Two particles execute SHM of same amplitude of 20 cm with same period along thesame line about the same equilibrium position.If the phase differnce between them is pi/3 find the maximum distance between them

Find:Maximum distances (R)Data:A1=A2=A=20 cm.Phase difference (d)= π÷3=60°.Formula:R=√A1^2+A2^2+A1A2cosd. ....^ means power.Solution:From formula,.R=√20^2+20^2+2×20×20×cos60°.R=√400+400+800÷0.5.R=√400+400+400.R=√3(400).R=20√3.R=20(1.73).R=34.6 cm.