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Optics: Finding the wave equation given position and amplitude information

  1. Jan 29, 2013 #1
    A harmonic wave traveling in +x-direction has, at t = 0, a displacement of 13 units at x = 0 and a displacement of -7.5 units at x = 3λ/4. Write the equation for the wave at t = 0.



    2. Relevant equations

    The equation for a harmonic wave is

    r = asin(kx-vt+θ)

    a being the amplitude
    k being the wave number k=2π/λ
    v being the velocity of the wave
    θ being the initial phase angle

    3. The attempt at a solution

    I set up the wave equations at both positions because we have two unknowns so we need two equations

    13 = asin(θ) & -7.5= asin((2π/λ)(3λ/4) + θ)
    13/sin(θ ) = a -7.5=asin(3π/2 + θ)

    Now I plugged in 13/sin(θ ) for a in the other equation and I ended up with

    -7.5=(13/sin(θ ))sin(3π/2 + θ)
    -7.5sin(θ ) = 13sin(3π/2 + θ)

    This is where I got stuck. Am I on the right track? I imagine there is a trig identity that will help me solve for θ and then I can easily solve for the amplitude. The answer according to the book is:

    15sin(kx+π/3)
     
  2. jcsd
  3. Jan 29, 2013 #2
    I solved it, but thanks to those who may have read it. For those that are curious you use the trig identity sin(u+v) = sin(u)cos(v) + sin(v)cos(u) and from there it is simple algebra
     
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