Wave Equation: Solve for Boundary Condition & Incident/Reflected Waves

[Kate2010]

Homework Statement

A stretched string occupies the semi-infinite interval -\infty<x\leq0.
y(x,t) := f(x-ct) + f(-x-ct) is a solution of the wave equation.

What boundary condition does y satisfy at x=0?

Describe what is going on in terms of incident and reflected waves.

Homework Equations

The Attempt at a Solution

Is the boundary condition just y(0,t) = 2f(-ct)?

At x=0, the displacement varies as a function of time so the end is not fixed. However, I'm unsure about how this relates to incident and reflected waves.

(I worked out that if y(x,t) = f(x-ct) - f(-x-ct) then f(x-ct) represented the incident wave and -f(-x-ct) represented the reflected wave.)

 

[vela]

Since the string is stretched, does that mean the end of the string is fixed at x=0?

 

[LCKurtz]

Homework Statement

A stretched string occupies the semi-infinite interval -\infty<x\leq0.
y(x,t) := f(x-ct) + f(-x-ct) is a solution of the wave equation.

That first term should be f(x+ct), right? And shouldn't the whole expression be divided by 2? And aren't you missing another term unless the string is released from rest?

What boundary condition does y satisfy at x=0?

Describe what is going on in terms of incident and reflected waves.

I suspect Vela's hint that your string is tied down at x = 0 is correct. The idea is to model a semi infinite string as one half of an infinite string. The trick is to have the initial condition be such that the infinite string has a node at x = 0. Consider what happens if f(x) is an odd function.

 

[Kate2010]

The solution I wrote is definitely the one I have been given to work with, and I checked and it does satisfy the wave equation.

So putting x = 0 I do get y(0,t) = f(-ct) +f(-ct).

If it is odd as you suggeest, -f(ct)=f(-ct), then y(0,t) = f(-ct) - f(ct), which isn't 0?

So I don't know how given this solution to the wave equation that it can have a node at x=0? Could it not have a light ring at x=0 on a wire perpendicular to the string and slide up and down on that?

 

[vela]

I think you're right about the end not being fixed. If the end is free to move, I seem to recall you get a boundary condition like dy/dx=0 or something of that sort.

 

[Kate2010]

Yes I agree with that :)

What I'm still a bit puzzled about is what this means in terms of incident and reflected waves, any ideas?

 

[vela]

At t=0, the incident wave is given by f(x) and the reflected wave by f(-x). If you were to plot these functions, the reflected wave would be the mirror image of the incident wave about the y axis. (The reflected wave doesn't really exist where x>0, but just pretend it does.) As t increases, the incident wave propagates to the right while the reflected wave propagates to the left.

How would you describe how the wave is reflected when it hits x=0, e.g. does the shape change and is there a phase change?

 

[Kate2010]

If dy/dx=0 does this mean that the end is not moving up and down? I don't think so as this doesn't make sense as I have established the vertical displacement changes with time. So does it just mean that it isn't moving right and left?

If we have the end moving up and down, there is no phase change in the reflected wave, it is just starting from a different point? Kind of like the soft boundary on this page? http://paws.kettering.edu/~drussell/Demos/reflect/reflect.html

 

[vela]

If dy/dx=0 does this mean that the end is not moving up and down? I don't think so as this doesn't make sense as I have established the vertical displacement changes with time. So does it just mean that it isn't moving right and left?

dy/dx is the slope of the shape of the string, so dy/dx=0 at x=0 means the end of the string is horizontal.

If we have the end moving up and down, there is no phase change in the reflected wave, it is just starting from a different point? Kind of like the soft boundary on this page? http://paws.kettering.edu/~drussell/Demos/reflect/reflect.html

Yes, this problem is exactly the soft boundary case on that page. When the end is free to move, the reflection looks like the incident wave flipped around horizontally and moving in the opposite direction.

 

[Kate2010]

Thank you!