Two particles A and B execute simple harmonic motion according to the equations y1 = 3 sin ωt and y2 = 4sin(ωt+π2)+3sin ωt. The phase difference between them is

Representing a quantity with phase (ωt) along X-axis any quantity with a phase (ωt+ϕ) can be represented by a vector making an angle ϕ with X-axis in the anticlockwise sense and any quantity with a phase (ωt-ϕ) can be represented by a vector making an angle ϕ with X-axis in the clockwise sense.Phase difference ϕ = tan-1(43)Alternatively, given y1 = 3 sin ωt-----(i)And y2 = 4 sin(ωt+π2)+3 sin ωt= 4 cos ωt+3 sin ωt = 5[45cos ωt+35sin ωt]=5[sin ϕ cos ωt+cosϕ sinωt]= 5 sin(ωt+ϕ)-----------------(ii)cos ϕ = 35; sin ϕ = 45 and tan ϕ = 43From (i) and (ii), the result follows.