- #1
MattMark'90
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Homework Statement
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.
Homework Equations
I think everything needed to solve this question is included in the problem description.
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.