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I Superposition (proof)

  1. Mar 31, 2017 #1
    I don't get the first part. why did he make the angle of sin equal to n pi.

    upload_2017-3-31_17-44-19.png
     

    Attached Files:

  2. jcsd
  3. Mar 31, 2017 #2
    A node is a location where the amplitude is zero. The sin function is zero when its argument is a multiple of π radians, nπ, where n = 0, 1, 2,...

    In degrees, the sin is zero at 0, 180, 360, etc.
     
  4. Mar 31, 2017 #3
    Ok I understood this part but which one is the amplitude "Y" or "A"?

    and can I take the angle of cos and make it equal to n pi/2 where n is odd number? It will also give me 0 in this case.
     
  5. Mar 31, 2017 #4
    I probably shouldn't have used the word "amplitude" for y. y is the displacement for a given x,t, whereas the amplitude is the maximum value of y.

    Those values of x that lead to the argument of sin being nπ will give y = 0 for all t. Will have to think about your question of setting the cos argument to nπ/2.
     
  6. Mar 31, 2017 #5
    Okay, check out the section Two sine waves traveling in opposite directions create a standing wave at: http://www.acs.psu.edu/drussell/Demos/superposition/superposition.html

    If the argument of the cos is nπ/2, then those values of t lead to y = 0 for all x. That is shown in the simulation on the referenced web page.
     
  7. Apr 3, 2017 #6

    PeterO

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    Homework Helper

    An over view:
    The argument of the sin includes an "x", leading to where (along the string) the function is zero.
    The argument of the cos includes a "t" leading to when (in time) the function is zero.
    The question related to where the nodes were, so work with the sin.
    In a standing wave, even points of antinode are periodically at zero displacement - when that happens is found by playing with the cos function
     
  8. Apr 3, 2017 #7
    That's shown in the simulation I referenced.
     
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