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Waves on Tight Strings: Boundary Condition Problem

  1. Apr 13, 2010 #1
    1. The problem statement, all variables and given/known data
    A string is attached to a ring of mass m which is free to move up and down a frictionless pole. The string is subject to tension T and its mass per unit length is [tex]\rho[/tex]. The displacement of the string from its equilibrium position along the x -axis is y(x,t).

    The boundary condition at the ring, provided [tex]\frac{\delta y}{\delta x}[/tex]<<1, is

    T [tex]\frac{\delta y}{\delta x}[/tex](0,t) = m [tex]\frac{\delta^{2} y}{\delta t^{2}}[/tex](0,t)​

    (i) This boundary condition is the consequence of which physical law?
    (ii) For the case of an incident wave y(x,t) = e[tex]^{i(-kx-\omega t)}[/tex], where [tex]\frac{\omega}{k}[/tex] = [tex]\sqrt{\frac{T}{\rho}}[/tex], write the total displacement as;

    y(x,t) = e[tex]^{i(-kx-\omega t)}[/tex] + re[tex]^{i(kx-\omega t)}[/tex]​

    Using the general boundary condition, find r as a function of k and m.

    2. Relevant equations

    I think everything needed to solve this question is included in the problem description.

    3. The attempt at a solution

    (i) I believe the boundary condition is a consequence of Newton's Second Law of Motion; the forces acting on a element of string are equated to mass times acceleration. The slope of the string is equal is equal to the tension, however, I do not understand why the left hand side is multiplied by Tension.

    (ii) I have calculated the first and second derivatives wrt to x and t respectively and at x=0, the boundary condition has the following form;

    T(-k +kr)ie[tex]^{i(-\omega t)}[/tex] = m(-[tex]\omega^{2}[/tex] -[tex]\omega^{2}r)[/tex]e[tex]^{i(-\omega t)[/tex]

    Cancelling the exponents this gives;
    T(-k +kr)i = m(-[tex]\omega^{2}[/tex] -[tex]\omega^{2}r)[/tex]

    From here I have tried using the relationship given in the problem statement to eliminate T but just end up introducing another unwanted variable. I cannot seem to form a relationship involving only k and m.
    Since I need a real answer I suppose I need to remove the complex numbers in the LHS of the equation but don't see how I could achieve this.
     
  2. jcsd
  3. Apr 13, 2010 #2
    (i) Well the simplest answer is that the derivative is unitless and you need T to give it units of force. More detailed answer is that you originally have tension T, and you describe dy/dx as the y-component of the force T. So the total force in the y-direction is T*dy/dx

    (ii) Can't help you here. I also get a [tex]\rho[/tex] term in the equation.
     
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