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Standing waves - Wave Equation

  1. Apr 6, 2013 #1
    I don't completely understand how equation 4.4.4 was derived and determined. I understand the derivation behind the basic wave equation 4.3.4 but not what happened in 4.4.4. Why is there a need for all the negative signs ? Would a simple phase change suffice ?

    Please do be a bit detailed in your explanation.... Pretend I'm an idiot. Thank you !




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  2. jcsd
  3. Apr 6, 2013 #2

    Redbelly98

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    Since the reflected wave travels in the opposite direction as the incident wave, you need to flip the sign on either the kx or the ωt term***. The author chose to reverse the sign of kx to make it -kx. There was already a - sign with -ωt. For the phase term, the choice of sign is arbitrary.

    Just changing the phase would express a wave traveling in the same direction as the incident wave, but shifted in phase.

    Hope that helps.

    EDIT added:
    *** This is because a rightward-traveling wave has the form f(kx-ωt) or f(-kx+ωt). A leftward-traveling wave has the form f(kx+ωt) or f(-kx-ωt).
     
    Last edited: Apr 6, 2013
  4. Apr 6, 2013 #3
    So I could equally have written kx+wt ? Why is there a need for the phase change ∅ ?
     
  5. Apr 6, 2013 #4

    Redbelly98

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    Yes; see the edit added to my earlier post.
    As the book says, the type of boundary will determine ∅. Usually the boundary is either fixed or has a maximum amplitude. Also, the location of the boundary plays a role in what ∅ is.

    Eg., for a fixed end located at x=x0:

    [tex]\cos(kx_0 + \phi /2) = 0[/tex]

    You'd solve that for ∅, given k and x0. Your book is taking the fixed end to be at x=0, so that simplifies things somewhat.
     
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