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Solving 1d Helmholtz with boundary conditions

  1. Feb 9, 2010 #1
    Hello all,
    This is to do with forced longitudinal vibration of a rod (bar).
    It's basically a problem to do with the linearised plane wave equation (1d).

    The rod is fixed firmly at one end, and excited at the other by a harmonic force.

    The wave equation (with constant rho/E instead of 1/c^2) is reduced to the helmholtz equation, which is fine. But the boundary conditions which exist (in this example) are at x=0, u=0 (u=displacement) and at x=L, (AE)*du(x,t)/dx=Fcos(wt)

    This leads to the solution of the plane wave equation which is:
    anyway, there is a time dependence there which I'm not really wanting.
    How do I remove this? Basically, i don't know if you noticed but I am lost!

    At the end of it all i'm looking for the point mobility.
    Last edited: Feb 9, 2010
  2. jcsd
  3. Feb 10, 2010 #2
    Sorry, but I think I don't understand your meaning.... you excite the end with an harmonic force, and you don't want the solution to depend on time?....
    Making some guessing, you say you get to a hemholtz equation. That amounts to assume your solution goes as [tex]u(x,t)=cos(\omega t)u(x)[/tex], and then, you can factor out the time dependency, both in the equation and in the boundary condition (AE u'(x)=F). The resulting system gives you u(x), with no time factor
  4. Feb 12, 2010 #3
    Ok I can probably make a bit more sense now hopefully...
    I have the general solution u(x,t)= Aexp(j(wt-kx))+Bexp(j(wt+kx))
    I can find the coefficients A and B of this equation by using the boundary conditions present, this is all fine.
    However, when the coefficients are put back in I am still left with wt. I know it's a stupid question but I'm having a prolonged mental block on what to do next!
    I just want it to depend on frequency, so what do I have to do to get this?
  5. Feb 12, 2010 #4


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    The bar is vibrating, with a cos(wt) time dependence. It sounds like you want the "amplitude" of the bar, which is obtained by saying cos(wt)=1 in your expression.
    It's not clear to me, from your expression, if there is a frequency dependence there.
  6. Feb 20, 2010 #5
    Ok, I finally found out where I was going wrong. In case anyone finds the information useful,
    You assume a solution of:
    U(x,t)=A*exp(jkx) +B*exp(-jkx)
    Then substitute in the boundary conditions ie at x=l, u(x)=0 since it is fixed at one end.
    The other boundary condition is at x=0, -F=EA*(partial(du/dx)).
    Differentiate u with respect to x, and let x=0.
    Doing this gets the values of A and B and the solution.
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