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Fourirer Analysis

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data

    String of length L and string mass-density of[tex]\mu[/tex] is fixed at both ends. at t=0
    y(x,t)=
    4xh/L 0<x<L/4
    2h-4xh/L L/4<x<L/2
    0 L/2<x<0
    Find the first four coefficients, is any one of them is zero? If so why is it?

    /\
    / \
    |/ \______|
    the peak is L/4 and the right corner of the triangle is L/2.
    2. Relevant equations

    2/L{[tex]\int[/tex](4xh/L) sin(nx[tex]\pi[/tex]/L) over 0<x<L/4 +
    [tex]\int[/tex](2h-4xh/L)sin(nx[tex]\pi[/tex]/L) over L/4<x<L/2 +
    [tex]\int[/tex] 0*sin(nx[tex]\pi[/tex]/L) over L/2<x<L}

    3. The attempt at a solution

    so the third integral is equal to zero, we are left with only two.
    After integrating the first two I've got the following result for An=
    (8h)/(n[tex]\pi[/tex])^2[2sin(n[tex]\pi[/tex]/4)-sin(n[tex]\pi[/tex]/2)]
    is this result correct? I've never seen Fourier series described by two sine terms. should I change the height to (h+1) so the last integral wont be zero and then integrate???
    any other ideas how to go about it?
     
  2. jcsd
  3. Oct 26, 2008 #2

    gabbagabbahey

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    Use the following trig identity: [itex]sin(2x)=2sin(x)cos(x)[/itex] to rewrite your second sine term as [itex]2sin(\frac{n \pi}{4})cos(\frac{n \pi}{4})[/itex]...what does that make A1,A2,A3 and A4?
     
  4. Oct 26, 2008 #3
    hmmmmm......I didnt think I can use cos as part of the Fourier series.
    is it ok? but I'm still going to have two terms of sin and cos--
    2sin(npi/4)(1-cos(npi/4)) it's still not uniformly zeros...
     
  5. Oct 26, 2008 #4
    every fourth term is a zero
     
  6. Oct 26, 2008 #5

    gabbagabbahey

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    Yes, every fourth term is zero....is there any physical interpretation for this result?
     
  7. Oct 26, 2008 #6
    I'm still trying to figure out how to graph the wave.
    But, I think that it means that the wave is symmetric about the x-axis and that's why it's zero, correct?
     
  8. Oct 26, 2008 #7

    gabbagabbahey

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    Not really; it means that every fourth harmonic is zero...in other words the initial wave pulse doesn't contain any harmonic of the frequency n*pi
     
  9. Oct 26, 2008 #8
    so the other half of the string never moves?
    i agree that every multiple of 4 will cause the amplitude to be zero, but i thought that the main notion from Fourier is that one should always consider harmonic frequency when using his analysis (i.e the frequency is the same for all=>harmonic)
     
  10. Oct 26, 2008 #9
    how would you go about graphing it? just choose arbitrary values?
     
  11. Oct 26, 2008 #10
    A1,A2 and A3 do contain the harmonic n*pi
     
  12. Oct 26, 2008 #11

    gabbagabbahey

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    A1,A2,A3 contain harmonics of npi/4, npi/2 and 3npi/4 not npi....this means that the fourier series of the initial waveform does not contain any terms with frequency (0,pi,2pi,....).

    To graph the function just do a 3D plot of y(x,t) with t as your z-axis.
     
  13. Oct 26, 2008 #12
    ohhh....got you. thanks
     
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