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Direct substitution problem

  1. Mar 12, 2008 #1

    ~christina~

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    1. The problem statement, all variables and given/known data
    verify by direct substitution that the wave function for a standing wave given in equation below is a solution to the general linear wave equatin.

    [tex] y= (2A sin kx)cos \omega t [/tex]

    [tex]\frac{\delta^2y} {\delta x^2}= \frac{1} {v^2} \frac{\delta^2 y} {\delta t^2} [/tex]

    2. Relevant equations (above)

    How do I do this by direct substittution ?

    I originally went and just proved it through partial differentiation but I've never heard of direct substitution before.

    Help please

    Thank you
     
  2. jcsd
  3. Mar 12, 2008 #2

    Hootenanny

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    I have a sneaking suspicion that you've verified the solution by direct substitution without knowing it. The method of direct substitution simply means substituting the proposed solution into the equation and seeing if it works. In other words, taking the second partial derivatives with respect to position and time and seeing if they fit the equation.
     
  4. Mar 12, 2008 #3

    tiny-tim

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    Yeah, sometimes examiners are really sneaky, and use words which you think are technical, but actually they're just obscure ordinary English. :frown:
     
  5. Mar 12, 2008 #4

    ~christina~

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    well not exactly. I couldn't get them to equal since I wasnt sure what to do after I got:

    [tex]\frac{\delta^2 y} {\delta x^2}|_t= -k^2(2A sin kx)cos \omega t) [/tex]

    and

    [tex]\frac{\delta^2 y} {\delta t^2}|_x= -\omega^2 2A cos\omega t(sin kx)[/tex]

    and if that's right...I'm not sure how to make them equal Is there a equation I can use to do that?

    [tex](-k^22A sin kx)(cos \omega t)= \frac{1} {v^2} (-\omega^2 cos\omega t)(2A sin kx)[/tex]

    Thanks
     
    Last edited: Mar 12, 2008
  6. Mar 12, 2008 #5

    tiny-tim

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    Hi christina!

    erm … they're the same! … except for a factor, which you can call v. :smile:
     
  7. Mar 12, 2008 #6

    ~christina~

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    Of course they are but I can't say that they are unless I show it first.
    The question is to verify by direct substitution...is this right ? (below).

    I'm thinking of dividing over the cos and sin and then canceling to just get

    [tex]\frac{\omega^2} {k^2} = \frac{1} {v^2} [/tex]

    but I'm trying to find a relationship in physics that allows that. If I'm not incorrect.
     
  8. Mar 12, 2008 #7

    ~christina~

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    I think I found the relationship after looking at one of my old posts.

    It's interesting though how I can't find this equation anywhere in my book but my lab teacher gave it to the class and it works here. ([tex]v= \omega /k[/tex])

    [tex](-k^22A)(sin kx)(cos \omega t)= \frac{1} {v^2} (-\omega^22A) (cos\omega t)(sin kx)[/tex]

    I'm thinking of dividing over the cos and sin to the left and then canceling to just get

    [tex]\frac{\omega^2} {k^2} = \frac{1} {v^2} [/tex]

    But I'm confused to as to when I use [tex]v= \omega /k [/tex]

    I end up with

    [tex]\frac{1} {v^2} = (\frac{k} {\omega})^2 [/tex] :confused:

    but is there another way to verify?

    Thank you
     
  9. Mar 12, 2008 #8

    ~christina~

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    Actually I think I found what I was supposed to get( above equation)...:biggrin:...right?
     
    Last edited: Mar 12, 2008
  10. Mar 12, 2008 #9

    HallsofIvy

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    Suppose you were asked to show that x= 1 is a solution to x5- 3x2+ 5x- 2= 0. Would you solve the equation? Of course not- you would just substitute 1 for x and show that the equation is true. That's "direct substitution".
     
  11. Mar 12, 2008 #10

    jtbell

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    Look up the definitions of [itex]k[/itex] and [itex]\omega[/itex] in terms of [itex]\lambda[/itex] and [itex]f[/itex] and substitute them into the equation above. You should get something that looks familiar. :smile:
     
  12. Mar 12, 2008 #11

    ~christina~

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    I understand. So this case however, it would be pluging an equation into the linear wave equation for direct substitution.

    Thank You HallsofIvy
     
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