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Wave equations

  1. May 24, 2005 #1
    Consider a traveling wave described by the formula

    y_1(x,t) = Asin(kx - wt).

    This function might represent the lateral displacement of a string, a local electric field, the position of the surface of a body of water, or any of a number of other physical manifestations of waves.

    The expression for a wave of the same amplitude that is traveling in the opposite direction is Asin(kx + wt).

    The sum of these 2 waves can be written in the form y_s(x,t) = y_e(x)*y_t(t). Where y_e only depends on displacement and y_t depends on the time.

    Find y_e(x) and y_t(t). Keep in mind that y_t(t) should be a trigonometric function of unit amplitude. Express your answers in terms of A, k, x, w, and t.

    I know I'm meant to use the identity sin(A-B) = sinAcosB - cosAsinB, but I don't know how to apply it.

    Any help would be great.

    Thank you.
     
  2. jcsd
  3. May 24, 2005 #2

    dextercioby

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    [tex] A\sin\left(kx+\omega t\right)+A\sin\left(kx-\omega t\right) [/tex]

    Factor "A" and use the identity

    [tex]\sin\left(\alpha+\beta\right)=\sin\alpha\cos\beta+\sin\beta\cos\alpha [/tex]

    for

    [tex] \left\{\begin{array}{cc}\alpha = kx & \beta=\omega t\\ \alpha=kx & \beta=-\omega t\end{array}\right [/tex]

    Daniel.
     
  4. May 24, 2005 #3
    Ok, I got A[2sinkxcoswt + 2coskxsinwt]. How do you then go from this to finding y_e(x) and y_t(t)?
     
  5. May 24, 2005 #4

    dextercioby

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    Nope.Use the fact that

    [tex]\sin \left(-\beta\right)=-\sin\beta [/tex]

    Post the result.

    Daniel.
     
  6. May 24, 2005 #5
    I end up with that answer and I am using that fact.

    For the first bit, Asin(kx-wt) = A[sinkxcoswt+sinwtcoskx]
    For the second bit Asin(kx-wt) = A[sinkxcos(-wt)-sin(-wt)coskx]

    cos(-wt) = coswt
    sin(-wt) = -sin(wt)

    Therefore, for the second bit, A[sinkxcoswt-(-sinwt)coskx] = A[sinkxcoswt+sinwtcoskx].

    I'm not sure what I'm getting wrong.
     
  7. May 24, 2005 #6

    dextercioby

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    Nope.

    [tex] A\sin\left(kx-\omega t\right)=A\sin\left(kx+\left(-\omega t\right)\right)=A\left(\sin kx\cos\omega t-\sin \omega t\cos kx\right) [/tex]

    Okay?

    Daniel.
     
  8. May 24, 2005 #7
    So it's 2Asinkxcoswt? How do you use that to find y_e(x) and y_t(t)?
     
  9. May 24, 2005 #8

    dextercioby

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    If add them,u'll get

    [tex]y_{e}(x)y_{t}(t)=2A\sin kx\cos\omega t [/tex]

    Do you see which is which...??

    Daniel.
     
  10. May 24, 2005 #9
    Yep, I've got it.

    Thanks :smile:
     
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