d
Equation of motion of particle 1 ,

\(x_1=A \sin \left(\omega t+\phi_1\right) \text {...1 }\)

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, \(\left(x_0+x_2-x_1\right)_{\max }=x_0+A\)

\(\Rightarrow\left(x_2-x_1\right)_{\max }=A\)

\(\text { To get }\left(x_2-x_1\right)_{\max } \text {, we }\)

\(\Rightarrow 2 A \sin \left(\frac{\phi_2-\phi_1}{2}\right)=A\)

\(\Rightarrow \sin \left(\frac{\phi_2-\phi_1}{2}\right)=\frac{1}{2} \)

\(\Rightarrow \frac{\phi_2-\phi_1}{2}=\frac{\pi}{6} \text { or } \frac{5 \pi}{6}\)

\(\Rightarrow \phi_2-\phi_1=\frac{\pi}{3} \text { or } \frac{5 \pi}{3}\)

\(\text { To get }\left(x_2-x_1\right)_{\max } \text {, we assume } \cos \left(\omega t+\frac{\phi_1+\phi_2}{2}\right)=1\)