Representation of traveling harmonic waves using PHASORS f f y horizontal rotating PHASOR A(θ) (a rotating segment of length A) as shown in the figure. f(x,t) The rotating segment of length A, together with its associated angle θ constitute a PHASOR f1 f2 Horizontal component of f1(x,t) Horizontal component of f2(x,t) f1(x,t) +f2(x,t) Horizontal component of A1(θ1) A1(θ1) A2(θ2) A2(θ2) In different applications, it is easier to add up the phasors. Once the total sum (the phasor sum) is found, its horizontal component will be f1 + f2 . y1 y2 f1 f2 A1(θ1) A2(θ2) f1 f2 Work in the phasors world A1(θ1) A2(θ2) this sum will have the form A(θ) f1 f2 A(θ) CASE: Adding two harmonic waves of the same frequency and the same wavelength = 2/k y1 = A1Cos(kx – t +1) = A1Cos(1) y2 = A2Cos(kx – t +2) = A2Cos(2) This implies constructing the corresponding phasors Phasor-1: A1(1) where 1= kx – t +1 Phasor-2: A2(2) where 2= kx – t +2 A2 A A1 1 2 k Reference axis kx -t It is clear, from the graph, that all the phasors rotate synchronously. Notice, the graph above reduces to A A1 A2 2 1 Reference axis Accordingly, the problem of adding the phasors A1(1) where 1= kx – t +1 and A2(2) where 2= kx – t +2 reduces to the addition of the simpler phasors A1(1) and A2(2) Such sum will have the form A() where A and to be determined in terms of A1, A2 and 2. 1, Finding A and From the previous figure (re-drawn below) A A2 A2 A 2 A1 1 Reference axis Similarly, also from the figure α A A1 1 A2 2 Reference axis f 1(x,t) A1=4 mm, and α1 = 0 f 2(x,t) A2=3 mm, and α2 = π/3 f 1(x,t) + f 2(x,t) f 1(x,t) f 2(x,t) We build a corresponding phasor A1(0) (see graph.) We build a corresponding phasor A2(π/3) (see graph.) A2(π/3) A1(0) We have found, A1(0) + A2(/3) = A() where A = 6.1 mm and = 0.44 rad A A1 2 A2 2 Reference axis 1 It means A1(kx -wt + 0) + A2(kx -wt + /3) = A(kx -wt + ) A A1 2 A2 2 kx -t Reference axis Going back to the real world, f1(x,t) + f2(x,t) = Horizontal component of{ A(kx -wt + )} = 6.1 mm Cos(kx -wt + 0.44 rad)