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Wave superposition.

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine the amplitude and phase of the luminous disturbance produced by the superposition of N waves of the same amplitude and phases which increase in an arithmetic progression ([itex]\delta[/itex],[itex]2\delta[/itex], ...[itex]n\delta[/itex])


    3. The attempt at a solution
    Using the trig identity cos(u+[itex]\delta[/itex]), where [itex] u=(kr-\omega t) [/itex] I rewrite the resulting wave(with asterisks), which is a linear combination of n waves with different phases. Associating the coefficients I get the following 2 equalities:

    [itex]A^*cos\delta^* = A \sum cos\delta_n[/itex]

    [itex]A^*sin\delta^* = A \sum sin\delta_n[/itex]

    Beyond that it gets ugly if I try to solve for A* or δ*, for example squaring both and adding gives me:

    [itex]A^* = \sqrt(A^2 ( \sum cos\delta_n)^2 + ( \sum sin\delta_n)^2)) [/itex]

    is there another way to do this?
     
    Last edited: Oct 2, 2011
  2. jcsd
  3. Oct 2, 2011 #2

    ehild

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    Use the Euler form of the waves: B*e= A*∑einδ.

    ehild
     
  4. Oct 2, 2011 #3
    I also tried that but it leads me to the same set of equations, is my solution for A* correct? Cause I can't think of anything else to do with it.
     
  5. Oct 3, 2011 #4

    ehild

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    Aeiδn is element of a geometric sequence with quotient e and Ae as first element. The sum of N element is the resultant wave.

    [tex]B e^{i \theta}=A e^{i\delta} \frac{e^{i \delta N}-1}{e^{i\delta}-1}[/tex]

    Factor out eiδ N/2 from the numerator and eiδ/2 from the denominator:

    [tex]B e^{i \theta}=A e^{i\delta (N+1)/2} \frac{e^{i \delta N/2}-e^{-i \delta N/2}}{e^{i\delta/2}-e^{-i\delta/2}}=A e^{i\delta (N+1)/2}\frac{\sin(N\delta/2)}{\sin(\delta/2)}[/tex]

    ehild
     
    Last edited: Oct 3, 2011
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