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Adding waves to get resultant wave

  1. Jun 25, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to show that the waves: sin((kx)+(θ/2)) and sin((kx)-(θ/2)), differing in phase by θ, add to give a resultant wave 2sin(kx)cos(θ).
    But the answer I get is different so I'm not sure how to do this.

    2. Relevant equations

    sinA + sinB = 2sin((A+B)/2)cos((A-B)/2)

    3. The attempt at a solution

    I tried adding the waves together and so:

    A = ((kx)+(θ/2))
    B = ((kx)-(θ/2))

    (A+B) = ((kx) + (θ/2)) + ((kx)-(θ/2))
    (A+B) = (kx) + (θ/2) + (kx) - (θ/2)
    (A+B) = (kx) + (kx)
    (A+B) = 2(kx)

    (A-B) = ((kx)+(θ/2)) - ((kx)-(θ/2))
    (A-B) = (kx) + (θ/2) - (kx) + (θ/2)
    (A-B) = (θ/2) + (θ/2)
    (A-B) = θ

    sinA + sinB = 2sin((A+B)/2)cos((A-B)/2)
    sin((kx)+(θ/2)) + sin((kx)-(θ/2)) = 2sin((2kx)/2)cos((θ)/2)
    sin((kx)+(θ/2)) + sin((kx)-(θ/2)) = 2sin(kx)cos(θ/2)

    So I must have have gone wrong somewhere because my final answer is 2sin(kx)cos(θ/2), but it should be 2sin(kx)cos(θ).
     
  2. jcsd
  3. Jun 25, 2012 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Your answer is correct.

    ehild
     
  4. Jun 25, 2012 #3
    Hello Rococo,
    Your solutions is correct.Try θ=90 degrees .R.H.S. is zero right? But the L.H.S ain't. You can check your solutions by substitution.
    regards
    Yukoel
     
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