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Feynman Lectures Exercise Challenge

  1. Aug 26, 2011 #1
    Hello, all.

    My colleagues and I are currently working on an exercise book for The Feynman Lectures on Physics (FLP). This book will include about 1000 exercises from the original Feynman Lectures course as taught at Caltech, covering pretty much the entire range of topics in all three volumes of FLP. Almost all of these exercises have been published before by Caltech, however the old Caltech exercise books are long out of print, and the exercises for FLP Vols. II and III, most of which originated with Feynman himself, were unfortunately never published with answers. We will be publishing them with answers, and worked-out solutions in many cases.

    As a "teaser" and special challenge to Feynman physics buffs everywhere, I am posting one of the problems from the new exercise book.

    In one of the Review lectures Feynman gave to his freshman students, just before their first big exam, he advised them as follows (copied from Feynman's Tips on Physics, a problem-solving supplement to The Feynman Lectures on Physics):
    "Now, all these things you can feel. You don’t have to feel them; you can work them out by making diagrams and calculations, but as problems get more and more difficult, and as you try to understand nature in more and more complicated situations, the more you can guess at, feel, and understand without actually calculating, the much better off you are! So that’s what you should practice doing on the various problems: when you have time somewhere, and you’re not worried about getting the answer for a quiz or something, look the problem over and see if you can understand the way it behaves, roughly, when you change some of the numbers."

    The challenge is to solve the problem given below (originally homework for FLP Vol. I, chapter 23) in the spirit of Feynman's advice, above. It must be solved without using any calculus or differential equations or integral equations or difference equations, etc., without iterative numerical methods, nor any such other fancy mathematical tricks! You may use only algebra, geometry, trigonometry, dimensional analysis, and Newtonian mechanics, in your solution, which should be guided by your physical intuition (however note: all intuitions used in solutions must be justified)! Your answer does not have to be exact, but it should at least be a very close approximation. Here is the problem:
    The pivot point of a simple pendulum having a natural period of 1.00 second is moved laterally in a sinusoidal motion with an amplitude 1.00 cm and period 1.10 seconds. With what amplitude should the pendulum bob swing after a steady motion is attained?​
    The first person to solve the problem correctly, within the above-described constraints on their solution, will win a free copy (one of my author's copies) of the FLP exercise book when it is published (which we hope will happen late this year or early next year). The problem will also be posted on The Feynman Lectures Website with the winner's solution.

    The Feynman Lectures Website will be the sole and final judge of the acceptability and correctness of all submitted solutions. You may email your solutions to me directly (if you want to keep them private) or post them in this thread, so that other people can discuss them.

    One other thing: If you are one of the people with whom I have discussed this problem, or you have heard about it indirectly through me, then you are disqualified from this competition - please recuse yourself - sorry!

    Good luck!

    Best regards,
    Mike Gottlieb
    mg@feynmanlectures.info

    www.feynmanlectures.info
    www.basicfeynman.com
     
    Last edited: Aug 26, 2011
  2. jcsd
  3. Aug 27, 2011 #2
    The outcome of the contest will be announced on January 1, 2012.
     
  4. Aug 27, 2011 #3

    K^2

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    That is a nice problem. Hope you don't mind me taking a shot.

    First, I'm assuming a linear oscillator. The envelope of bob's oscillation must have a period of 11 seconds, since that's the time it takes for the relative phase between two oscillation to go around. The envelope is sinusoidal, and must pass through the node at the same slope as the envelope of a resonance, since it's effectively at resonance at these points. (The phase of the support's displacement leads or lags bob's displacement by exactly 90° there.)

    So the only question is at what rate would amplitude of bob's oscillation grow if both periods were at 1s exactly. The kinetic energy of the bob as it passes through the node is (1/2)mv². The force due to bob displacement is F=kx for some k and x=1cm. We don't know what the k or m is, but k/m=ω², and that's available. Rate of change of pendulum's energy is therefore F*v. With total energy related to current amplitude by E=(1/2)ka², where a is amplitude. v=a/ω. Energy increases at the rate of kxa/ω. So a² is increasing at rate of 2xa/ω. Since a(t)=ct (resonance), a²=c²t² which is area under a triangle y=2c²t, and therefore increases at a rate of 2c²t = 2ac. So the rate of increase of a=x/ω=x/2π.

    So the solution is the amplitude of the sinusoidal wave with period of 11 seconds and passes through zero at slope of x/2π. Sin with period 2π will pass through zero at slope 1. So A=(x/2π)*(11/2π) =11*x = 11cm.
     
    Last edited: Aug 27, 2011
  5. Aug 27, 2011 #4
    So, we solve the problems for you and you have the credit for publishing. Is this how it goes?
     
  6. Aug 27, 2011 #5
    Thanks very much for offering, but we don't need anyone to solve problems for us at Caltech. (So, are you from MIT, or what? :wink:). This particular problem has been solved for a long time. It was first published by Caltech in 1963. It was published again in "Exercises for Introductory Physics" (EIP) by Leighton & Vogt in 1969.

    Our new exercise book will include all the exercises in EIP, including this one, for which a numeric answer will be provided, but no solution. However, in the books (past and future), where this problem appears, there are no constraints on the form of the solution. I added the "no highfalutin math" constraint for this challenge, in the spirit of Feynman's philosophy of solving physics problem using physical reasoning and intuition wherever possible.

    The person who wins the contest, in addition to getting a free book, will have their solution posted with the problem at http://www.feynmanlectures.info" [Broken], so if you notice any, please let us know.)

    Best regards,
    Mike Gottlieb
    www.feynmanlectures.info
     
    Last edited by a moderator: May 5, 2017
  7. Aug 27, 2011 #6
    Of course I don't mind you taking a shot, K^2. However, you only get one!

    In general, I don't plan to comment on people's solutions to this problem, because I don't want to risk inadvertently giving away clues. So, I will not comment on your method of solution, but, since you posted the problem here, I will comment on a couple of things I notice:
    You have (implicitly) defined v as the speed of the bob, a as the amplitude of the bob (which is a distance), and ω as the natural (angular) frequency of the pendulum (= 2π radians/second). Yet you write "v=a/ω," in which the left-hand side has units of distance/time, while the right-hand side has units of distance*time. Similarly, you have "the rate of increase of a=x/ω=x/2π" which has units of distance/time on the left, and distance*time on the right. These equations can not be correct.
    There seems to be a small problem with your arithmetic, but perhaps you meant (11*2π) instead of (11/2π).
     
    Last edited: Aug 27, 2011
  8. Aug 27, 2011 #7

    K^2

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    You're right on both counts. It never ceases to amaze me how often one can make two mistakes in one problem and still get the right answer. I guess I should have checked the units at every step of the way.

    I do feel like there should be a more elegant way to go about it than using energy flow. (Seemed like a good idea at the time, but it got messy.) So hopefully somebody will come up with a better method.
     
  9. Aug 28, 2011 #8
    K^2,

    You never cease to amuse!
    Thank you.
    In order for two mistakes to occur in a (otherwise correct) solution, which amazingly produces the right answer, the mistakes have to be "opposite" in the sense that one undoes the other. That is not so in the case of your solution to this problem, however. One of your mistakes is that you are using an equation "v=a/ω" which is simply nonsense. It doesn't mean anything and can not be true any more than apples can be oranges. (Speeds can not have units of time*distance!). Your other nonsense equation, "a=x/ω=x/2π" followed from this one - so those two mistakes did not "cancel": one implied the other. The third (apparent) mistake I pointed out in your solution was in your arithmetic, which I suggested correcting by changing one of the terms, but that does not appear to "undo" your previous (dimensional) mistake. Your answer (with my arithmetic correction) is "A=(x/2π)*(11*2π) =11*x = 11cm." Where x/2π=x/ω has units of distance*time, and 11 has units of time ("The envelope of bob's oscillation must have a period of 11 seconds") So in order for your answer to have units of distance, the "2π" in the term "(11*2π)" would have to have units of time^-2. I am not sure where this "2π" comes from, but if it is merely radians/cycle (as it appears to be) then it is dimensionless and your answer has units of distance*time^2, whereas if it is ω (which equals 2π sec^-1 in your solution), then your answer has units of distance*time. So, in my opinion, if your answer is numerically correct (and I am not saying that it is), it is not because you have made mistakes that cancel each other.
    It would have sufficed to check the units of your answer.
    With this, I agree completely!
    Just to be clear (in case anyone other than K^2 and I are reading this): I am, in general, not commenting on people's solutions or answers to this problem because I don't want to give away any clues. So, if someone claims they have the right answer, and I don't make any comment about that, my silence is not a silent agreement, it is merely silence.​
     
    Last edited: Aug 28, 2011
  10. Aug 28, 2011 #9

    K^2

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    Oh, I saw what happened with that 2π as soon as you mentioned it. What amazes me is how these things happen by random chance to give you identical factors, to within a change of power, in different places of the problem. Of course, what stacks the odds is the choice of "simple" numbers for the problem. In this case, if the period was anything other than 1s, I'd spot the difference immediately. (Not that I'm blaming the problem. This one's great.)
     
  11. Aug 28, 2011 #10
    Well, I don't see "what happened with that 2π." So far as I can see, your answer is given in the wrong units. Can you please enlighten us by explaining how "random chance [gave] you identical factors, to within a change of power, in different places of the problem" (specifically what places, what powers, etc.) and how that, when accounted for properly by suitable adjustments to your original solution, leaves the answer (11) unchanged and in the correct units (distance)?
     
    Last edited: Aug 28, 2011
  12. Aug 28, 2011 #11

    K^2

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    (1/2)ka²=(1/2)mv² yields v=aω, not a/ω. That's me dividing by 2π instead of multiplying by it. And naturally, changing period of sinusoidal curve from 2π to 11 changes slope by factor of 11/2π, rather than 11*2π. That's me multiplying by 2π instead of dividing by it. Effectively, I include an extra factor of (2π)² and (2π)-2 in two different places, which canceled each other out in the final result. The physics there is solid. It's only the algebra that's messed up.
     
  13. Aug 28, 2011 #12
    K^2,

    You know... authors of academic texts don't make much money, and they don't get many (free) author's copies, so before I make any "decisions" I would like to make sure I understand your solution.

    Call the natural period of the pendulum P, the period of the pivot point's motion p, and assume (as per the original problem) that p>P and that p/P <= 1.10 . Call the amplitude of the pivot point's motion a, and the amplitude of the bob A. So that, for example, in the problem as originally stated, P=1.00 sec, p = 1.10 sec, a = 1.00 cm, and we are asked to find A.] Please express A as a function of the variables p, P, and a, and show that it produces your answer of 11 cm.
     
    Last edited: Aug 28, 2011
  14. Aug 28, 2011 #13

    K^2

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    Sure. I called amplitude of the pivot's motion x, so that becomes a. Relative period (period of the envelope) is T=1/(1/P-1/p). ω is 2π/P. With these substitutions:

    A = (a*ω)*(T/2π) = a(T/P) = a/(1-P/p)

    For p/P = 1.1, 1/(1-P/p)=11. This should work for any ratio p/P, with divergence at p=P corresponding to resonance.
     
  15. Aug 28, 2011 #14
    Thanks, K^2.
     
  16. Aug 28, 2011 #15
    That is not the correct answer. The correct answer should read (for small oscillations of the pendulum, small meaning the amplitude of oscillaiton measured by the maxium angle of deflection from the vertical is much smaller than 1 rad):

    [tex]
    A = \frac{a}{\left(\frac{T_{\mathrm{pivot}}}{T_{ \mathrm{pend.} }}\right)^{2} - 1}
    [/tex]

    When we substitute the numerical values:

    [tex]
    A = \frac{1.00 \, \mathrm{cm}}{\left(\frac{1.10 \, \mathrm{s}}{1.00 \, \mathrm{s}}\right)^{2} - 1} = \frac{1.00 \, \mathrm{cm}}{1.21 - 1} = \frac{1.00 \, \mathrm{cm}}{0.21} = 4.8 \, \mathrm{cm}
    [/tex]
     
    Last edited: Aug 28, 2011
  17. Aug 29, 2011 #16

    K^2

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    {Mathematica code follows.}

    ω = 2 Pi;

    a = 1.0;

    s = NDSolve[{x''[t] == -ω^2 (x[t] - a Sin[ω t/1.1]), x[0] == 0, x'[0] == 0}, x[t], {t, 0, 11}];

    Plot[x[t] /. s, {t, 0, 11}]

    {Output and notebook attached.}


    Just saying.
     

    Attached Files:

  18. Aug 29, 2011 #17
    K^2,

    That's an interesting graph. It shows that the amplitude of the pendulum grows and diminishes periodically, every 11 seconds. (I have attached another graph generated using your notebook but for t=0 to 88 instead of t=0 to 11.)

    I have a suggestion for you: make a pendulum from a piece of string with a weight tied to one end, hold the other end in you hand, and move it slowly back and forth with a steady motion, a little slower than the natural period of the pendulum. Keep it up until the bob attains a steady motion (as per the stated problem). Then compare the amplitude of the bob as a function of time to your graph.
     

    Attached Files:

  19. Aug 29, 2011 #18

    K^2

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    The solution I show assumes no damping. That results in a periodic envelope. Nothing that can be done about that.

    If you take a real pendulum, there is going to be damping. The envelope, then, has a rather complex shape. Try running the same notebook with the following equation:

    x''[t] == -ω^2 (x[t] - a Sin[ω t/1.1]) - 2 ω c x'[t]

    For different values of c, there will be entirely different solutions.


    HOWEVER, one may ask what the amplitude is as c->0 and t->Inf. That is closer in spirit to the question asked, and it does have a solution. I did not consider that. Neither mine nor Dickfore's equations have anything with a solution to THAT problem. If that's what you want solved, that one's still open.
     
  20. Aug 29, 2011 #19
    K^2, the equation you are solving does not correspond to the situation described in the problem.
     
  21. Aug 29, 2011 #20

    K^2

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    Really? Would you like to write out the diff-eq that does?
     
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