y=sin(4x-3t), y=sin(4x-3t+3), what is the amplitude of the resultant wave?

Answer 1

1 in both cases. There will be a difference only in phase. But you may have stated the problem incorrectly.

Answer 2

Each is a wave with amplitude unity (the coefficient before the sin term). For example, at x and t=0, the amplitude of the first wave is zero and that of the second wave close to -1 (sin(3) = sin(pi) nearly). At given values of x and t, one wave will be so offset from the other, so I would guess this is the maximum value of the resultant wave. Is this a trick?

Answer 3

y1 = sin(4x-3t), y2 = sin

Amplitude of y1 = amplitude of y2 = 1

y1 + y2 = sin(4x - 3t) + sin(4x - 3t + 3)

Let u = 4x - 3t + 3/2

Thus 4x - 3t = u - 3/2 and 4x - 3t + 3 = u + 3/2

So y = y1 + y2 = sin(u - 3/2) + sin (u + 3/2)

= 2 sin [½((u + 3/2) + (u - 3/2))] cos [½((u + 3/2) - (u - 3/2))]

= 2sinucos(3/2)

= [2cos(3/2)]sinu

dy/du = 2cos(3/2) cosu = 0 for stationary points

So u = π/2, 3π/2 (for 0 ≤ u ≤ 2π)

When u = π/2 sin u = 1 and when u = 3π/2 sinu = -1

Whence y1 + y1= sin(u - 3/2) + sin (u + 3/2)

and sin (3π/2 - 3/2) + sin (3π/2 + 3/2) ≤ y1 + y2 ≤ sin (π/2 - 3/2) + sin (π/2 + 3/2)

ie -0.1414744.... ≤ y1 + y2 ≤ 0.1414744....


So its amplitude is 0.1414744.... (= 2 cos (3/2))

Answer 4

sin(4x-3t) + sin(4x-3t+3) = 2sin((8x-6t+3)/2)cos(-3/2)

now if we compare with Asin(wt+kx) we have:

A = 2cos(-3/2)=2cos(3/2)