# Talk:Pendulum (mathematics)

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## Pendulum with bob traveling in horizontal circle

Just for fun, I derived the equations describing a pendulum whose bob is traveling in a horizontal circle, i.e. a circle whose plane is parallel to the floor, instead of traveling back and forth. I neglect friction of all kinds, so the bob orbits a point directly beneath the suspension point, and the wire tension, bob speed, and bob acceleration magnitude are all constant. If θ is the constant angle of the wire relative to the vertical, I get:

${\displaystyle t=2\pi {\sqrt {\frac {L\cos \theta }{g}}}}$

Note that in the limit as θ approaches zero, we get the same formula as the small angle limit in a pendulum traveling back and forth. With simple experiments at home using a stopwatch, a 12-ounce plumb bob, and 2.7 meters of nylon twine, I confirmed that the period of horizontal circular motion are within measurement error of the back-and-forth period.

I thought this was interesting, but I have not seen this result anywhere "official". Does anyone know of a usable reference for this result? I would add it to the article if it weren't just my original research. CosineKitty (talk) 21:25, 26 March 2009 (UTC)

Yes, an interesting observation. Perhaps it belongs in our article on conical pendulum? Dbfirs 07:19, 27 March 2009 (UTC)
It's in the History section of this article, but it's kind of hard to find. There are citations to references there, and more in the article conical pendulum. You got the same equation for the period as is derived in that article, so it must be right, congrats. Maybe you could improve the derivation there, it is a bit convoluted in the later part. --ChetvornoTALK 07:43, 27 March 2009 (UTC)
Thanks, guys! Now that I know it is called a conical pendulum, I know what to search for. The mathematical derivation there is exactly what I did, only I agree it could use a diagram or two, plus maybe a couple of extra steps to make it clearer, along with references if possible. I will see what I can do. CosineKitty (talk) 20:11, 27 March 2009 (UTC)
Below is a test of my first attempt at a diagram for the conical pendulum article. I am new to using Inkscape, so please be gentle, but constructive criticism is welcome. (Especially if you know how to do it in Inkscape, not just that the diagram is ugly!) Note that I would like to rename the variable l to L in the text of the article so it looks less like the numeral 1. CosineKitty (talk) 21:40, 27 March 2009 (UTC)

This pendulum motion has the desirable feature of a constant velocity of the mass, and thus a constant centrifugal force vector (although a variable force direction) and thus a constant mass-energy constant value, unlike the back and forth pendulum, where the velocity value varies.These variations must have been neglected in the analysis?WFPM (talk) 18:56, 9 March 2012 (UTC)

As stated elsewhere, the "centrifugal force vector" is useful only in the frame of reference of an insect on the pendulum bob. It doesn't exist in the frame of reference of the fixed point from which the pendulum is suspended. The centripetal force vector has only its magnitude constant. Where do you get the idea from that increase of mass with velocity is significant? Dbfirs 13:03, 10 March 2012 (UTC)
We are comparing a perpetual motion concept and I was comparing the relationship of factors between the oscillatory pendulum and the circular moving pendulum and noting that in the circular moving pendulum the velocity is constant, whereas in the oscillatory it varies from zero to some value. And if there were some factor that related to the pendulum velocity there should be a difference.WFPM (talk) 14:53, 10 March 2012 (UTC) Note that in the circular moving pendulum you don't have the same energy conversion process going on as in the oscillatory process, but merely a conservation of angular momentum process.WFPM (talk) 15:13, 10 March 2012 (UTC) That's the trouble with mathematical concepts. Once they are involved, the idiosyncrasies of the individual processes get lost in the shuffle.WFPM (talk) 00:14, 11 March 2012 (UTC)
With the usual meanings of the words in Physics and Mathematics, the speed is constant for the conical pendulum, whereas the velocity is constantly changing (in direction). We agree, then, that mass can be regarded as independent of speed in the treatment of both pendulums? Dbfirs 06:54, 11 March 2012 (UTC)
I wouldn't want to say that mass is independent of speed, because in the oscillatory pendulum we're back to the centrifugal force factor being related to the instantaneous speed/velocity of the plumbob through the arc, plus that the downward force in the tension structure would be a variable. Also note that the (perceived)lateral motion of an oscillatory pendulum can be 1: An actual angular momentum conserving process or 2; A non angular momentum process due to unnoticeable rotation of the directional scale vector under the pendulum. So a pendulum can evidently be said to have a certain amount of oscillatory motion in 1 direction plus a variable amount of angular momentum which results in a periodic amount of lateral motion in the orthogonal direction. And the conical pendulum is the one with a constant amount of angular momentum. Which looks to me like the more stable situation.WFPM (talk) 17:24, 11 March 2012 (UTC) I apologize for when I said velocity up above, when I should have said speed, because I mentally don't associate the word velocity with a direction.WFPM (talk) 17:36, 11 March 2012 (UTC) Wouldn't that be an example of the term "a distinction without a difference?WFPM (talk) 17:45, 11 March 2012 (UTC)
Everyday usage often takes velocity to be "a posh word for speed", but this is the talk page of Pendulum (mathematics). I'm puzzled about why you think that a (fictitious or reactionary) centrifugal force means that mass changes with speed. Dbfirs 09:55, 13 March 2012 (UTC)
I'm not a mathematician. I look at it from a physicist viewpoint and try to work my way down from the ideal to less desirable situations. The ideal seems to be the conical or circular motion concept, with a different cone for each level of stored angular momentum. Then we get an irregularity, resulting in an oscillatory motion in one direction, and a consequential amount of change in "speed". That involves in the energy conversion process with its associated factors. And we should then question the "speed" factor's significance in this. And we wind up with the centrifugal force value being a function of the square of the speed factor. So the question becomes as to what mother nature thinks about a process that is concentrating kinetic energy (=mass?) into a certain portion of the process. She's a lot smarter than I am about these things.WFPM (talk) 18:03, 13 March 2012 (UTC) On second thought, maybe you can't concentrate kinetic energy into a part of a process, so what else has to happen?WFPM (talk) 00:33, 14 March 2012 (UTC)
I'm rather surprised that, as a physicist, you confuse speed with velocity, and that you believe that "Mother Nature" "thinks", but perhaps I am misinterpreting your responses. You seem to be overlaying a simple model based on Newtonian Mechanics with a misunderstanding of special relativity, an outdated concept of "centrifugal force", and some sort of mystical belief. Apologies if this is not the case. Dbfirs 06:54, 14 March 2012 (UTC)
I'm not a physicist, but just an engineer, who tries to use whatever tools of thought he can find to apply to a physical problem at hand. I just summarize factors into a "Mother nature" higher authority concept as do lots of subject matter authorities without intent to demean anything. Here we have the problem of how the dissipation of the stored kinetic energy of the pendulum is being managed, and we're trying to pin the blame on some factor we can recognize. And if we can't solve the problem then we need to reevaluate our system of analysis and assumptions. And mathematics is partly to blame for this. They create a formula and "voila", they now think they know exactly what is going on in the situation. And Plato proved that it is better to know that you don't know than not to know that you don't know. So I'm arguing that the concept of mathematical solutions tends to make us too confident in our abilities to know about science problems and their solutions should be considered in that frame of mind.WFPM (talk) 19:02, 14 March 2012 (UTC)
In that case, we are in agreement, though I don't quite see where the problem lies with kinetic energy being converted to of from gravitational potential energy. As far as I know, the whole theory works and gives consistent results. Dbfirs 19:27, 14 March 2012 (UTC)
That's fine. But we know that the KE will ultimately dissipate and the oscillation of the pendulum will stop. Am I correct in assuming that it has to do with some process related to the square of the highest "speed" in the process? Or is that in violation of some mathematical principle?WFPM (talk) 21:46, 14 March 2012 (UTC) Note that a friction factor would be related to a (centrifugal?) force vector which would be emphasized by squaring the speed.WFPM (talk) 22:03, 14 March 2012 (UTC)
And there's another way of looking at this problem. Suppose we have 2 identical pendulums, and we start one with a certain level and contained amount of stored gravitational energy and let oscillate horizontally. Then we take the other and give it a horizontal circular rotation motion with the same amount of combined stored gravitational energy plus angular kinetic energy. Can we then use mathematics to predict which motion will decay faster and reduce the motion of the pendulum action? I don't know. Do you?WFPM (talk) 23:36, 14 March 2012 (UTC)
The mathematics of decay is indeed muddy. The cause of decay is mainly air resistance and friction at the pivot. The reactive centrifugal (I prefer to say centripetal) force is a major determinant of the loss in energy at the pivot both from friction and because the slight movement of real-life pivots transfers some energy to the support. Some resistive factors are proportional to the square of speed and others are not. We can set up differential equations using given assumptions, but reality will usually be more complex than our equation, so the best we can expect from solving a differential equation is an approximate solution. Many pendulums are designed so that the resistive forces are minimal, or exactly balanced by a regular input of energy to compensate, thus simplifying and regulating the behaviour of the pendulum. Dbfirs 11:05, 15 March 2012 (UTC)
Agreed! And so the mathematics is only so good at modeling the situation and after that we're on our own. And that.s after 1000 years of looking at a problem. And 2 friction possibilities, both related to the square of "speed", but not modeled. And we still don't know which is more efficient, the circulatory rotation or the lateral oscillation process. And how you can periodically build up and then dissipate an amount of kinetic energy as part of a process. And we make a "to do" about the direction of a velocity vector as if it had something to do with the operation of the process. But it does bring up the question of what nature thinks about a change of direction of a force vector, and it is my opinion that nature is indifferent to direction. She works with volumes within a 3 dimensional spacial continuum.WFPM (talk) 16:27, 15 March 2012 (UTC)
Incidentally, a pendulum is only a special situation of what I would call a Tethered mass. And since we now have the ISS, with practically zero gravity, are we mathematically prepared to explain what would happen to the kinetic energy stored within a tethered mass either inside or even outside the ISS? WFPM (talk) 17:24, 15 March 2012 (UTC)
It is always possible to improve the mathematical model and thus improve the accuracy of the predictions. It is easy to set up differential equations with terms proportional to speed and to the square of speed. Efficiency (in the sense of avoiding energy loss) depends on the design of the pivot, and, in the case of the conical pendulum in a vacuum, it is the changing direction of the velocity vector that is the main cause of the loss in kinetic energy, so "nature" cannot be "indifferent" to direction. Einstein's theories claim that "she" works with a four-dimensional space-time continuum in which direction is important. You are, of course, correct in claiming that both an oscillating pendulum and a conical pendulum turn into simple motion in a circle in the absence of gravity. Simple application of Newton's second law explains the conversions between kinetic and potential energy in the oscillating pendulum. Dbfirs 21:46, 15 March 2012 (UTC)
I don't think I am agreeing with you. Once nature identifies the nature of the identity she is dealing with, she is then indifferent about the direction of the next increment of force. And as the forces add vectorially, the kinetic energy goes up in proportion to the square of the speed. I would say she is moving a volume of something, and doesn't care which way she pushes it. Of course if she continues to push in the same direction, the distance she pushes gets larger and stores up more energy. And the amount of stored energy must be related to the volume.WFPM (talk) 00:48, 16 March 2012 (UTC)
My Neodymium magnets that I play with hang on a thread and don't have a pivot. But I guess you'd have to have a pivoting tether in a 3 dimensional spacial motion experiment. But I don't see how you could get into an energy conversion process, and thus the velocity vector would always be changing direction as it decreased.WFPM (talk) 01:17, 16 March 2012 (UTC)
I prefer not to comment any further on these opinions. Dbfirs 17:25, 16 March 2012 (UTC)
Appreciate your comment and understand. The understanding of tethered mass physics can have far reaching ramifications. Sorry I cant help with the pendulum physics article.WFPM (talk) 18:39, 16 March 2012 (UTC)

## Animations

Well, during Wikimania I promised pt Wikipedia user Lechatjaune I'd make a few animations for the pendulum articles, and here they are. I decided to make several ones, to better illustrate each different mode of oscillation based on the initial angle. I'd like if you guys picked the better ones to include in the article, since I don't feel I have the technical eloquence to properly describe the phenomenom in encyclopedic terms. Anyway, here they are. Let me know if there's anything wrong! :) — Kieff | Talk 10:26, 3 September 2009 (UTC)

And, of course, the equilibrium states:

These are fantastic! I used some of them in the Pendulum article. Thanks, Kieff --ChetvornoTALK 16:12, 29 December 2011 (UTC)

Note that the first 8 of these animations model the retracement of an arc of a path of the plumbob, whereas the last 2 model a vertical circular path.WFPM (talk) 00:53, 14 March 2012 (UTC)

"Fantastic", or even "incredible" they may well be ("ingenious" might be more apt), but what exactly are the red lines ? The main article describes them as "phase portraits" but this seems to be an example of a metaphor obscuring meaning rather than illuminating it. A plain description of what is plotted horizontally, and what vertically should be given if these lines are to have any meaning. The main article is not in any way enhanced by the inclusion of clever animations which are decorative but meaningless. G4OEP

## Merge Pendulum (derivations) in

Both Pendulum (derivations) and Pendulum (mathematics) seem to have been split off from the main Pendulum article (on 29 May 2006 and 12 July 2006 respectively), and rather than having two articles it seems sensible to combine the mathematical content in a single article. Certainly the derivation of the the formula deserves to be placed here. --Salix (talk): 09:37, 2 August 2010 (UTC)

I know this is a bit of an old post but I agree with the "merge". Pendulum (derivations) is pretty much a stub anyways and most if not all of the information there is already in Pendulum (mathematics). Considering this, it might be more of a "delete" than a "merge".Phancy Physicist (talk) 16:04, 29 May 2011 (UTC)
Agree to the merge. --ChetvornoTALK 21:57, 2 June 2011 (UTC)
I believe I have successfully added the derivations from Pendulum (derivations) into the article. I am going nominate the Pendulum (derivations) page for deletion.
Phancy Physicist (talk) 04:48, 9 June 2011 (UTC)
Please go to Talk:Pendulum (derivations) to weigh in on this merge/deletion discussion.
Phancy Physicist (talk) 03:42, 10 June 2011 (UTC)

## Posts from Talk:Pendulum (derivations)

Since the merge is in progress I am moving the talk to here so that anyone can address the talk if they want.

I think it'd be good to add a way to derive the Pendulum using Lagrangian formulation.

Using Lagrangian mechanics the pendulum can be derived through the use of the Euler-Lagrange equation of motion. Defining Lagrangian as the difference between kinetic and potential energies,

${\displaystyle L=T-V\,}$

where T is kinetic energy and V is potential.

${\displaystyle T={\frac {m}{2}}\left({\dot {x}}^{2}+{\dot {y}}^{2}\right)}$
${\displaystyle V=mgh=mgl(1-\cos \theta )\,}$
${\displaystyle L={\frac {m}{2}}\left({\dot {x}}^{2}+{\dot {y}}^{2}\right)-mgl(1-\cos \theta )}$
${\displaystyle x=l\sin \theta }$

where x is the displacement in the horizontal direction

${\displaystyle y=l\cos \theta }$

y is the displacement in the vertical direction

${\displaystyle {\dot {x}}^{2}=l^{2}{\dot {\theta }}^{2}\cos ^{2}\theta }$
${\displaystyle {\dot {y}}^{2}=l^{2}{\dot {\theta }}^{2}\sin ^{2}\theta }$
${\displaystyle L={\frac {m}{2}}\left(l^{2}{\dot {\theta }}^{2}\cos ^{2}\theta +l^{2}{\dot {\theta }}^{2}\sin ^{2}\theta \right)-mgl(1-\cos \theta )}$

Using Lagrange's Equation

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\theta }}}}={\frac {\partial L}{\partial \theta }}}$

we get

${\displaystyle {\frac {d}{dt}}{\frac {\partial L}{\partial {\dot {\theta }}}}=m\left(l^{2}{\ddot {\theta }}^{2}\cos ^{2}\theta +l^{2}{\ddot {\theta }}^{2}\sin ^{2}\theta \right)=ml^{2}{\ddot {\theta }}^{2}}$
${\displaystyle {\frac {\partial L}{\partial {\theta }}}=-mgl\sin \theta }$
${\displaystyle ml^{2}{\ddot {\theta }}=-mgl\sin \theta }$
${\displaystyle {\ddot {\theta }}=-{\frac {g}{l}}\sin \theta }$

--Nefreat

### Wrong negative signal in some formulas of the derivations

The explanation for the negative signal in equations:

${\displaystyle F=-mg\sin \theta \,}$
${\displaystyle a=-g\sin \theta \,}$

namely, that ${\displaystyle g\,}$ is negative because it is pointing downward, is simply not true.
If the formulas refer to absolute values, then there should be no minus signal.

The reason behind these negative signs is the authors wish to obtain the correct signal in equation:

${\displaystyle \ell {d^{2}\theta \over dt^{2}}=-g\sin \theta \,}$

My argument for this minus signal is as follows. The starting equation is

${\displaystyle a_{t}=g\sin \theta \,}$

where ${\displaystyle a_{t}\,}$ is the tangencial acceleration.
Now, in Figure 2, let us suppose that the bob is going UP. In this case we should write:

${\displaystyle a_{t}=-{dv \over dt}}$

because the tangencial acceleration and the velocity are in opposite directions. And,

${\displaystyle v={ds \over dt}}$,

because velocity is in the direction of the growing of the arc ${\displaystyle s\,}$. This gives:

${\displaystyle a_{t}=-{dv \over dt}=-{d^{2}s \over dt^{2}}}$

and since ${\displaystyle s=\ell \theta \,}$, we finally have

${\displaystyle a_{t}=-\ell {d^{2}s \over dt^{2}}}$

The same argument goes when the bob is going DOWN. In this case we have:

${\displaystyle a_{t}={dv \over dt},}$ and ${\displaystyle v=-{ds \over dt}}$

But the final result is, as before,

${\displaystyle a_{t}=-\ell {d^{2}s \over dt^{2}}}$

Rui Ferreira 18:05, 12 April 2007 (UTC)

### Symbols used

I am a person who is just beginning to understand the concepts described in this article. I find the information in the article to be very valuable. However, one minor point was a bit confusing to me. The "L" symbol used for the length between the pivot point and the center of mass in the diagram and the one used in the equations are different. It took me a bit to grasp that the two different symbols meant the same thing. It's a minor point, I know, but other burgeoning physicists like myself might find this a bit confusing. Lxman (talk) 21:27, 23 December 2010 (UTC)

I was taking a look at the infinitesimal navbox which guided me to the section titled Infinitesimal oscillations of a pendulum. I found this to be something of a misnomer because, at least in sources I have seen this is usually called the small angle approximation. Glancing at the history I see this was changed back in April, here. I was wondering if it could be changed back. Thenub314 (talk) 19:15, 15 December 2011 (UTC)

Yes, I agree that most people call this the small angle approximation. Although the formulae are perfectly accurate only for infinitesimal oscillations, I've never seen an infinitesimal oscillation, for obvious reasons. Does anyone object to changing the heading back? Dbfirs
It has been a few days, I will take this as a no. Thenub314 (talk) 19:05, 19 December 2011 (UTC)
Felix Klein calls it "infinitesimal oscillations of a pendulum". "Small angle approximation" refers to a broader range of phenomena, such as approximating cosine x by 1-x^2/2. Since we are dealing with oscillations we might as well call a spade a spade. I object to the name change back to the generic "small angle". Tkuvho (talk) 14:30, 29 December 2011 (UTC)
I haven't read Felix Klein, but most other people seem to call this "Small angle approximation". All such oscillation formulae are approximations, since infinitesimal oscillations are impossible to observe (and are hardly "spades"!) Isn't approximating cosine x by 1-x^2/2 actually a narrower condition, giving 99% accuracy up to about 19 degrees compared with less than 7 degrees for sine? The cosine approximation is used for potential energy calculations. Dbfirs 15:20, 29 December 2011 (UTC)
Thanks for the clarification. If I understand you correctly, in physics literature the term "small oscillation" is used more often than Klein's term. If so, the subsection should be titled accordingly. However, for the benefit of a mathematically trained reader I would like to include a mention of Klein's term with a reference to Klein's book (elementary mathematics from an advanced viewpoint) where this is discussed, if you think this is appropriate. Tkuvho (talk) 15:24, 29 December 2011 (UTC)
Could you possibly explain how Klein's viewpoint differs from the usual treatment in both physics and mathematics? I don't see how an accurate equation for a non-existent phenomenon is inherently "better" than an approximate equation for a real phenomenon, but I agree that it would be worth mentioning that the error tends to zero as the oscillations reduce in amplitude. Dbfirs 15:30, 29 December 2011 (UTC)
Every mathematical model is an idealisation of the physical phenomenon, and the pendulum is no exception. The infinitesimal approach to the pendulum is a convenient way of formalizing the phenomenon, where mathematical infinitesimals correspond to physically small oscillations. Historically physicists have talked about infinitesimals all the time from Newton onwards. I don't think it would be correct to say that Newton dealt with non-existent phenomena. At any rate, we are merely discussing terminology here; it is my impression that the term "infinitesimal oscillations of the pendulum" is more familiar to a mathematician. Tkuvho (talk) 15:39, 29 December 2011 (UTC)
Incidentally, even if the oscillation is infinitesimal, the error is still nonzero though it is an infinitesimal of a higher order. This would be the most common approach to infinitesimals. There are other approaches featuring nilpotent infinitesimals where the error would actually disappear, but they are less commonly used. Tkuvho (talk) 15:41, 29 December 2011 (UTC)
I challenge the claim that the term "infinitesimal oscillations of the pendulum" is more familiar to a mathematician, but my only knowledge of Felix Klein is his bottle. Perhaps we disagree over the meaning of the term infinitesimal. Dbfirs 19:09, 29 December 2011 (UTC)
That's interesting. What does "infinitesimal" mean to you? Notice that I mentioned several options above. In your view, if e is infinitesimal, is e squared necessarily zero? Tkuvho (talk) 09:36, 30 December 2011 (UTC)
Of course not, just much closer to zero. How is Felix Klein's treatment of small quantities tending to zero different from Newton's? Dbfirs 18:01, 1 January 2012 (UTC)
Felix Klein was writing before Abraham Robinson and therefore did not have the hyperreal numbers at his disposal, so his discussion is heuristic to a certain extent, which should be acceptable by physics standards as far as I understand. He certainly admired the classical heuristic infinitesimals of Newton and especially Leibniz. Note that if you have a real number smaller than, say, the tiniest string in M-theory, for all practical purposes it is indistinguishable from an infinitesimal, so there is no gain in "reality" here. Tkuvho (talk) 18:07, 1 January 2012 (UTC)
With all due respect to Klein it is a fairly obscure term for this subject, and as such not appropriate for the section title. For example, a quick google test turns up around 6,000 results for the infinitesimal oscillations and pendulum, many of which are sites which reproduce wikipedia's content. On the other hand, a search for small angle approximations and pendulums gives around 57,000 results. I stand by my claim this is not a very widely used phrase in mathematics. Thenub314 (talk) 20:28, 1 January 2012 (UTC)
Being a mathematician I can say with honesty I had never encountered the term until reading it here. Thenub314 (talk) 19:59, 29 December 2011 (UTC)
Infinitesimals used to be a bigger subject matter and the 9th edition of the EB has an approximately 100 page article on the calculus of infinitesimals (Book#8).WFPM (talk) 18:44, 12 March 2012 (UTC) But now we have calculus to calculate things like the area under the curve y/x squared.WFPM (talk) 18:52, 12 March 2012 (UTC)
The editors above seem to have disagreed intensely as to the appropriateness of following Felix Klein in describing this problem as "infinitesimal oscillations of a pendulum" How do you see this? Tkuvho (talk) 19:14, 12 March 2012 (UTC)
It's kind of like the subject matter Relativity, where the mathematics is one thing and the actuality is another.WFPM (talk) 21:08, 12 March 2012 (UTC)
I, too, have my own personal beliefs about the nature of "reality", but they don't lead me to reject the validity of the mathematical models that have been thoroughly tested in that they lead to predictable consistent results. Where there is a choice of models, I prefer to adopt the most efficient, which is often the simplest. For Wikipedia, it is probably best to report the most common model in standard texts. Dbfirs 07:01, 14 March 2012 (UTC)
That's not really the issue here. We are discussing the term "infinitesimal oscillations of a pendulum" as it was used by Felix Klein several decades before mathematically reliable models of infinitesimals were developed (notably by Abraham Robinson). The question is whether the term is in the literature or not. People have referred to this as "infinitesimal oscillations" in the sense of naive heuristic infinitesimals rather than any specific formal model. We are all certainly in agreement that the real numbers are in far wider use than the hyperreal numbers. Tkuvho (talk) 11:58, 14 March 2012 (UTC)
I've changed the template to remove this link, since one particular use of the small-angle approximation doesn't really relate to infinitesismals. Scientific29 (talk) 23:50, 23 April 2014 (UTC)

## Is this (a special case of) Mathieu's equation?

According to Mathews and Walker Mathieu's equation is a linear equation while equation (1) is nonlinear. For instance, d^2 theta/dt^2+theta sin(at)=0 is a case of Mathieu's equation. I propose to eliminate the sentence. Naudts (talk) 15:29, 20 February 2012 (UTC)

You are right. There are forced oscillators whose behaviour is described by Mathieu functions but this simple harmonic oscillator is not. Removing the references to the Mathieu function now.
Phancy Physicist (talk) 06:02, 11 March 2012 (UTC)

## Simpler and better?

The words "simpler and better" to describe an alternative formula seem to be a matter of opinion. Our anonymous Russian friend seems to insist on his exact wording. What does anyone else think? Dbfirs 16:19, 22 September 2012 (UTC)

Regarding simpler vs. shorter vs. alternate: The simplicity of a formula is POV. It depends on how familiar a particular reader is with its building blocks. The comparison of the shortness of both expressions depends on whether K(k) and/or M(x) are assumed to be known to the reader or whether their definition is considered part of the formula. In the former case both expressions are of similar length, so this is again POV. I would therefore consider alternate the most neutral and thus appropriate choice.
Regarding faster vs. better. That depends on what is meant by faster and better, respectively: in terms of Rate of convergence, is it the order or the rate of convergence, or something else? I would associate faster with rate, but wouldn't know how to interpret better. I don't know about the convergence behavior of both expressions, so I can't tell if faster fits, but either way I don't think better is well-defined enough to be useful here.
Even though I'm not in favor of the current phrasing, I strongly suggest to leave the article alone until the discussion is finished. If needed, more input can be asked for at Wikipedia talk:WikiProject Mathematics. — HHHIPPO 19:07, 24 September 2012 (UTC)
I read the reference as demanded by our anonymous friend, and it contains neither "simpler" nor "better". I've no objection to the alternative formula, in fact I quite like it, but it does rely on an obscure iterative process (that ought to be better-known because it is powerful). The words "simpler and better" seem to be against the policy of Wikipedia:No original research. Dbfirs 20:22, 24 September 2012 (UTC)
(later note) On your advice above, I've restored the opinionated version for others to comment on and adjust as necessary. I'll withdraw from the "war" over the words and await the outcome of discussions. Dbfirs 20:33, 24 September 2012 (UTC)
I have read the provided reference for that equation as well. I support the alternate over shorter or simpler because as HHHIPPO said, it is POV to say one is simpler than the other depending on what you assume the reader knows.
I also agree with not using better because it promotes a preference that is not widely excepted or at least has not been proven to be widely excepted by the scientific community.
Phancy Physicist (talk) 01:42, 25 September 2012 (UTC)
Yes, I wonder whether User:Syrmath is another alias of the author of the article. I don't blame him for pushing his own article, and we will certainly retain his formula here, but he'll have to learn not to keep reverting against consensus in the basic article (Pendulum). We have retained his wording here, even though he probably means "faster-converging" rather than an opinion of "better". Dbfirs 15:12, 4 October 2012 (UTC)
They were blocked as a sock: User_talk:Syrmath a13ean (talk) 19:21, 4 October 2012 (UTC)

## Simpler and better converging!

The preceding discussion was carried out under the wrong title. I prefer providing clarifications under the right title. The exact formula given in the referenced article is certainly much simpler than the power series formula and is, indeed, much better converging. The referenced arithmetic-geometric mean is described in another wikipedia article, where its quadratic convergence is dispalyed. — Preceding unsigned comment added by 193.233.212.18 (talk) 11:00, 26 September 2012 (UTC)

Is the formula: ${\displaystyle {\frac {2\pi }{M(\cos(\theta _{0}/2))}}{\sqrt {\frac {\ell }{g}}}}$ simpler than the formula: ${\displaystyle 4{\sqrt {\ell \over g}}Ksin(\theta _{0}/2)}$ (given that you are equally familiar with M and K)?
Do you mean that your formula converges faster? If so, then I agree (obviously), and that is part of the point I was trying to make. Dbfirs 07:21, 27 September 2012 (UTC)
On the assumption that you meant "faster-converging", I've changed your text to clarify the meaning you intended. Please discuss here (under one of your aliases) if this is not what you meant. Dbfirs 18:37, 9 October 2012 (UTC)
The whole "faster-converging" bit is somewhat confusing since we're comparing a recursively-defined function with a taylor expansion. a13ean (talk) 19:14, 9 October 2012 (UTC)
... so should we just have "fast-converging", so that no comparison is implied? Dbfirs 07:44, 28 October 2012 (UTC)

## Clarification

If my memory serves me well, I was taught that the equation y" = –sin y has no explicit solution, only series approximations, and the article seems to confirm this. I'm wondering if it would be more accurate to replace the phrase "not easy to solve", with "has no explicit solution", or something to that effect.77Mike77 (talk) 18:53, 18 March 2013 (UTC)

It depends whether you consider elliptic integrals as explicit. They used to appear in books of tabulated values like sine and cosine. Another way to solve the equation is via the Arithmetic–geometric mean, but these are all convergence methods. Do you consider a sine or cosine function to be explicit? They have series approximations. Dbfirs 22:32, 18 March 2013 (UTC)

I was thinking of "explicit" meaning "can be written in the form y = f(x) where f(x) is made from elementary functions, as defined here --> http://en.wikipedia.org/wiki/Elementary_function ". So sine would be elementary, even though there's a series approximation for it, but elliptic integrals wouldn't be. But if that notion of "explicit" is not universally accepted, I won't argue the point. I don't have the textbook anymore that stated that. We were solving a variety of ordinary differential equations, such as Bessel equations and Legendre polynomials, and focused a lot on existence theorems as well, then the second part of the course dealt more with series approximations, and also Fourier Series. I'm sure you know what you're talking about, because of how you solved the pendulum swinging in a circle in 3-D; I don't think I could have solved that myself without huge effort. A relative of mine got a poor mark from the high school science teacher on a pendulum lab with bobs on a short string swinging wildly. He noticed that it didn't follow the Law Of The Pendulum, and was penalized for innacuracy. I tried to solve it (before the internet), and learned that there were only series solutions, but that my relative's observations were correct. Anyway, I think this is an excellent article overall.77Mike77 (talk) 00:06, 20 March 2013 (UTC)

Yes, you are correct that the equation has no solution in terms of "elementary functions". Apologies for the quibble, it's just that the so-called "elementary functions" are the ones that are traditionally tabulated and included in calculators and spreadsheets, as opposed to the "non-elementary" (rarer) ones that were only occasionally tabulated (and are not included in my calculators or spreadsheets). If you want to add a comment in the article about no solution in terms of elementary functions, then please go ahead. Dbfirs 09:34, 20 March 2013 (UTC)

## Remark/Question on Main Article's 1.Simple Gravity Pendulum.

Hi, I was reading the derivation of a pendulum's differential equation from the conservation of energy principle. I agree with the section:

It can also be obtained via the conservation of mechanical energy principle: any object falling a vertical distance h would acquire kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by

${\displaystyle \Delta U=mgh\,}$

change in kinetic energy (body started from rest) is given by

${\displaystyle \Delta K={1 \over 2}mv^{2}}$

I am not certain we can say the following:

Since no energy is lost, those two must be equal

${\displaystyle {1 \over 2}mv^{2}=mgh}$
${\displaystyle v={\sqrt {2gh}}\,}$

I believe we can only say that at all moments:

${\displaystyle mgh+{1 \over 2}mv^{2}=TotalMechanicalEnergy=mg*h(hMaximum)={1 \over 2}m*v^{2}(vMaximum)}$

I think we could however say that the derivatives of the kinetic energy and the potential energy are equal, which i believe is what the proof goes on to do (so the final result is correct),

Or was the article expressing: Max KE = Max GPE. and I simply misunderstood.

Thank you, please correct me If I am wrong. — Preceding unsigned comment added by 134.214.58.239 (talk) 17:21, 4 March 2015 (UTC)

If no energy is lost, why do you doubt that loss in PE is equal to gain in KE? Where else could the energy go? Dbfirs 21:37, 20 March 2015 (UTC)

## Transition from oscillatory to rotary motion and Abrarov's critical solution

A formula of motion of the pendulum from lower stable equilibrium to upper unstable equilibrium is authoritatively referenced, so if Sigwald is unable to argue against it, since he does not understand it, he better refrain from taping with a verifiable addition. So far, he has been given 3 days without a convincing explanation. The primality of the source is not a serious obstacle in this case, since the formula itself is not posted in wikipedia.Expresser (talk) 17:55, 10 March 2015 (UTC)

It would be useful if we could find a secondary source for this. Dbfirs 21:33, 10 March 2015 (UTC)

As I understand, the given source is not quite primary. Unfortunately, the primary source seems to be given in the publications presented here http://www.ccas.ru/depart/mechanics/TUMUS/Abrarov/Abrarov.html, none of which is in English. Yet, that is not a problem given automated translators. The given source is more accessible, though. I think this important case of transition from oscillations to revolutions must be verified before being presented in wikipedia. In by-itself it involves elementary hyperbolic trigonometric functions, so it must be accessible to many editors of this page, myself including.Expresser (talk) 15:17, 11 March 2015 (UTC)

Author of the same edits (User:SupremeFormula) already have been banned in ruwiki for several days because of edit warring. The manner of discussion allows me to assume that Expresser is the same user. And he still has not answered of who is Adlaj (author of cited publication) and in what reviewed journal this article have been published. --Sigwald (talk) 22:49, 11 March 2015 (UTC)

I in no way endorse "my" version. I was just trying to assume good faith and hoping that someone could find the equations in published texts. Perhaps we could have details of any publications? Is it Zii︠a︡utdin Sirazhiddinovich Abrarov? Dbfirs 18:22, 13 March 2015 (UTC)
You can keep your threatening and insults to yourself. Wikipedia articles must refer to reliable sources. What is a source of this article and why should we count Adlai as an expert? Maybe he isn't even PhD? --Sigwald (talk) 20:35, 16 March 2015 (UTC)

## Hamiltonian

Apologies to 169.231.11.245 if I've reverted more than I should have done, but it really isn't necessary to introduce the Hamiltonian in this simple treatment. The principle of energy is adequate as stated (if no other energy change occurs, then kinetic energy lost must be equal to potential energy gained, or vice versa). Is there some university text that insists on doing the calculation your way? Another anon editor (134.214.58.239) also doubted the principle. It is widely taught in the UK. Is it doubted elsewhere? Dbfirs 14:36, 28 July 2015 (UTC)

I agree; introducing the Hamiltonian is ridiculous. It is unnecessarily complicating a simple derivation.
It's funny. There are numerous important topics on pendulum mathematics missing from this article that could be added: corrections for buoyancy and viscosity of air, and elasticity of a pendulum string, Q_factor, resonance width and accuracy of a pendulum as a timekeeper, the Airy condition and how to drive a pendulum isochronically, the mathematics of gridiron and mercury temperature-compensated pendulums. Yet editors just keep banging away at the same old subjects that are already adequately covered; the derivation and large angle approximations. --ChetvornoTALK 18:32, 28 July 2015 (UTC)
...later ... I've restored the section that 169.231.11.245 removed after the above discussion. The derivation from the principle of energy is commonly found in standard texts. If there's something wrong with it, please discuss here. Dbfirs 07:19, 11 August 2015 (UTC)

## misprint

the argument of K in eq.3 is not inside () and thus seems a multiplicative factor pietro

188.14.97.4 (talk) 10:03, 4 October 2016 (UTC)