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Homework Help: Combination of two waves.

  1. Jul 27, 2011 #1
    I'm getting so confused about this question, any help would be great. :)

    1. The problem statement, all variables and given/known data

    Two infinite waves Ψ1, Ψ2 have the same wavelength and polarisation and have amplitudes of E1 = 3 and E2 = 7 units. They are added together with a phase difference of 125 degrees.

    1) What will be the relative intensity of the resultant wave? Assume initial phase of 0 degrees for Ψ1.

    2) In the same conditions, what will be the phase of the resultant wave (in degrees)?



    2. Relevant equations

    N/A

    3. The attempt at a solution


    So far I've got Ψ1=3sin(kx-ωt) and Ψ2=7sin(kx-ωt+2.182)
    And I know that I is proportional to A squared.

    Now I'm pretty stuck!
     
  2. jcsd
  3. Jul 27, 2011 #2

    ehild

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    Homework Helper

  4. Jul 27, 2011 #3
    Thanks for the help, ehild!

    I've found a formula in my notes for resultant amplitude of 2 waves: (A3)^2 = (A1)^2 + (A2)^2 + 2(A1)(A2)cos(δ)

    So using this, I get an (A3)^2 of 34, which I suppose is the relative Intensity. Not sure how I should answer this on the question though, I haven't really specified the relative intensities of A1 and A2.

    As far as part 2 goes, when I add up Ψ1 and Ψ2, I get stuck with some terms I can't get rid of, and I certainly can't put it into a nice 'A sin(kx-wt +φ)' form.
    I'm also getting a bit confused about usage of δ, φ, α, ε, phase, phase difference, phase constant, etc. In my notes I found the following for resultant phase difference (I think):

    tanα=(A1sinα1+A2sinα2)/(A1cosα1+A2cosα2)

    Not sure if this is of any use.

    When I add together the 2 waveforms I get Ψ3=3+7cosδ(sin(kx-ωt))+7sinδ(cos(kx-ωt)).

    Thanks again for the response!
     
  5. Jul 27, 2011 #4

    ehild

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    You get the relative intensities of the original waves as the square of the amplitudes.
    The sum of the two waves is equal with a third one, which is
    A3 sin(kx-wt +α) with the terms in your notes. The whole argument of the sine function is the phase of the wave. You see, it changes both with time t and place x. The constant term , 2.182 radian in case of Ψ2 is the phase constant. It is zero in case of Ψ1. The phase constant is the phase at t=0 and x =0, the problem refers to it as initial phase.

    The resultant amplitude is given for waves with phase difference δ which is the same as the difference of the phase constants:
    δ=(wt-kx+2.182)-(wt-kx)=2.182.

    The phase constant of the resultant wave is given in your notes for two waves: one with amplitude A1 and phase constant α1, the other with amplitude A2 and phase constant α2. A1=3 and the phase constant of the first wave α1=0 in the problem, and A2=7, α2 = 2.182 or 125°. From these, you get the phase constant α of the resultant wave.

    ehild
     
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