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Finding the time period using the potential

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data
    A particle of unit mass moves in one dimension with potential V(x) = ½μ2x2 + εx4 (ε>0). Discuss the motion of the particle.
    If the particle released from rest at x=a (a>0) express the time period T for the particle to return to a in the form of an integral and show that when ε is small, T is reduced by approximately 3επa23.

    2. Relevant equations
    v = dx/dt = ±√{(2/m) * (E - V(x))}
    Time period taken for a particle to move from point x1 to x2:
    T = ∫1/{v} dx

    3. The attempt at a solution
    Discuss the motion:

    If we sketch V(x) it will look alike to parabola with V(0) = 0. Depending on the value of E that the particle has, it will have a simple harmonic motion and will oscillate in the region enclosed between E and the x-axis.

    Time period to return to a:
    I obtained this and it matched with the answers (sorry for the long expression I wanted to upload a picture but it did not work)
    T=(4/μ) ∫(dx)/(√{(a2-x2) +(2/μ2)ε(a4-x4)})

    Show that T is reduced by approximately 3επa23
    In this part I am not sure how the fact that ε is small will affect the integral. I am not sure how small the numerator will become and whether the denominator will be greater than numerator. In this case should I just integrate?

    Thank you for your help!
     
  2. jcsd
  3. Oct 31, 2016 #2

    TSny

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    ##\varepsilon## is assumed to be small enough such that ##\frac{2}{\mu^2} \varepsilon (a^4 - x^4)## is small compared to ##a^2 - x^2##. Expand ##\left[a^2 - x^2 + \frac{2}{\mu^2} \varepsilon (a^4 - x^4) \right]^{-1/2}## to first order in ##\varepsilon##.
     
  4. Oct 31, 2016 #3

    TSny

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    V(x) will look approximately like a parabola, but it is not a true parabola. So, although the particle will oscillate, is the motion simple harmonic motion?
     
  5. Oct 31, 2016 #4
    Hi, thank you for your comments! Regarding the part with ε, could you please specify what do you mean by "expand to first order in ε"? Do you mean to expand (2/μ2) * ε(a4−x4) using Taylor series and then take the first two terms?
    As for the description of the motion, I thought that the particle will oscillate between two points depending where it starts on the graph and so will oscillate in a similar way to a pendulum. However, I am not sure if it's SHM or not since when oscillating the particle will be likely to lose some of its energy and so the amplitude of its oscillations will reduce. At the same time the particle will lose some of its energy due to external forces and the question does not mention any and so I am not sure if I can assume that or not.
    Thank you once again!
     
  6. Oct 31, 2016 #5

    haruspex

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    No, expand
     
  7. Oct 31, 2016 #6
    I am sorry but I do not quite understand whether you mean to expand using Taylor series or not and if yes, about what x_0? I tried expanding it around x_0=0 and my first derivative gave me 0 leaving me with only f(0) as an approximation, and obviously this cannot be the case since we will not be able to get pi in the final expression. I also tried opening the brackets and completing the square. Then I used arcsin to solve the integral and I got a very long scary expression which I can't really simplify :(
    I am sorry for being slow but I can't really see what else I can do.
     
  8. Oct 31, 2016 #7

    PeroK

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    I took a look at this, but I must confess I don't understand the question. If ##\epsilon## is small, then ##T## is reduced by something compared with what?

    Edit: is it compared with ##\epsilon = 0##?
     
  9. Oct 31, 2016 #8

    ehild

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    Yes, it is compared with the time period when ##\epsilon = 0##.
     
  10. Oct 31, 2016 #9

    PeroK

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    Thanks @ehild

    @green-beans I would take out a factor of ##(a^2 - x^2)##, if you haven't already. Also, it's easier to use the Binomial expansion than try to differentiate your function for the Taylor series.
     
  11. Oct 31, 2016 #10

    ehild

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    You have to expand with respect to ε and stop at the first-order term.
    As @PeroK suggested, first take out the factor a2-x2 from ##
    \left[a^2 - x^2 + \frac{2}{\mu^2} \varepsilon (a^4 - x^4) \right]^{-1/2}##
    , and expand the other factor with respect to ε. Note that a4-x4=(a2-x2)(a2+x2).
     
  12. Oct 31, 2016 #11
    So, by saying to expand with respect to ε, do you mean to take x_0 = ε?
     
  13. Oct 31, 2016 #12

    ehild

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    No. Take ε_0=0.
     
  14. Oct 31, 2016 #13
    Oh, so you differentiate as well in respect to epsilon and regard it as a function of ε and expand it using Taylor series around epsilon =0?
    Sorry for asking so many questions. It's just I tried expanding it, and I did not get the right answer
     
  15. Oct 31, 2016 #14

    PeroK

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    That's another way to do it. But, the most obvious way is just to think of ##\epsilon## as some small number and find the difference between the two integrals: one with ##\epsilon = 0## and the other with some small number. You should recognise the set-up for using the Binomial expansion.
     
  16. Oct 31, 2016 #15
    Thank you for your help! I still keep on getting something close to the answer but I can't figure out what I am doing wrong. Here I attached the link to my solution: https://drive.google.com/open?id=0BzewpRKo76bDWWNld1U3SHNBVEk
     
  17. Oct 31, 2016 #16

    PeroK

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    It looks like you just have a couple of algebraic lapses in there somewhere. Easily done and hard to spot!

    I can't immediately pin down where you've gone wrong.

    I got:

    ##T = \frac{2}{\mu} \int_{-a}^{a} (a^2 - x^2)^{-\frac12} [1 + \frac{2\epsilon}{\mu^2}(a^2 + x^2)]^{-\frac12} dx##

    Do you get that?
     
  18. Oct 31, 2016 #17
    Oh, I think I copied the initial integral incorrectly. I'll redo it again and I'll get back here :) thank you once again
     
  19. Oct 31, 2016 #18

    PeroK

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    Just get that integral above and hit it with the Binomial! It's much simpler.
     
  20. Oct 31, 2016 #19
    So, I got a similar integral but in front of it I got 4/μ and the limit between 0 and a. So, that should be the same thing. When using Taylor series for ε I then obtain two integrals. The first one is 1/sqrt(a^2-x^2)^(-1/2). The second one (-ε/μ^2)*(a^2+x^2)/sqrt(a^2-x^2). When solving the integral, I still get the wrong answer which is (4/μ) (pi/2 - (ε/μ^2)*(3pia^2/4) which would have been the answer if I did not get pi/2 for the first term. So, I am not sure what I am doing wrong since I redid the integral several times :(
     
  21. Oct 31, 2016 #20

    PeroK

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    The first integral is the period when ##\epsilon = 0##, so only the second integral is the difference when ##\epsilon \ne 0##.
     
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