Waves on Tight Strings: Boundary Condition Problem

[MattMark'90]

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 \rho. The displacement of the string from its equilibrium position along the x -axis is y(x,t).

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

T \frac{\delta y}{\delta x}(0,t) = m \frac{\delta^{2} y}{\delta t^{2}}(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^{i(-kx-\omega t)}, where \frac{\omega}{k} = \sqrt{\frac{T}{\rho}}, write the total displacement as;

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

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^{i(-\omega t)} = m(-\omega^{2} -\omega^{2}r)e^{i(-\omega t)

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

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.

 

[nickjer]

(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 \rho term in the equation.