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Lowest frequency non-uniform string

  1. May 2, 2014 #1
    1. The problem statement, all variables and given/known data
    See attachment (stuck with part b at the moment)


    2. Relevant equations



    3. The attempt at a solution

    [tex]\phi=D(x)T(t)[/tex]
    so
    [tex](1+bx)D''(x)T(t)-D(x)T''(t)=0[/tex]
    [tex](1+bx)\frac{D''(x)}{D(x)}=\frac{T''(t)}{T(t)}[/tex]

    let
    [tex]\frac{T''}{T}=\sigma[/tex] (1)
    use trial solution [tex]T=be^{rt}[/tex]

    subbing into (1) and solve for r.
    [tex]r=\pm\sqrt{\sigma}[/tex]

    use same trial solution and repeat steps for
    [tex](1+bx)\frac{D''}{D}=\sigma[/tex]

    [tex]r=\pm\sqrt{\frac{\sigma}{1+bx}}[/tex]

    from principle of superposition

    [tex]D(x)=a_1e^{\sqrt{\frac{\sigma}{1+bx}}x}+a_2e^{-\sqrt{\frac{\sigma}{1+bx}}x}[/tex]

    [tex]T(t)=b_1e^{\sqrt{\sigma}t}+b_2e^{-\sqrt{\sigma}t}[/tex]

    Then I get confused with boundary conditions can someone let me know if I am on the right lines so far and give me any advice for proceeding?

    Thanks
     

    Attached Files:

    Last edited: May 2, 2014
  2. jcsd
  3. May 2, 2014 #2
    Your solution for D is not correct. In particular, since r turned out to be a function of x, your calculation of D'' must take that into consideration. Also, σ turns out to be negative. You must be aware of that when looking for the ansatz solutions.
     
  4. May 2, 2014 #3
    not really sure I understand, am I correct in using [tex]D=ae^{rx}[/tex] and subbing that into[tex](1+bx)\frac{D''}{D}=\sigma[/tex] to find r? Or is my r incorrect?

    EDIT: I see what you mean nevermind..

    EDIT: I don't suppose you could give me a hint on how to find the trial solution? I have not come across an ODE in the form of f(x)y''-ay=0, what is this type of differential equation called?
     
    Last edited: May 2, 2014
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