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Wave equation for a string

  1. Feb 6, 2005 #1
    Assume the well known PDE of an infinite length string

    D^2(y)/Dt^2 = c^2* ( D^2(y)/Dx^2)

    where y=y(x,t) is the transverse displacement of the string.
    D/Dx= partial derivative with respect to x
    D/Dt= partial derivative with respect to t
    c= velocity of the wave

    According to Morse and Ingard's Theoretical Acoustics (page 97), if the shape of the string is described by the function y(x,t)=F(x-ct), then Dy/Dt= -c*F'(x-ct)= -c*Dy/Dx (Where F'(z)= the derivative of F with respect to z).

    I found the last statement a little bit confusing. Could anyone explain why F'(x-ct)=Dy/Dx ? Obviously if F'(x-ct)=Dy/Dt then a constrain is put on c which is not correct...
     
  2. jcsd
  3. Feb 6, 2005 #2

    Galileo

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    What is [itex]\frac{\partial}{\partial x}F(x-ct)[/itex]?

    Does it equal [itex]F'(x-ct)[/itex]?
     
  4. Feb 6, 2005 #3
    I have concluded that Dy/Dx=F'(x-ct) after working with some examples.
    So I wrote down several functions with (x-ct) as an argument and I took the partial derivative with respect to x. Then I took the derivative with respect to the argument as a whole (z=(x-ct) ) to form F'(x-ct). They are equal. Done.
     
  5. Feb 6, 2005 #4

    dextercioby

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    Did u write everything in one line like i did?? :wink: :tongue2:

    [tex] \frac{\partial F}{\partial x}=\frac{dF(x-vt)}{d(x-vt)}\frac{\partial (x-vt)}{\partial x}=\frac{dF(x-vt)}{d(x-vt)} [/tex]

    Daniel.
     
  6. Feb 6, 2005 #5
    Exactly!

    I have done the same both for Dy/Dx and Dy/Dt to verify that Dy/Dt=-c*Dy/Dx

    Thanks.
     
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