Waves on Tight Strings: Boundary Condition Problem

But you should end up with a r that involves only k and m, not \rho (density).In summary, the conversation discusses a string attached to a ring, subject to tension T and with a mass per unit length of \rho. The displacement of the string from its equilibrium position is described by y(x,t). The boundary condition at the ring is provided by the equation T\frac{\delta y}{\delta x}(0,t) = m \frac{\delta^{2} y}{\delta t^{2}}(0,t) and is a consequence of Newton's Second Law of Motion. In the case of an incident wave y(x,t) = e^{i(-kx-\omega t)}, the total displacement is
  • #1
MattMark'90
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0

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.
 
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  • #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.
 

1. What is the "boundary condition problem" in the context of waves on tight strings?

The boundary condition problem refers to the specific conditions that must be met at the ends of a tight string in order for waves to propagate along it. These conditions include the string being fixed or free at the ends, as well as the tension and displacement of the string being continuous at the boundaries.

2. How do the boundary conditions affect the behavior of waves on tight strings?

The boundary conditions play a crucial role in determining the types of waves that can propagate along a tight string. For example, a fixed end will result in the formation of standing waves, while a free end will allow for the propagation of both standing and traveling waves.

3. What are the solutions to the boundary condition problem for waves on tight strings?

The solutions to the boundary condition problem depend on the specific conditions at the ends of the string. In general, the solutions can be expressed as a combination of standing and traveling waves with specific amplitudes and frequencies that satisfy the boundary conditions.

4. How does the tension of the string affect the boundary condition problem?

The tension of the string is a key factor in determining the behavior of waves on tight strings and therefore affects the boundary condition problem. A higher tension will result in a higher wave speed and shorter wavelengths, while a lower tension will result in a slower wave speed and longer wavelengths.

5. Are there any real-world applications of the boundary condition problem for waves on tight strings?

Yes, there are many real-world applications of this problem, such as in musical instruments like guitars and violins, where the strings must be fixed at the ends and the boundary conditions must be carefully controlled to produce specific frequencies and harmonics. This problem also applies to engineering structures like suspension bridges, where the tension and boundary conditions of the cables must be carefully designed to withstand different types of waves and loads.

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