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Homework Help: PDE - Need help getting started.

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known dataDerive the differential equation governing the longitudinal vibration of a thin cone which has uniform density p, show that it is
    1/x/SUP] d/dx(x du/dx) = (1/c) d u/dt

    Hint: The tensile force sigma = E du/dx where E is the Young's modulus (a constant), u is the longitudinal displacement and x is the longitudinal coordinate.

    2. Relevant equations

    3. The attempt at a solution I wasn't sure where to start so if someone give help me understand how to start this problem I would appreciate it.
  2. jcsd
  3. Sep 6, 2009 #2
    Write [itex] F = ma [/tex] for a longitudinal element of the rod of length [itex]dx[/itex]. The force on the left is the tensile stress times the area, i.e.,
    E A(x) du/dx
    pointing to the left.
    Can you do the rest?
  4. Sep 7, 2009 #3
    Okay now I have the following:
    F = ma
    F = EA(x)U(x)
    = d/dx[EA(x)d/dx)] = m(x) d^2u/dt^2
    u(x,t) = U(x)T(t)
    d/dx[EA(x) d/dx[U(x)T(t)]] = d^2/dt^2 [m(x)T(t)]
    t(t) d/dx [EA(x) d/dx [U(x)T(t)] = 1/T(t) d^2/dt^2 T(t) = -w^2
    d/dx[EA(x) d/dx U(x)] + w^2m(x)U(x) = 0

    I need to get to

    1/x^2 d/dx[x^2 U(x)] = 1/c^2 U(tt)

    but not sure if I'm on the right path and how I get from my result to the required result.


  5. Sep 7, 2009 #4
    OK, don't introduce the separation of variables form [itex]u(x, t) = T(t) U(x)[/itex]. At this stage we are merely trying to derive the PDE, not solve it.

    The force on the left face of the differential element is
    F(x) = EA(x)\frac{\partial u}{\partial x}
    pointing to the left.

    The force on the right face will be (using a Taylor series)
    F(x + dx) = F(x) + \frac{\partial F}{\partial x}dx[/tex]
    pointing to the right.

    Hence the net force is: (you write it down).

    Next, the mass of the element is [itex]m = \rho A(x) dx[/itex], where [itex]\rho[/itex] is the density.
    The acceleration of the element is [itex]\partial^2 u/\partial t^2[/itex].

    So [itex]F = ma[/itex] becomes (you write it down).

    Next introduce [itex]c^2 = E/\rho[/itex] into the above equation. This gives (you write it down).

    Next write [itex]A(x) = \pi R^2(x)[/itex], where for a cone [itex]R(x) = \alpha x[/itex], [itex]\alpha[/itex] being the slope of the cone and [itex]x[/itex] is measured from the tip of the cone. Introduce this into the equation. This gives the answer. Done?
  6. Sep 7, 2009 #5
    Thank you for your patience and help in explaining this in a manner that I was able to understand.
  7. Sep 3, 2011 #6
    hey how do u get the net force in the above question?
    I know t is the FR-FL. But what would it be?
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