# Waves on Tight Strings: Boundary Condition Problem

1. Apr 13, 2010

### MattMark'90

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 $$\rho$$. 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.

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$$^{i(-\omega t)}$$ = 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.

2. Apr 13, 2010

### 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.