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Superposition of waves

  1. Sep 25, 2005 #1
    How do I find the sum of these 2 waves, Asin(kx-wt) and Asin(kx+wt)?

    I have no clue how to add 2 sins with diffent phases.


    Thank you for your help!
     
  2. jcsd
  3. Sep 25, 2005 #2

    robphy

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    Can you re-express sin(kx+wt) using a trig identity?
     
  4. Sep 26, 2005 #3
    Humm would I use this identity sin(A + B)= sinA cosB + cosA sinB

    so Asin(kx-wt) = -AsinkxAcoswt -AcoskxAsinwt
    Asin(kx+wt) = AsinkxAcoswt +AcoskxAsinwt

    But wouldnt that equal to zero?
     
  5. Sep 26, 2005 #4

    robphy

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    Your identity is fine.
    Check your use of it. Check the signs you wrote down.
     
  6. Sep 26, 2005 #5
    oh i saw where i made my mistake
    the answer should be 2Asin(kx)cos(wt)?


    Now after I get that how would I find the

    ye(x), called the envelope, depends only on position

    and yt(x) depends only on time

    yt(x) should be a trigonometric function of unit amplitude.

    I need to express ye(x) and yt(x) in terms of A k omega x and t.


    Is there partial derviative involved?

    Thank you for your help!!
     
    Last edited: Sep 26, 2005
  7. Sep 27, 2005 #6
    anyone please?
     
  8. Sep 28, 2005 #7

    Galileo

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    Look at the function and see what it means. You've seperated the time and position variables. You could view the motion of each point as a function of time as an harmonic oscillator. (Some points don't move at all). Plot the function to see what it looks like. If you get some insight in the equation, the questions are very easy.
     
  9. Sep 28, 2005 #8
    After finding the superposition forumla for Asin(kx-wt) and Asin(kx+wt), which is 2Asin(kx)cos(wt), How do I tell which direction is the wave traveling?
     
  10. Sep 28, 2005 #9

    robphy

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    Plot your superposed function for t=0. Notice where the x-positions of the peaks are.
    Plot your superposed function for (say) t=0.01. Notice where the x-positions of the peaks are now.
    Which way did it shift? If 0.01 is too complicated, try t=(1/12)*(2*pi/w).
    If it's not clear, try doubling the value of t you just used.
     
  11. Sep 28, 2005 #10
    Humm from what i see, it seems that the wave its moving vertically, oscillating. Is that correct? I am not too sure how to graph these.

    I graphed sin(x)cos(0.01) and then sin(x)cos((1/12)*(2pi))
     
  12. Sep 28, 2005 #11

    robphy

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    Presumably, you understand the meaning of "which way the sin(kx-wt) part travels"... and likewise for the sin(kx+wt). It can be seen as the direction along the x-axis of the motion of a peak. Apply the same reasoning to the superposed wave. You've practically got it... You just have to give the answer in the form stated above.

    The second expression is easy to plot a graph of
    "amplitude sin(x)cos((1/12)*(2pi)) [at time (1/12)*(2pi)] vs position x".
    Can you numerically evaluate cos((1/12)*(2pi)), where pi=[itex]\pi[/itex].
     
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