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Mechanical wave

  1. Mar 10, 2007 #1
    consider a pulse traveling down a stretched rope. what makes the pulse to travel in a particular direction (let say to the right) and not the other?
     
  2. jcsd
  3. Mar 10, 2007 #2
    Draw it out; essentially when you draw it you should setup a force body diagram on what's exactly happening on the rope.

    After, you will then realize why it travels in a particular direction.

    If you're still having troubles; make a rough sketch or something of what you think is happening with the force body diagram. However, I can write it down/scan it and post it up if you cannot and you can reaffirm your thoughts.
     
  4. Mar 10, 2007 #3

    russ_watters

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    Unless I'm misunderstanding the question, an oscillation travels in both directions unless it is started at the end of the rope.
     
  5. Mar 10, 2007 #4

    Doc Al

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    As AngeloG was alluding to, the forces on each side of a traveling pulse are not the same. On the leading edge, the forces accelerate the rope away from the undisturbed position; on the trailing edge, back.

    And as Russ says, if you just pluck the rope symmetrically, you'll get pulses in both directions.
     
  6. Mar 10, 2007 #5
    how can i post my diagram if i have no scanner?
    which program should i use to draw it? AngeloG, can you post your diagram? thank you.

    oh, so if i pluck the rope anywhere (not exactly to be in the middle) but not the ends of the rope, the pulses will be in both direction? but how to explain it? what is the mechanism in it?

    i m not quite understand what do you mean here. are you saying that the forces on both sides of the pulse are different? what makes it different?

    thank you for answering me
     
  7. Mar 11, 2007 #6
    Yes, if you pluck a rope somewhere other than at one of the ends, the wave will travel in both directions. It's because there's no preferred direction for the wave to travel. Furthermore, the wave equation allows for disturbances in both directions. The only reason a disturbance at one end of a string will cause a disturbance in a single direction is because the wave has nowhere to go in the other.

    This actually screwed me up for quite awhile when I was doing my senior thesis. At one point I had to write a program to get a wave to propagate in one direction down a string. For some reason, the wave propagated in both directions. It turns out that it was because I had simply put the wave equation into my program without recognizing that disturbances can travel in both directions. I had to factor the wave equation into two equations. Each factor is called a "baby wave equation," and represents wave motion in one direction. I've attached the calculation below as an image, so that you can see how this works.
     

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    Last edited: Mar 11, 2007
  8. Mar 11, 2007 #7

    vanesch

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    The point is that dynamics in mechanical systems is described by second-order differential equations in time. This means that the "information" that is needed (the initial conditions) for the system to "know" what to do next, is of first order in time: so not only the immediate POSITION (the form of the rope) counts, but also its first derivative wrt time (the velocity of the different parts of the rope).

    This is why, when you start with a STILL situation (pluck the rope) and have a first derivative wrt time implicitly set to 0, that you get a different solution, than when you consider a wave that "was already travelling" (and which has, when it takes on the same immediate position, does not have the first derivative wrt time equal to 0).

    So, although the immediate position (form, f(x)) at a given time t0 can be the same, the solution can be different simply because the other half of the initial conditions needed (the first time derivative) will be different. It is this difference (which would give nevertheless the same photo of the rope at t0), which makes a travelling wave, eh, travel in a certain direction, and a plucked rope send out travelling waves in both directions.
     
  9. Mar 11, 2007 #8

    russ_watters

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    well, what happens if you pluck a string quickly? You make a little symmetrical dimple in it, with each little element of the string pulling on the section next to it. And that's how a transverse wave propagates.
     
  10. Mar 12, 2007 #9
    Thank you for your explanation. I get some picture after reading it (although i still need time to understand it completely).

    Most of what i found on here are explaining the mechanism of PLUCKING the string or INITIATING the pulse. I have understand that part, thank you. However, i still wondering that what is the mechanism of a TRAVELLING pulse that make it continue to travel in a particular direction.

    Just like Russ_Watters say here. When the pulse is travelling to the right of a string, why is that the little element at the peak only pull the section next to it (on its right) but never pull the section before it (on its left)? How to explain it in physical view (although vanesch has explained it by methametical means that the 1st time derivatives, which is the velocity of each element, also contributes to the determining of its motion)? thank you.
     
    Last edited: Mar 12, 2007
  11. Mar 12, 2007 #10

    vanesch

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    Well, a travelling pulse, at any moment t1, will have:
    1) a certain form f(x)
    2) a certain time derivative of the form d/dt f(x)

    (these are independently specified in initial conditions, but have a linked relationship in a pulse "that is already travelling in one direction").

    These initial conditions are exactly what is necessary to have as a dynamical evolution at t1 + dt:
    1) a certain form f(x - v.dt)
    2) the time derivative d/dt f(x - v.dt)

    In other words, the pulse is now displaced over v.dt to the right, and the initial conditions are exactly the same there now. So this continues for ever.

    If you think that it is kind of a miracle that these initial conditions come out EXACTLY as it should, it isn't in fact. The solution set of the wave equation is just so, that the most general solution can be written as:
    f(x-vt) + g(x+vt) with f and g arbitrary functions. If you *already have* a travelling wave which means that you only have the first term in your solution, then this is exactly what is needed to satisfy the above "miraculous" conditions.

    So the "miracle" is in your specification of already having a pulse travelling in one direction. Once it does (and hence takes on the form f(x-vt)), then the wave equation is simply such that it continues to do so, and that, at any moment t1, there's the correct link between the instantaneous form and its time derivative, to serve as the right initial conditions to continue the movement as it was set off.
     
  12. Mar 12, 2007 #11

    russ_watters

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    As I said, the deformation of the string is symmetrical. So it has to travel in both directions.
     
  13. Mar 13, 2007 #12
    I think he means once the wave has already formed, what keeps one pulse from reversing direction.

    The answer to that question is conservation of momentum, the wave carries with it some momentume, and a such it has to keep going in whatever direction it wa going in originally.
     
  14. Mar 13, 2007 #13
    Thank you Vanesch. I get what you want to say. You are trying to say that, when the pulse is formed, from the general solution, f(x-vt) + g(x+vt), we know that the pulse will travel in both ways, just like russ watters has told me. However, when the pulse is travelling in one direction, the instantaneous f(x) and d/dt f(x) allow it to continue to travel in the same direction. The pattern of f(x) and d/dt f(x) would not change but they are just looked like moving with the pulse. Hence, they keep the pulse moving on that direction. Am i correct?
    But these all only valid if we can describe the system by 2nd derivative in time, right? I m not sure what do you mean by this, is it the wave equation (d'Almbert's equation, the first equation that in arunma's calculation)? Or any other equation? Can you please tell me what is the equation? thank you.

    CPL.Luke, ya, you get what i meant. But in classical view, does a pulse carries a momentum? How is it? Can you explain it further? thank you.
     
  15. Mar 13, 2007 #14

    russ_watters

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    Actually, no, a longitudinal wave does not carry momentum in the direction it moves.
     
  16. Mar 13, 2007 #15

    Mentz114

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    I can't make out who's talking about transverse waves ( plucked strings) and who's talking about longitudinal waves ( pulses ?).

    If you shake one end of a rope you can get a transverse wave travelling in one direction.
     
  17. Mar 13, 2007 #16

    Doc Al

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    Traveling pulses along a stretched rope are transverse waves.
     
  18. Mar 13, 2007 #17

    russ_watters

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    Sorry, that was an error in my post. I'll leave it since several people replied, but that should have said there is no momentum carried in transverse waves. Really, though, I wouldn't say that either type of wave carries momentum. Not in the same way as light, for example, carries momentum. Any particle on or influenced by the wave will oscillate and have a net displacement of zero.
     
    Last edited: Mar 13, 2007
  19. Mar 14, 2007 #18

    vanesch

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    Yes. What I meant to say was, that if the pulse is already travelling in the right direction, then if you take a specific instant t1, and you look at the dynamical situation at that moment, and consider that as the "initial conditions" for the time evolution that will follow after t1, then the relationship between "configuration" (that's f(x)) and "momentum" (the time derivative of it) in these initial conditions will be such that the pulse will simply travel on in the same direction.

    Yes, it is the standard wave equation, the first one in arunma's calculation.

    Note that this is very similar to a much simpler problem. A ball is thrown in space, and has a certain velocity and position. Now, imagine we look at the ball's motion at t1. It is at a certain position. But that by itself doesn't give you the right motion at the next instant ! You also need the right velocity (or momentum) at t1. Now, in this case, this is trivial because it is a constant momentum. But if you take, at t1, as "initial conditions" the position of the ball at t1, and the momentum of the ball at t1 (which is constant...), then, oh miracle, the ball will continue EXACTLY the same motion as it was doing up to then. So what made the ball move "in the right direction" ? Its initial conditions containing both position and momentum.

    It is the fact that Newton's equation is second-order in time that makes us need initial conditions which are both position and velocity. If Newton's equations were first order in time, position would have been enough, and there wouldn't have been a means to "transmit" the momentum to the next instant.
     
  20. Mar 17, 2007 #19
    Oh, Ok. thank you very much. I understand already. thank you.
     
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