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Some more wave confusion

  1. May 1, 2005 #1

    quasar987

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    First, consult the figure at the end of the first page (Problem 3):

    http://www.iro.umontreal.ca/~montanom/cours/phy/1620_tp_8.pdf

    The problem reads: "A rope of lenght L and mass M is stretched with a tension T between the 2 rings of negligible mass allowed to slide w/o friction along the lines parallel to the y axis. Initially, the rings are maintained at y = 0, while we give the rope a shape of the form y(x, t=0) = Asin²(pi x/L) as illustrated. We then let go of the rope and the rings at time t = 0.

    a) Draw the periodical function corresponding to the definition of the problem and give its spatial period.

    b) Give the complete expression of the motion of the rope as a function of time in terms of its Fourier components."

    The answer to b) is

    [tex]y(x,t) = \left( \frac{A}{2} - \frac{A}{2}cos\left(\frac{2\pi x}{L}\right)\right) cos(\omega t)[/tex]

    Where the part in parenthesis is just the fourier expansion of Asin²(pi x/L). From this answer I read 2 surprising things

    1) Judging from the term cos(xt), the rope oscillate in a normal mode.

    2) At x = 0 and x = L, the displacement y is 0 at all times. So the resulting motion is exactly the same as if the rope was glued to a wall.

    My question is:

    How am I suppose to know what will happen to this rope when I release it? I.e. how was I supposed to know that I all I needed to do was multiply the fourier term by cos(wt)?! Stated otherwise: given an initial configuration of the rope, how can I predict how it will behave when released? Does any deformation on a rope whatesoever vibrate in a normal mode when released?!
     
    Last edited: May 1, 2005
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  3. May 1, 2005 #2

    robphy

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    Check your "answer to b)". [It's dimensionally incorrect.]
    With the rings free to move as shown, this is like a pipe with open ends.
    ( d=A ? )
     
  4. May 1, 2005 #3

    Integral

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    This is not correct, when they are released at t=0, the rings are free to move.

    You are given the initial condition for x, you need to supply the rest of the initial conditions which would be:

    yt(x,0)=0

    and the boundary conditions

    yx(0,t)=yx(L,t)=0

    this is the boundary condition to simulate ends free to move.
    Now simply complete the solution to the wave equation
     
  5. May 1, 2005 #4

    quasar987

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    You've got quite the eye. I fixed that. I also replaced 'd' by 'A'.

    Integral, the solution I have posted is the one suplied by the teacher, not one I have found myself. I know the ends are free to move. But his solution says that they don't. For when I plug x = 0 or x = L, cos = 1 ==> y = 0 regardless of t.

    I'm not sure what that means, but it surely implies that you think the resulting motion is a solution of the wave equation. What makes you think that?

    Thanks.
     
  6. May 1, 2005 #5

    OlderDan

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    Assuming you can neglect gravity (otherwise the whole thing slides down, or if the lines are horizontal the massive rope gets pulled out of the plane), there is no net external force acting on the system. It's center of mass must be stationary. The symmetry of the initial displacement must be preserved, so whatever happens is going to happen symmetrically about the midpoint between the the lines at x = 0 and x = L. The center of mass is located halfway from the line between the rings to the crest of the displacement, at the point (L/2, A/2) and there it must stay.

    I suspect the equation is written incorrectly. The A/2 term should be a constant representing the center of mass position, with all motion relative to that.
     
    Last edited: May 1, 2005
  7. May 1, 2005 #6

    Integral

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    Aren't you dealing with solutions of the wave equation? What else do you think it would be? Let's see, you have a string wiht a given mass and tension, you have a prescribed initial distribution and boundry condions. This is a forumlation for the wave equation.

    The question is, does the given solution satisfy the initial and boundry conditions. If it does then it is correct, if it does not then it is incorrect. You need to verify that it does. The actual problem seems to be that YOU should find the solution and verify that the given solution is what you get.

    The fact that the ends are free to move does not mean they must. Consider a standing wave.
     
  8. May 1, 2005 #7

    quasar987

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    Actually the solution he gave us is

    [tex]y(x,t) = \frac{A}{2} - \frac{A}{2}cos\left(\frac{2\pi x}{L}\right) cos(\omega t)[/tex]

    I was just so sure he had made a typo because all he does is find the fourier expansion of the wave and then BAM, give us this final answer. No boundary thingy, no mention of the center of mass, nothing. It's very frustrating! :mad:

    I would have to say: only if A is small, such that the angle of the slope of the string is small. See https://www.physicsforums.com/showthread.php?t=73743 .

    Is it like the uniqueness theorems of electrostatics? If I find a solution that satisfies the wave equation, and whose boundary conditions match the actual problem, then necessarily, this is THE solution?
     
  9. May 1, 2005 #8

    OlderDan

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    As you can see, that gives a whole different picture to the motion of the ends of the string.

    Small A is a necessary assumption by the statement of the problem. You are to assume constant mass per unit length and constant tension in the rope. If the rope actually looked like that picture the first movement of the rope would eliminate tension. Even if the rope were ideally elastic, the first motion would result in decreased tension, and a change in mass per unit length. Make the reasonable assumptions and treat the problem as a wave problem.
     
  10. May 1, 2005 #9

    quasar987

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    Okay, okay, this is starting to get much clearer!

    There are still a few more points though.

    Suppose the boundary conditions are that the ends are glued to the wall. I have verified that

    [tex]y(x,t) = \left( \frac{A}{2} - \frac{A}{2}cos\left(\frac{2\pi x}{L}\right)\right) cos(\omega t)[/tex]

    satisfies the wave equation as well as the 3 boundary conditions. Now this equation describes a normal mode of vibration. And when the rope will pass through the equilibrium position (what you called the horizontal line), its CM will be located at (L/2,0). So it will have moved. Find the error. :grumpy:
     
  11. May 1, 2005 #10

    OlderDan

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    In the problem you presented, the ends are not glued to the wall. If they were, the center of mass of the rope could move because there would be a net external force acting. In your problem, the initial conditions are the shape of the rope at time zero. At that moment in time you are allowing the ends to move freely. What happened before that time to give the rope the shape it has at time zero is no longer relevant. After time zero, how does the shape of the rope evolve? The center of mass from that time on will remain at (L/2, A/2) The revised solution you posted earlier satisfies this condition.
     
  12. May 2, 2005 #11

    quasar987

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    Oh yes, the force of the suposedly immovable wall acting on the ends of the rope..

    Last thing: in finding the Fourier representation of the function Asin²(pi x/L), how does one determine the shape of the function as a whole, i.e. over many cycles? From what I clues I have gathered, the function must be even around free ends of the rope and odd around fixed ends, or something to that effect.

    What is the whole logic behind this? What is the physical significance (if any) of this "extended spatial wave form"?
     
  13. May 2, 2005 #12

    OlderDan

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    Symmetry is an important consideration, and symmetry depends on which point you choose for the center of symmetry. If you choose x = 0 at one end as shown in your diagram, then your odd (even) for free (fixed) ends applies. If you choose the center between the the ends as the center of symmetry, the wave must be even in both cases.

    The logic of the extended spacial waveform is that the mathematical representation of the wave between the lines cannot stop there. The domains of the functions that represent the waveform in the 0<x<L region extend from minus infinity to plus infinity. The good news is that those functions are all periodic with period L (many are periodic with periods that are integer subdivision of L, but that means they are also periodic over length L). Every length L of the extended waveform looks exactly like every other length L, including the length L that is the region of interest.

    The solutions to the wave equation are all sinusoidal functions of x. A combination of those functions will produce the shape of the initial waveform, and their time evolution will give the shape of the waveform at all later times.
     
  14. May 2, 2005 #13

    quasar987

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    I've organized my thoughts in a "theory". Tell me if this is right.

    Fourier's theorem allows us to represent any periodic function as an infinite sum of sine and cosine. If I'd like to represent the shape of a wave at a time t_0 with a Fourier serie, I must first pretend that its shape is part of a periodic function f(x) extending to infinity on both sides. In principle, ANY periodic function f(x) that looks like my waveform between x=0 and x=L will do the trick and I just have to work out its fourier expansion. However, in view of the fact that A_n = 0 for odd function and B_n = 0 for even functions (about the origin), it will be convenient to chose f(x) to be even or odd. And it will be even more convenient to get an f(x) of period L or 2L because it will make it easier to evaluate the integral of the A_n or B_n.
     
  15. May 3, 2005 #14

    OlderDan

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    This is pretty close. You have a function defined in the interval 0<x<L, or with x = 0 in the middle, in which case the usual choice is to make the range (-L/2)<x<(L/2) so that the period of the function is L. If the value of y(x) is the same at both boundaries, then the extended function is continuous and is often continuously differentiable. If not, you get a big contribution from high frequency Fourier components that approximate the discontinuity as you cross the period boundaries. The extended function is determined by the Fourier compents obtained by decomposing the waveform in the one interval you have. it's not really a matter of you picking an extended waveform. It is a matter of deriving the coefficients of the extended function to match the function in the interval you have. There are extensions of Fourier analysis where you don't always have a perfectly defined interval and some extension techniques are needed, but those are usually to force the extended function to a constant (often zero) outside the region of interest. And then there are things that don't really have a finite interval of length L, and are assumed to extend to infinity. Things like a tsunami wave and similar "wave packets" would involve such analysis.

    Symmetry is great when you have it, but you cannot create it if you don't. A nice example of a wave form that is neither odd nor even is a string being "plucked" off center. Imagine a string fixed at both ends as in a musical instrument being plucked at an arbitrary position. The intial wave form is then two straight line segments from the pluck position to the ends. The extended waveform will be that same shape repeated over and over again. It will be neither odd nor even. The Fourier components will be different, depending on where the string is plucked, even though the fundamental frequency will be the same. A good classical guitarist exploits this phenomenon to produce tones of different quality by plucking the strings in different places.
     
    Last edited: May 3, 2005
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