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Phase of resultant wave

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data
    Two harmonic waves are described by (pic attached). What is the phase of the resultant wave when x = t = 0?

    A. 3
    B. 1
    C. 4
    D. 2

    2. Relevant equations
    [tex]y=Asin(kx-\omega(t)+\phi)[/tex] (Single wave equation)
    [tex]y=2Asin(kx)cos(\omega(t))[/tex] (Standing wave equation)



    3. The attempt at a solution
    I originally chose D because the phase angle in the second equation was -2. However, then I thought about that fact that the question asked for the resultant wave. Thus, I was thinking that since both x and t are 0, the first wave equation reduces to 0 and the second to 7m sin(-2). Thus, the resultant wave equation should be only 7m sin(-2). Where do I got from there? (Is that even the correct thing to do?) :confused:
     

    Attached Files:

  2. jcsd
  3. Feb 11, 2010 #2

    rl.bhat

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    Find the resultant y1 and y2. While doing it you have to use some trigonometric simplification.
    sinC +sinD = 2*sin[(C+D)/2]*cos[(C-D)/2]
    Then decide the phase difference.
     
  4. Feb 11, 2010 #3
    What would be the phase difference after I've done that? (I haven't solved for the resultant wave yet, I'm just wondering how I get the phase difference after I do so)
     
  5. Feb 11, 2010 #4

    rl.bhat

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    Just solve it. The phase difference is 1.
     
  6. Feb 11, 2010 #5
    Thank you very much :smile:, but what I meant was, how I get the phase difference from the resultant wave function (in other words, in [tex]y=Asin(kx-\omega(t)+\phi)[/tex] I know that the phase difference is [tex]\phi[/tex], but I wasn't quite so sure what the phase is in the resultant wave equation).

    I think the resultant wave equation is [tex](14m)sin(5\frac{x}{m}-100\frac{t}{s}-1)cos(1)[/tex]. I see two one's in that equation, so I still am not completely sure how to find the phase difference. Also, the question asks for the phase when x and t are both 0. How do time and position come into play? I thought the phase was a constant?

    More help would be much appreciated :wink:
     
  7. Feb 11, 2010 #6

    rl.bhat

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    When you write
    y = A*sin(kx- ωt + φ), φ is the starting point of the oscillation. And that is called phase. In that sense the phase of the resultant wave is 1 rad. It is not the phase difference. Even in the problem they have asked phase.
    We consider phase difference when we compare two different waves. In that case, in the given problem, one wave has zero phase and other has 2 rad. The phase difference between them is 2 rad.
     
  8. Feb 11, 2010 #7
    I'm all confused now :cry:

    So what is, and how do I find, the phase of the resultant equation I'm trying to find?
     
  9. Feb 11, 2010 #8

    rl.bhat

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    In my post#4, I should have mentioned it as phase, not the phase difference.
     
  10. Feb 11, 2010 #9
    Thank you... can you clarify how to calculate or to find that for the phase?
     
  11. Feb 11, 2010 #10

    rl.bhat

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    I don't know what you want to calculate.
     
  12. Feb 11, 2010 #11
    The phase.
     
  13. Feb 11, 2010 #12

    rl.bhat

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    Your resultant wave is
    y = 14*sin(5x/m - 100t/s -1)
    When x = 0, t = 0. phase is -1 rad, and displacement is y = -14*sin(1 rad) = -11.78 m.
     
  14. Feb 12, 2010 #13
    Why is there no cosine part to the resultant wave equation? Is phase -1 or +1 rad?
     
  15. Feb 12, 2010 #14

    rl.bhat

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    Yes. You are right. When x = 0 and t = 0, the amplitude of the resultant wave is 14*sin(1 rad)*cos(1 rad)
    And phase is -1 rad.
     
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