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Homework Help: Unusual Harmonic Motion

  1. Jun 7, 2013 #1
    1. The problem statement, all variables and given/known data
    The potential energy of a particle varies as U=K|X|3, it is oscillating and the amplitude is 'A' then find out the time period's variance with 'A'


    2. Relevant equations

    F=-dU/dx
    a=F/m

    3. The attempt at a solution

    none..
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jun 7, 2013 #2
    That's a good start. Now form these equations.
     
  4. Jun 7, 2013 #3

    rude man

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    Just posting to get on the thread.
     
  5. Jun 7, 2013 #4

    haruspex

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    You will need to be careful with the sign when you write out the expression for F as a function of x.
     
  6. Jun 7, 2013 #5
    A bit offtopic though. You can use the "Subscribe to this Thread" found under the "Thread Tools" (top right in the OP) menu.
     
  7. Jun 7, 2013 #6

    ehild

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    You may use conservation of energy. The energy can be obtained from the amplitude. Find the velocity v=dx/dt as function of x, and integrate for a quarter period.

    ehild
     
    Last edited: Jun 7, 2013
  8. Jun 8, 2013 #7
    the equation for v is coming such that it is a cubic in x .. unable to solve it
     
  9. Jun 8, 2013 #8

    ehild

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    There is no solution in closed form, but the definite integral depends somehow on the amplitude. Show what you got.


    ehild
     
  10. Jun 8, 2013 #9

    rude man

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    Thanks. Didn't know.
     
  11. Jun 8, 2013 #10
    @rude man: I see the energy conservation method. But cant it be solved using the equation a= F/m, and then solving the differential equation?
     
  12. Jun 8, 2013 #11

    rude man

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    That was ehild's post that recommended the energy approach.

    I tried what you did, wound up with mx'' + 3k/m x^2 = 0 and could not solve it either. I tried a "guessed" solution of x = Asin(wt) which was disastrous. Evidently, the oscillations are not pure sinusoids, neither should we expect them to be in view of the fact that the ODE is nonlinear.
     
  13. Jun 8, 2013 #12
    Sorry, dint see it was ehild. I didn't expect that the equation would come out like this mx'' + 3k/m x^2 = 0. Nor do i know how to solve this equation(never studied any method to solve non linear equations).
    Now i an can understand why ehild recommended the energy method :D.
     
  14. Jun 8, 2013 #13
    The energy method should yield an equation of the kind dt =f(x)dx, from which one could get the period, if not explicitly, then as a quadrature.
     
  15. Jun 8, 2013 #14
    How about a numerical solution? In my attachment, I have (I think) calculated the horizontal acceleration to be..
    a = (3Agx^2)/(1+9A^2x^4)
    This seems reasonable to me since the horizontal acceleration, to the left or -x direction should be zero when x = inf, and zero when x = 0. So, from a starting point, say (1,1) assuming y = x^3 can we find the time until the acceleration = zero? I haven't done this as yet but it might work.
     

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  16. Jun 8, 2013 #15

    rude man

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    I tried that, starting with mv2/2 + kx3 = kA3, confining to x > 0.

    That gives v(x). Then I said dt = dx/v(x) as I think you suggested. Then I said T/2 = ∫[x/v(x)]dx from 0 to A where T = period.

    What a disaster that resulted in ...
     
  17. Jun 8, 2013 #16

    ehild

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    That gives the magnitude of v(x)... |v(x)|=2k/m√(A3-x3)

    What is that extra x in the integral??? And a quarter period elapses while going from x=0 to x=A.

    Integrating v(x), it leads to [tex]\int_0^A{\frac{dx}{\sqrt{A^3-x^3}}}=\int_0^{T/4}{\frac{2k}{m}dt}[/tex]

    Factor out A3/2 from the square root,and introduce the variable z=x/A, then you get an integral which is dimensionless, a simple number (consult Wolframalpha), multiplied by some function of A.

    ehild
     
  18. Jun 8, 2013 #17

    haruspex

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    I don't think you need to solve the DE. You just need to find the form of T = f(K, A, m). The function f may include a definite integral which could be calculated in principle. E.g. ##\int_0^1\frac{du}{\sqrt(1-u^3)}##
    Edit: I see ehild beat me to it
     
  19. Jun 9, 2013 #18

    rude man

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    Don't know where I picked up that extra x either. Senior moment I guess. And yes, 0 to A is only 1/4 period, not half. Otherwise that's exactly what I did. Thanks.
     
  20. Jun 9, 2013 #19
    maybe we can work on an analogy.... A ball (of negligible radius {to cancel rotational energy})that rolls down a slope of equation y=x^3.. though i really don't see how it helps
     
  21. Jun 9, 2013 #20

    rude man

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    A simple number?

    Doing what you suggest you wind up with the integral as A5/2∫dz/(1 - z3)1/2 with limits 0 and 1 which is stil a disaster, with elliptic functions, imaginary terms, and singularities for z = 1 (and possibly elsewhere), corresponding to x = A.

    Perhaps you could show us the rest ...

    EDIT: Ok, with the integral evaluated at z = 0 and z = 1 I see from the plots that we get a real number, and the imaginary part, being constant over that interval, evaluates to zero.

    All I can say is, thank God for the graphic displays of wolfram alpha!
     
    Last edited: Jun 9, 2013
  22. Jun 9, 2013 #21

    rude man

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    Good analogy, but I join you in not knowing how it helps solve the problem.

    Using wolfram alpha and its graphic results, we see that the impossible-looking integral evaluates to a finite real and zero imaginary number. As far as I'm concerned the problem is solved. Do you have any questions about the energy approach ehild suggested way back in post #6?
     
    Last edited: Jun 9, 2013
  23. Jun 9, 2013 #22

    ehild

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    Why is A in the numerator?????


    [tex]\int_0^A{\frac{dx}{\sqrt{A^3-x^3}}}=\int_0^A{\frac{dx}{A^{3/2}\sqrt{1-(x/A))^3}}}=\frac{1}{A^{1/2}}\int_0^1{\frac{dz}{\sqrt{1-z^3}}}[/tex]
    (z=x/A).

    According to Wolframalpha, the last integral is about 1.402.

    http://www5a.wolframalpha.com/Calculate/MSP/MSP18621f124d5ch6cg4i0c00001c2h70h366557i3g?MSPStoreType=image/gif&s=33&w=258.&h=58 [Broken].

    ehild
     
    Last edited by a moderator: May 6, 2017
  24. Jun 9, 2013 #23

    rude man

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    Another senior moment. It's still in the numerator, as A-1/2. :smile:

    Without something as powerful as wolfram alpha this would not have been solvable, far as I'm concerned. I wonder what the question poser had in mind.

    PS how do you do definite integrals in wolfram alpha?
     
    Last edited by a moderator: May 6, 2017
  25. Jun 9, 2013 #24

    ehild

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    I think the question was how the time period depended on the amplitude. And it is proportional to A-1/2.

    I wrote in the definite integral to Wolframalpha as

    integral _0^1 (dz/sqrt(1-z^3))

    and that was the result:

    [Broken]

    ehild
     
    Last edited by a moderator: May 6, 2017
  26. Jun 9, 2013 #25
    @ehild: I tried solving the integral on my I own my but just got stuck. Is there a method?
     
    Last edited by a moderator: May 6, 2017
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