Can somebody explain boundary conditions, for normal modes, on a wire?

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I don't really understand boundary conditions and i've been trying to research it for ages now but to no real avail. I understand what boundary conditions are, I think. You need them along with the initial conditions of a wire/string in order to describe the shape of motion of the string. I guess giving a question example would reveal my lack of knowledge better than me trying to explain it:

1)A rope is held under tension, F. It lies along the x-axis when undisturbed. The mass
per unit length is l for x < 0 and r for x > 0. A wave-like disturbance is incident
from the right.

(i) Give and explain the origin of the boundary conditions on the displacement,
y(x), at x = 0.

I'm not entirely sure here? It is simply just: y(x=0,t) = 0? What about this x < 0 and x > 0 business, what does that explicitly mean? Does it give insight into whether or not the wire is fixed at both ends, fixed at only one end, attached to another wire etc etc?

What about for a string fixed at one end but not the other? Supposedly the answer is:

y(0; t) = [itex]\frac{\delta y}{\delta x}[/itex](L, t) = 0

What does this mean?

Thank you for your time.
 

Answers and Replies

  • #2
Simon Bridge
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Boundary conditions are like initial values - except they constrain the differential equations in space.
So, for instance, for a string fixed at both ends, the initial condition will be the initial displacement frunction of the string: y(x,0), the boundary conditions would be that y(0,t)=y(L,t)=0 - because the endpoints are fixed: they don't move.

For your example (1)
The point x=0 is not fixed in place - y(0,t)=0 is not true for all time - when the wave passes that point, that part of the string will lift right?

But what does have to happen is that y(x<0,t) has to match smoothly with y(x>0,t) - there can be no sudden breaks in the wire, and no sharp corners either.

Per your second example:
For a wire fixed at only one end, you have y(0,t)=0 as normal, but you'd commonly have $$\left.\frac{\partial}{\partial x}y(x,t)\right|_{x=L}=0$$ ... which puzzles you.

The derivative of y wrt x is the slope - a slope of zero is "horizontal" - so the equations is saying that the very end of the wire must be horizontal at all times.

IRL the end will make a flapping motion and may vary it's slope a bit - but for a lot of situations that boundary conditions gives a good approximation.
 
  • #3
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I think I understand the second part now, but for the first: y(0,t)=y(L,t)=0 is therefore the answer for the first question? It says that both ends of the wire are fixed doesn't it? x = 0 and x = L?

Okay I think that makes sense now it has been explained, thanks a lot!
 
  • #4
jtbell
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I think I understand the second part now, but for the first: y(0,t)=y(L,t)=0 is therefore the answer for the first question? It says that both ends of the wire are fixed doesn't it? x = 0 and x = L?
Correct, y(0,t)=0 means y = 0, for x = 0 and any value of t; and y(L,t) = 0 means y = 0, for x = L and any value of t.
 
  • #5
Simon Bridge
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I think I understand the second part now, but for the first: y(0,t)=y(L,t)=0 is therefore the answer for the first question?
Um ... I used a string/wire fixed at both ends as an example for how a boundary condition is different from an initial condition.
Your first question does not mention the ends of the wire at all ... implies that the string is infinitely long.

Therefore you only have boundary conditions for x=0.
The way to cope with this sort of thing is to divide the wire into two regions:
$$y(x,t)=\begin{cases}y_1(x,t)&:x < 0\\ y_2(x,t)&: x > 0\end{cases}$$... notice I left off what happens at x=0? That is what the boundary conditions are for.
There's two - one relates the amplitudes of y1 and y2, and the other relates the gradients of y1 and y2.

Have you seen:
http://www.acs.psu.edu/drussell/Demos/reflect/reflect.html
http://www.animations.physics.unsw.edu.au/jw/waves_superposition_reflection.htm
... the last one is good for the demos at the end.

It says that both ends of the wire are fixed doesn't it? x = 0 and x = L?
That's right - and as jtbell points out - if the displacement y at position x and time t is given by y(x,t), then y(a,t)=b is a way of writing: "the displacement at x=a is y=b, for all time".
I could equally write y(x=a,-∞<t<∞)=b

Similarly, if I wrote: y(x,a)=b, then I'd be saying that the displacement is y=b everywhere at time t=a.

It's a good idea to get used to treating maths as a language.
 

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