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Deriving wave equation with single string and small amplitude wave.

  1. Apr 8, 2014 #1
    1.This equation in the link below refers to the small angle approximation regarding deriving the wave equation from newtons laws from small amplitude waves in a single string with fixed tension.



    2.http://imgur.com/NGSwzcl



    3. I'm a bit rusty on the maths and have no idea how these first order derivatives suddenly equal a second order derivative. could someone explain please? The notes I'm using are http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/AnalyzingWaves.htm
     
  2. jcsd
  3. Apr 8, 2014 #2
    This just follows from the formal definition of a derivative:

    f'(x) = [itex]\stackrel{lim}{Δx→0}[/itex] = [itex]\frac{f(x+Δx)-f(x)}{Δx}[/itex]

    So if f(x) = g'(x) then f'(x) = g''(x).
     
    Last edited: Apr 8, 2014
  4. Apr 10, 2014 #3
    Thanks so much, such a simple thing I told you I was rusty. I now fully understand how to derive the wave equation thank you! also I was wondering if you could help me with getting from the second last line to the last line if it makes sense to you in the following image. apologies for my lecturers handwriting. Again I know it's my maths skills I really need to get a grip on. http://imgur.com/twm2QpQ
     
  5. Apr 11, 2014 #4
    Hard to tell the difference from his x's and t's but he is just substituting from the definitions:

    c2 = [itex]\frac{T}{ρ}[/itex]

    [itex]\frac{∂^{2}y}{∂t^{2}}[/itex] = c2 [itex]\frac{∂^{2}y}{∂x^{2}}[/itex]
     
    Last edited: Apr 11, 2014
  6. Apr 13, 2014 #5
    I can see how that will come into it but I realized its the third last line going into the second last line I don't quite understand.
    Where does the factor of half go?
    Where does the dy/dx in the left side come from?

    The right side I understand you just square it out add the d/dt.
    And I understand the substituting between the second last and last line, the only thing about the last line is the factor dy/dt.

    If I'm irritating you with my ignorance please let me know, thank you.
     
  7. Apr 13, 2014 #6
    No worries, we've all been there.

    I think he made a mistake and left out the 2.

    I think the dy/dx term you are referring to is supposed to be dy/dt. Like I said, it's hard to tell the difference between his t's and x's.
     
  8. Apr 21, 2014 #7
    I think it is dy/dt because KE=∫[itex]\frac{1}{2}[/itex]ρ([itex]\frac{dy}{dt}[/itex])[itex]^{2}[/itex]dx.

    and PE =∫[itex]\frac{1}{2}[/itex]ρc[itex]^{2}[/itex]([itex]\frac{dy}{dx}[/itex])[itex]^{2}[/itex]dx.

    Any thoughts?

    P.S.
    sorry for the delay was finishing my allsky camera report.
     
  9. Apr 21, 2014 #8
    I think it is dy/dt because:

    KE=∫[itex]\frac{1}{2}[/itex]ρ([itex]\frac{dy}{dt}[/itex])[itex]^{2}[/itex]dx.

    and PE =∫[itex]\frac{1}{2}[/itex]ρc[itex]^{2}[/itex]([itex]\frac{dy}{dx}[/itex])[itex]^{2}[/itex]dx.

    Any thoughts?

    P.S.
    sorry for the delay was finishing my allsky camera report.
     
  10. Apr 22, 2014 #9
    Those are the definitions.
     
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