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Homework Help: One dimensional wave equation

  1. Sep 12, 2006 #1
    Could someone help me out on the following questions?

    Q. Consider the free vibrations of a string of length L clamped at x = 0 and constrained at x = L such that [tex]u_x \left( {L,t} \right) = - ku\left( {L,t} \right),k > 0[/tex].

    (a) Show that the eigenvalues are given by the positive roots of: [tex]k\tan \left( {\omega L} \right) + \omega = 0[/tex]. How many positive solutions for omega does this equation have?

    (b) The string is released from rest so that [tex]u_t \left( {x,0} \right) = 0[/tex], and the string is given an initial deformation given by u(x,0) = (hx)/L, where h is constant. Use the principle of linear superposition to find the solution u(x,t).

    Wave equation: [tex]\frac{{\partial ^2 u}}{{\partial t^2 }} = c^2 \frac{{\partial ^2 u}}{{\partial x^2 }}[/tex]

    For part (a) I just tried a separable solution [tex]u\left( {x,t} \right) = T\left( t \right)\phi \left( x \right)[/tex]. Where lambda is a constant I obtain the two ODEs:

    [tex]
    T''\left( t \right) + \lambda c^2 T\left( t \right) = 0
    [/tex]

    [tex]
    \phi ''\left( x \right) + \lambda \phi \left( x \right) = 0
    [/tex]

    The boundary conditions (the string being clamped at x = 0 and constrained at x = L in the specified way) results in the following ODE problem.

    [tex]
    \frac{{d^2 \phi }}{{dt^2 }} + \lambda \phi = 0,\phi \left( 0 \right) = 0,\phi '\left( L \right) + k\phi \left( L \right) = 0
    [/tex]

    I found that if lambda is not positive then the ODE has trivial solutions so I take lambda to be positive.

    [tex]
    \phi \left( x \right) = A\cos \left( {\omega x} \right) + b\sin \left( {\omega x} \right),\lambda = \omega ^2 > 0
    [/tex]

    Plugging in the boundary conditions of the ODE above I get a A = 0 and [tex]F\left( {\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)} \right) = 0[/tex]. I take F not equal to zero to avoid a trivial solution to the ODE. So I get

    [tex]
    {\sin \left( {\omega L} \right) + k\omega \cos \left( {\omega L} \right)}
    [/tex] = 0

    which does not correspond to the given result. I also can't figure out how many positive solutions for omega the equation that I needed to derive in part (a) has.

    I don't think I can do part (b) without having completed part (a) because I need to sum over a sequence of functions whose arguments contain omega to find the solution to the PDE.

    Can someone give me some help? Thanks.
     
    Last edited: Sep 12, 2006
  2. jcsd
  3. Sep 12, 2006 #2

    HallsofIvy

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    Science Advisor

    You were good up to this point but I think perhaps you got [itex]\phi[/itex] and [itex]\phi'[/itex] confused. The general solution to your equation is, as you say
    [tex]
    \phi \left( x \right) = A\cos \left( {\omega x} \right) + b\sin \left( {\omega x} \right),\lambda = \omega ^2 > 0
    [/tex]
    (Any reason for using b instead of B here?)
    Certainly, since
    [tex]\phi(0)= A= 0[/tex]
    we have
    [tex]
    \phi \left( x \right) = b\sin \left( {\omega x} \right)
    [/tex]
    so that
    [tex]
    \phi' \left( x \right) = b\cos \left( {\omega x} \right)
    [/tex]
    [tex]\phi'(L)+ k\phi(L)= b\omega cos(\omega L)+ kbsin(\omega L)[/tex]
    [tex]= b(\omega cos(\omega L)+ k sin(\omega L))= 0[/tex]
    In order that b not be 0, so that [itex]\phi[/itex] be non-trivial, you must have
    [tex]\omega cos(\omega L)+ k sin(\omega L)= 0[/tex]
    or, dividing by [itex]cos(\omega L)[/itex],
    [tex]\omega+ k tan(\omega L)= 0[/tex]
    exactly the condition given.
     
  4. Sep 13, 2006 #3
    It looks I just pulled an "F" from out of nowhere but I think I just misread my working while I was transferring it from my work book and onto PF. So there was no particular reason for choosing b, I just wasn't thinking straight, not enough sleep.

    Anyway thanks for pointing out my mistake. I see how to get to that equation now.

    [tex]
    k\tan \left( {\omega L} \right) + \omega = 0
    [/tex]

    [tex]
    \tan \left( {\omega L} \right) = - \frac{\omega }{k}
    [/tex]

    I'm not sure how to determine how many positive solutions there are for omega. I think I need to actually find an explicit solution for omega to solve the PDE, is it possible or ncessary to solve for omega to solve the PDE?

    One more thing, for a PDE what's the difference, if any, between a boundary condition and an initial condition? Any help would be great thanks.
     
  5. Sep 13, 2006 #4

    HallsofIvy

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    Science Advisor

    tangent goes to negative infinity at all odd multiples of [itex]\pi[/itex] so the graph of [itex]y= tan(\omega L)[/itex] will cross the straightline graph of [itex]y= -\frac{\omega}{k}[/itex] infinitely often.
     
  6. Sep 14, 2006 #5
    Ok but I don't know how I can isolate omega. To construct an expression for omega I tried considering if there were any special values of x such that tan(x) = x but I couldn't recall any such values. I also tried considering the equation in terms of cosine and sine...wcos(wL) + ksin(wL) = 0. If wL = (3/4)pi then I sort of get the LHS = zero. (off by some constant multiples of course) But apart from that I haven't gotten anywhere.

    Edit: Nevermind, it's impossible to find an explicit expression for omega.
     
    Last edited: Sep 14, 2006
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