# Solving 1D wave equation for a flag (not attatched at both ends)

1. Nov 7, 2013

### Flucky

Hi all, I have the question:

Consider a flag blowing in the wind. Assume the transverse wave propagating along the flag is one dimensional.

Solve the wave equation for the wave on the flag, assuming the displacement of the flag is zero at the flag pole and the other end of the flag is free. The initial condition of the free end of the flag has maximum displacement ymax and zero velocity.

All the derivations I find online are for strings attached at both ends (looking specifically at this site here). In this derivation it considers tension in the string to be going both ways but if it's a flag it will only be acting one way.

Any pointers in how to go about this?

Cheers

2. Nov 7, 2013

### Simon Bridge

You write down the wave-equations and solve it as a boundary-value problem.

You need to make a decision about how to define the boundary values.
i.e. flags are usually not very stretchy, so the overall length won't change ... but you may not be advanced enough to handle that sort of thing. The question reads to me like you are expected to treat it as "a string with one free end". So google for that.

3. Nov 8, 2013

### BruceW

That website does give the correct equation for your problem. There will still be tension, even though one end is free. Also, if there is tension in one direction, there will also be tension in the other direction, that's pretty much how tension is defined. (In your flag situation, there will be tension in both directions). So you've got the right equation, now you need to implement the boundary conditions. The 'free at one end' boundary condition is not completely obvious. (I could not guess what it was when I first tried this kind of problem). Maybe have a think about it. Or look around on the interwebs :)

4. Nov 8, 2013

### Flucky

Thanks both I'll look into the boundary conditions today and get back to you.

5. Nov 8, 2013

### AlephZero

This is not the same situation with uniform tension in a string. The tension will vary from a maximum at the flag pole to zero at the free end.

Without some more context, we don't know at what level of answer is expected to the question in your OP. You could make some fairly simple assumptions, or it could turn into a research project in non-lnear coupled fluid-solid interaction

6. Nov 8, 2013

### BruceW

I don't see why the tension would go to zero at the free end. I think the string equation is probably an OK approximation, so therefore the tension would not go to zero at the free end. I'd guess the main thing that would make it a bad approximation is due to the flag drooping over. But Flucky is told that the problem is one dimensional, so I suppose he doesn't need to worry about the drooping of the flag. You're right though, there is a lot of stuff he could do with this problem, rather than going for the simple answer.

7. Nov 8, 2013

### AlephZero

The forces on the flag must be consistent with Newton's laws. If the tension at the free end is non-zero, what is the equal and opposite reaction force?

The tension along the length of the flag is similar to the tension in a heavy string hanging downwards. The drag force from the air flow distributed along the length is analogous to the weight of the heavy string.

8. Nov 8, 2013

### Simon Bridge

AlephZero is correct. However,I suspect that, if this were, indeed, an exercise in a course in non-linear dynamics, the question would be phrased somewhat differently.
Assignments often take advantage of the education level to save typing.

9. Nov 9, 2013

### BruceW

this is what I meant about the boundary condition for the free end being 'not obvious'. Say $u(x,t)$ is the vertical displacement of the string from equilibrium at position $x$ and time $t$. And let's say the string is fixed at $x=0$ and free at $x=L$. Then the boundary condition at $x=L$ is that the component of tension in the string in the vertical direction must be zero at $x=L$. But the component of tension in the horizontal direction at $x=L$ is generally nonzero. So the tension in the string at $x=L$ is going to be generally non-zero for the boundary condition of a 'free end'.

edit: and since the tension is simply in the direction of the tangent to the string, this means the string must be horizontal at $x=L$, and it has generally nonzero tension here.

edit again: I guess physically, the tension of the string right at the end of the string is zero. But inside the string, the tension is nonzero. So there is going to be a discontinuous change in the tension of the string at exactly the end of the piece of string. for sure, the tension in the string does not gradually go to zero as you approach the end of the string. (at least in the simple string model which the OP'er is using).

Last edited: Nov 9, 2013
10. Nov 9, 2013

### AlephZero

I have no idea what model the OP is using, or is supposed to be using (the two might be different!)

I agree the question as stated in the OP is poorly defined. Inventing an interpretation of it that matches one's one ideas/prejudices about what the answer should be isn't likely to be much help, though.

11. Nov 9, 2013

### Simon Bridge

The approach needed is going to be determined by the broader context of the coursework already completed.
We don't know what that is. We've done what we can. Lets wait for Flucky to get back to us.

12. Nov 11, 2013

### Flucky

Hi chaps sorry for the delay, had a stab at the question-

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13. Nov 11, 2013

### Simon Bridge

Please explain why you chose the boundary conditions on page 1.
(And what is that symbol you are using for the function - looks a bit like a lower-case delta?)

14. Nov 11, 2013

### Flucky

The function is meant to be an f. Got to admit I'm not entirely sure on boundary conditions but I assumed the flag was length L (lowercase) for it

15. Nov 11, 2013

### Simon Bridge

<squints> Oh I see it: you do your effs the opposite to how I do.
You wrote an upper-case L though... I just figured...
To make sure your boundary conditions make sense - just write them back out in plain language.
Guessing that f(x,t) is the transverse displacement of the fabric of the flag from position x at time t, you said:

1. $f(0,t)=f(L,t)$ says: the displacement of the part of the flag close to the flagpole (x=0) is always the same as the displacement of the far end of the flag.

2. $f(x,0)\neq 0$ just says that no part of the flag (x) has zero displacement at the start (t=0).
The entire flag is displaced - and no part of it crosses the f=0 line!
That could happen - be if you just shifted the flagpole to one side.

3. $\partial_t f(x,0)=0$ says that the initial velocity of everywhere in the flag is zero.
The whole flag is rigid and stationary.
(I'm guessing you meant the partial wrt t there ... it looks like a k to me though.)

Did you read the discussion above at all?
Usually, if you put a curve in a flag, the distance from flagpole to tip gets smaller.
But I'm guessing from your response that you are not expected to take that into account.

To get sensible boundary conditions - first set up your coordinates: how you are describing the flag:

You appear to be modelling the flag as waves on a string - the end of the flag stays at x=L, and has a transverse displacement with time according to y=f(L,t). The displacement of part x in the flag is f(x,t).
... you should normally be more careful than that - the important thing is to make sure that every variable has a definition in normal words to say what it is. Assume the reader is too thick to figure out what f stands for or that L is the unstretched length of the flag.

Next: identify the words in the question that inform you about the boundary conditions ... write them out separately, as a list, then translate those words into the math symbols you have just defined.
This is actually the physics part of the problem - the rest is just maths.

The question gives you information on:
1. the displacement of the part closest the flagpole for all time: $f(0,t)=$
2. the initial displacement of the end farthest from the flagpole: $f(L,0)=$
3. the initial velocity of the end farthest from the flagpole: $\partial_t f(L,0)=$

If you are familiar with different kids of solutions to the wave equation, it is actually giving you more information than that. The choice of model (waves on a string) also restricts the kinds of solutions you can get. Whatever, for a second order PDE you need a certain number of initial conditions in order to find a unique solution.

Last edited: Nov 11, 2013
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