1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Time-averages of superposition of waves.

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider the superposition of two waves;

    [itex]\zeta_1 + \zeta_2 = \zeta_{01} e^{i(kr_1 - wt)} + \zeta_{02} e^{i(kr_2 - wt + ∅)} [/itex]

    where [itex] ∅ [/itex] is a phase difference that varies randomly with time. Show that the time-averages satisfy;

    [itex]<|\zeta_1 + \zeta_2|^2> = <|\zeta_1|^2> + <|\zeta_2|^2> [/itex]

    2. Relevant equations

    (1) If it wasn't clear, The two waves are;

    [itex] \zeta_1 = \zeta_{01} e^{i(kr_1 - wt)} [/itex] and
    [itex]\zeta_2 = \zeta_{02} e^{i(kr_2 - wt + ∅)} [/itex]

    3. The attempt at a solution

    Unless I have my definition of time-average wrong. I can't seem to get this to work.

    [itex]|\zeta_1 + \zeta_2|^2 = (\zeta_1 + \zeta_2)(\zeta_1^* + \zeta_2^*) = |\zeta_1|^2 + |\zeta_2|^2 + \zeta_1\zeta_2^* + \zeta_2\zeta_1^* = |\zeta_1|^2 + |\zeta_2|^2 + 2\zeta_{01}\zeta_{02}cos(k(r_1 - r_2) - ∅)[/itex]

    Then, I believe, the time average is given by;

    [itex]\frac{1}{T}\int^T_0 {|\zeta_1|^2 + |\zeta_2|^2 + 2\zeta_{01}\zeta_{02}cos(k(r_1 - r_2) - ∅)} dt[/itex]

    However, I don't see how this turns in to the form I desire. It would require that the last term (containing the cosine) is time-averaged to zero. Can this be the case? Also, can [itex] ∅ [/itex] still even be considered a function of time when it varies RANDOMLY?
  2. jcsd
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted