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Math Help - Wave equation - vibrations of a beam

  1. #1
    Junior Member
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    Oct 2008
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    Question Wave equation - vibrations of a beam

    Hi, I have this question:

    The longitudinal displacement u(x,t) for longitudinal vibrations of a beam of uniform cross-section satisfies the wave equation:

    v u_{xx} - p u_{tt} = 0

    where v is the modulus of elasticity of the beam and p is the mass density. For a beam of length L with the end x = 0 fixed and the end x = L free, the boundary conditions are:

    u(0,t) = 0 , u_x(L,t) = 0.

    Determine the natural angular frequencies of vibration of such a beam.

    Using separation of variables u(x,t) = F(x)G(t) I get two decoupled first order equations:

    F"(x) - (c/v)F(x) = 0 and G"(t) - (c/p)G(t) = 0.

    I'm not sure where to go from here though. In all my examples from lectures, both ends have been fixed so the form of F(x) is known to be a function of sine and cosine. But I have no idea how to deal with the free end.

    Please help,
    Katy
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  2. #2
    Newbie
    Joined
    Apr 2009
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    Hi, I'm unsure where the c has come from in your decoupled equations so starting with

    v u_{xx} - p u_{tt} = 0

    and following with u = F(x)G(t) then substituting into the first equation

    \frac{F''}{F} = \frac{p}{v}\frac{G''}{G}

    and as both sides are independant of each other can be written equal to a constant, -k^2 say. This could be any constant, but the square allows for neatness in a minute, the negative allows for a solution in the cosine / sine form rather than hyperbolic solutions.

    You would then get a solution of the form

    F(x) = A \cos(kx) + B \sin(kx)

    As I said I don't quite follow where the c comes from but assuming it's important I'm guessingyour lecture notes lead to something similar to

    F(x) = A \cos(\sqrt{c/v}x) + B \sin(\sqrt{c/v}x)?

    Regardless the G function can be solved similarly but it's t dependant and t is unimportant so leave it as G.

    u(x,t) = \left(A \cos(kx) + B \sin(kx)\right)G(t)

    Boundary condition at x=0 leads to B=0

    u(x,t) = A \cos(kx) G(t)

    Then differentiate u wrt x for the second BC

    u_x = -Ak \sin(kx) G(t)

    This equals zero for x=L, either G=0 which would be trivial so

    -Ak\sin(kL)=0

    k can't be equal to zero so it is set so the sine function is zero for all L,
    i.e.
    kL = n\pi
    k = n\pi / L

    Then
    u = A \cos(\frac{n\pi}{L}x)G(t)

    and (I think) you have the answer assuming you have \sqrt{c/v} instead of my k?
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