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KE and PE of a wave in a string

  1. Jan 1, 2014 #1

    Saw

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    I am reading in Halliday that the kinetic energy and potential energy carried by a wave in a string are both maximal at y = 0 (equlibrium position) and zero at y = A or r (maximal elongation).

    This sounds strange to me because usually KE decreases as PE increases and vice versa. Besides, if at any time both types of energy were zero, shouldn't the wave stop?
     
  2. jcsd
  3. Jan 1, 2014 #2
    That doesn't sound right to me! PE should be a minimum at the equilibrium position.
     
  4. Jan 1, 2014 #3

    jtbell

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    Can you give us a page reference (or chapter+section), and an exact quote of the paragraph, so people can see what you read?
     
  5. Jan 1, 2014 #4

    Saw

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    Well, this is what he literally says:

    In fact, if one part were wrong, I tend to think it would not be the part of PE, because actually (as a drawing in the book shows) the string is more stretched at y = 0.
     
  6. Jan 1, 2014 #5

    Saw

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    Fundamentals of Physics, 9th edition, Halliday and Resnick, Jearl Walker, Section 16-7, pages 421 and 422.
     
  7. Jan 1, 2014 #6

    Saw

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    I have investigated a little. This article sheds some light:

    - Actually Halliday, like other text books, purposefully provides this explanation (no typing error): "in a traveling wave the potential and kinetic energies are in phase: they are both maximal together, and they vanish together too. Specifically, at a crest or a trough the kinetic energy vanishes as the string element is momentarily stationary, and the potential energy of the string element vanishes because the string element is horizontal. At zero displacement both are maximal, as both the string transverse velocity and the string’s slope are maximal".
    -As to why the wave does not stop at a crest or a trough, the standard explanation seems to be "That energy has propagated along the string to a material element that had zero energy before, and so on".
    - However, the autor objects that such approach does not fit with standing waves, where precisely there is no energy transfer, no energy propagation.
    - In the author's opinion, the standard derivation fails because it "is based on the stretching amount of the local string element, without worrying what the two endpoints of the strings are doing". After taking this into account, the article concludes that the energy of the wave is always constant and alternates as usual between KE and PE.

    This looks appealing to me, but I do not understand well how considering the end points leads to such conclusion. Can anyone comment?
     
  8. Jan 1, 2014 #7
    Heavens, Why would they be equating tension (stretching) in the string the potential energy? To a first approximation I thought always the tension was assumed to be constant along the string (T) if the ends are fixed.
     
  9. Jan 1, 2014 #8

    AlephZero

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    I don't have the book, but it would have been very helpful if the OP had said the question was about traveling waves, not standing waves.

    I haven't read every word of the arxiv link in detail, but it seems to be settting up a straw-man paradox about tthe difference between the total amount of energy in the wave and the energy in a fixed (small) segment of the string, and then knocking it down again. It seems self evident (to me anyway) that for a traveling wave, (1) the total energy in the wave is constant, but (2) the energy travels at the same speed as the wave.

    You have to do work to rotate an element of the string which is under tension, i.e. change its slope.

    If you isolate an small element of string, there are two ways to find the amount of work. One way (probably the most "obvious" way for students who have not studied continuum mechanics yet) is to find the "force x distance" work done by the tension forces applied to the ends of the string element.

    The other way is to notice that the length of the string segment changes as the slope changes (that is obvious if you think about a large slope, for example 45 degrees) and calculate the change in internal energy. If you know about Green's strain tensor etc in continuum mechanics, this s just "plug and chug", compared with the first way, of drawing a free body diagram and having to think about what is going on.

    Both ways give the same answer, and if the amplitude of the wave is small enough to ignore terms above first order, they both assume the string tension is constant.
     
  10. Jan 1, 2014 #9

    Saw

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    The OP is about both traveling and standing waves. I don’t think that the issue at hand (are KE and PE descending and augmenting in phase or do they alternate?) changes because we focus on traveling or standing waves, does it? In fact, the arxiv that I mentioned assumes so and precisely notes that Hallyday’s explanation does not work in standing waves and uses that as an argument to claim that such explanation is not valid, either, for the travelling ones.

    Sure. The question is only that Halliday, while accepting those postulates, presents a shocking construction where (given that -in their opinion- the components of total energy, KE and PE, move in phase), the energy somehow vanishes at some instants, only to resuscitate later with the same constant value…

    Yes, we can assume that the string’s tension is constant, but the fact is that for the wave to propagate we need the string’s segment affected at each instant by the wave motion to be stretched, don’t we?

    Given this, focusing on the subject of the OP, which approach are you supporting, (i) Halliday’s or (i) the arxiv’s: (i) does the stretching (and hence the EP) go up and down in phase with velocity (hence with EK) or (ii) do they alternate?

    I think they have to alternate, although I do not follow what the arxiv says about the string ends.

    I am thinking that simply the string is communicated a velocity v (maximal KE, PE still zero) and this motion stretches the string, although this effort progressively consumes the v, until the maximum amplitude is reached at the crest (KE zero, maximal PE) and then the cycle continues with compression entailing that the PE is consumed to the benefit of KE, until upon return at equilibrium we get again maximum KE and PE zero…

    Halliday book’s problem may be that they rely on the below drawing as if it were (in their own words) a "snapshot" (status at an instant). That is why they say that at equilibrium position (y = 0) the string element is at the maximum of its stretching. However, this is not a snapshot, but a graphic. In reality the string only reaches maximum stretching when it has consumed all the KE, i.e. when the wave reaches the crest.

    StringHalliday.JPG

    Does it make sense?
     
  11. Jan 2, 2014 #10

    Saw

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    I have checked that in the case of the standing wave, it is generally admitted that the energy sloshes between KE and PE, just as in the spring-mass system, for example. Why should the travelling wave be different? After all, what we call a "standing wave" is just "two travelling waves overlapping". And this overlapping may be relevant for other purposes, but it should not change anything in this particular respect we are concerned with.

    This suggests that the Halliday passage would be wrong, would you agree on that? But what would then be the right explanation, the above mentioned arxiv's or a simpler one as commented in my previous post?
     
  12. Jan 2, 2014 #11
    We might consider what the situation is with a string with one free end where you give it a sharp up and down pull at the fixed end to create a single disturbance travelling along the length. In that circumstance you could regard the point of greatest deflection to be under the most tension with the rest of the string having zero PE and zero KE until the deflection gets there.
     
  13. Jan 2, 2014 #12

    Saw

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    Jliang, I am not sure I follow your thought experiment. But, yes, certainly, the more I think of it, "the point of greatest deflection" (y = maximum) looks like that "that under most tension" (PE = maximum, KE = 0). However, the fact is that textbooks say the opposite (y = maximum --> PE and KE = 0), which really sounds awkward. It would be nice if we got some guidance from the experts.
     
  14. Jan 3, 2014 #13
    I found this article which is very good and has cleared up the confusion for me:
    http://faculty.ifmo.ru/butikov/WaveEnergyPS.pdf
    The potential energy of the string as a whole is a mimimum when the standing wave passes through the the equilibrium position, but it is not so for each little segment. Thanks for raising this issue - it seems there is a lot of confusion about it from the articles!
     
  15. Jan 3, 2014 #14

    Saw

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    Thank you Jilang for finding this article. The author, like most sources, supports the idea that KE and PE travel in phase in a string wave. He even mentions and criticizes the other article I lad linked to. So I suppose that I should accept such general view, although I confess that I still find it hard to assimilate.

    I can understand that one thing is total energy and another energy density at a given point or segment. But then I need some explanation as to how energy keeps travelling, anyhow, at least if energy is said to become ZERO (!) at a given point. What I find in the books is that the key lies in the fact that energy is transferred. Well, yes, it may be that at y = maximum the relevant point or segment (P) has now energy zero necause it has already transferred its energy to the next element (P+1). But if P has zero, it is because energy was dwindling before. So what P may have transferred is what it received from P-1, that is to say, very little. How is it that, in spite of everything, the energy builds up again, so as to become maximum at the equilibrium point?

    I cannot understand how mechanical energy can oscilate between different values and still be conserved. Saying that it is conserved over the cycle does not fix the problem for me. That is only a statement, but I would need an explanation of the physical mechanism ensuring that the cycle is actually made possible. The oscilation between PE and KE (with ME remaining as constant) is such explanation. If that does not apply, I need an alternative.

    By the way: this applies also to EM waves. The E and B fields are also said to travel in phase. How can that be and, what is more, how can they create each other this way? How can something that is zero create anything?

    [Note: Trying to be proactive in finding an answer, I have thought that the following may be an explanation. With the transfer explanation, we have ensured that P+1 at least gets a little energy packet. From then on, the energy builds up because the little packet travels more slowly than its chasers and it is progressively caught up by the latter. Can that be the right track?]
     
  16. Jan 3, 2014 #15

    sophiecentaur

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    I would tend to agree with that. The speed at which any tension along the string (tangential to the curve) would be much higher (bulk modulus and density would imply the speed would be that of sound in steel) than the transverse wave speed (a few m/s). It could surely be taken as instantaneous, imo and, hence, the same all the way along. And is it not obvious that the identifiable Potential Energy of a section of string is the work done in getting it to its position - which makes it a maximum when it's furthest from the equilibrium position: in quadrature with the KE, as expected.
     
  17. Jan 3, 2014 #16
    Reading the article earlier It seems that there is a net transfer of energy between the nodes and the antinodes throughout the cycle. This makes sense as at maximum displacement there is most energy at the nodes and at minimum displacement most energy is at the antinodes. So it sloshes backwards and forwards by pi/2 in each direction. Looking forward to finding out more about the E B situation over the weekend! A delving we will go....
     
  18. Jan 3, 2014 #17

    sophiecentaur

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    I think there is confusion between the progressive wave situation and the standing wave situation. At the node (this is only on a standing wave) there is never any PE or any KE. In the antinode, energy is exchanged between PE and KE - just like with a mass bouncing on a spring. The string is not stationary at an antinode (obvious I know) but what you see is a 'stationary' envelope and that can be confusing when an explanation is needed.
     
  19. Jan 3, 2014 #18

    AlephZero

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    True.

    False, but I don't feel inclined to write a chapter of a textbook on continuum mechanics right now, to explain why.

    That is confusingly ambiguous, if it's not clear whether you are talking about traveling waves or standing waves, and whether you mean in quadrature with respect to position along the string, or with respect to time at a fixed position.
     
  20. Jan 3, 2014 #19
    Sophie, glad you could make it. I saw you were very busy with Standing Waves and Travelling Waves, but hoped this thread would keep going long enough for you to get here! Are we on the right track?
     
  21. Jan 3, 2014 #20

    Saw

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    Well, that PE is maximum when y = ym. i.e. when amplitude is maximum, is what I also tended to think. But I am trying to convince myself of the opposite because all consulted sources say that PE is maximum at equilibrium position. i.e. when y = 0. In any case, what seems out of discussion is that such PE is associated to elastic stretching.

    I will quote the full relevant passage in Halliday & Resnick for reference (fig. 16-9 is the one I copied before), but most other sources say the same:

     
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