Solving 1d Helmholtz with boundary conditions

  • Thread starter ptptaylor
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  • #1
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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:
u(x,t)=Fsin(ax)cos(wt)/(AEcos(aL))
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.
 
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Answers and Replies

  • #2
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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
 
  • #3
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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?
 
  • #4
Redbelly98
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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.
 
  • #5
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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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