Wave Interference and superposition

[Destrio]

Consider two point sources S1 and S2 which emit waves of the same frequency and amplitude A. The waves start in the same phase, and this phase relation at the sources is maintained throughout time. Consider point P at which r1 is nearly equal to r2.

a) Show that the superposition of these two waves gives a wave whole amplitude Y varies with the position P approximately according to:

Y = (2A/r)cos(k/2)(r1-r2)

in which r = (r1+r2)/2.

b) Then show that total cancellation occurs when (r1-r2)=(n+.5)λ and total reinforcement occurs when r1-r2 = nλ


so initially we have
W1 = Asin(kx-wt-r1)

W2 = Asin(kx-wt-r2)

and we can use sinB + sinC = 2sin(.5)(B+C)cos(.5)(B-C)

to make them look like: [2Acos((r2-r1)/2)]sin(kx-wt-(r1+r2)/2)

but i believe that only works if they are always in phase

any help would but much appreciated, I've been stuck on this for a long time

thanks

 

[member 553083]

!Answer: a) The superposition of the two waves is given by adding the two amplitudes together: Y = A*sin(kx-wt-r1) + A*sin(kx-wt-r2)= A*[sin(kx-wt-r1) + sin(kx-wt-r2)]Using the formula sinB+sinC = 2sin(.5)(B+C)cos(.5)(B-C), we can rewrite this as Y = 2A*sin(.5)((kx-wt-r1) + (kx-wt-r2))cos(.5)((kx-wt-r1) - (kx-wt-r2))Since we are considering a point P at which r1 is nearly equal to r2, we can substitute r1 and r2 with the average distance from the sources, which is equal to r = (r1+r2)/2: Y = 2A*sin(.5)(2(kx-wt-r))cos(.5)(2(kx-wt-r))Simplifying, we get Y = (2A/r)cos(k/2)(r1-r2)b) For total cancellation to occur, (r1-r2)=(n+.5)λ, and for total reinforcement to occur, r1-r2 = nλ.