \left(\frac{\phi_2-\phi_1}{2}\right)=A$
$\Rightarrow \sin \left(\frac{\phi_2-\phi_1}{2}\right)=\frac{1}{2} $
$\Rightarrow \frac{\phi_2-\phi_1}{2}=\frac{\pi}{6} \text { or } \frac{5 \pi}{6}$
$\Rightarrow \phi_2-\phi_1=\frac{\pi}{3} \text { or } \frac{5 \pi}{3}$
$\text { To get }\left(x_2-x_1\right)_{\max } \text {, we assume } \cos \left(\omega t+\frac{\phi_1+\phi_2}{2}\right)=1$
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