Two particles start moving simultaneously from the same point and execute simple harmonic motion along the same line. Distances of the particle from their equilibrium position are ${\mathrm{x}}_1=4\cos 4\mathrm{\pi t}\mathrm{cm}\mathrm{and}{\mathrm{x}}_2=4\cos \left(4\mathrm{\pi t}+\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\mathrm{cm}$. Then maximum separation between the particles is 

Particles are oscillating with same frequency and the phase difference between them is $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$3$}\right.$. From the phasor diagram it is evident that separation between the particle M and N is maximum when their respective reference particles P and Q are symmetrically placed about the y-axis.

$\therefore$ Maximum separation $=\mathrm{OM}+\mathrm{ON}=\left(4\sin \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.+4\sin \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right.\right)\mathrm{cm}=4\mathrm{cm}$