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Harmonic motion

  1. Oct 29, 2016 #1
    1. The problem statement, all variables and given/known data
    A potential energy function for a particle moving in one-dimension is given as:
    V (x) =k1x^2/(2)+k2/x

    (a) Locate all the equilibrium points.
    (b) Show that the motion is always periodic for any amount of total energy.
    (c) What is the frequency f the motion if the amplitude of oscillation is very small?

    2. Relevant equations

    3. The attempt at a solution:
    I figured out part a, and I couldn't figure out how to do c. Any advice on how to do it would be appreciated.
    My explanation for b was as follows:
    since the second derivative of the potential is positive, the particle will undergo periodic motion since no amount of energy can make it "escape", rather, it will be trapped forever and undergo periodic motion since the potential energy graph is positive.

    Is this a correct explanation?
     
    Last edited by a moderator: Oct 30, 2016
  2. jcsd
  3. Oct 29, 2016 #2

    Simon Bridge

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    (a) equilibrium points are the stationary points of V(x) right - they are stable if the point is concave up. This what you said?
    (b) V(x) is concave up everywhere in x, therefore there is no energy higher than V. This is basically what you said.
    (c) You have to use the definition of periodic motion, frequency etc. ... maybe the equation of motion will help here? F = -V' = ma
     
  4. Oct 30, 2016 #3
    Yup that's what I said for a and b. Not sure how F=-V' will help for part c? What formula do I use, and how do I use the fact that the amplitude is extremely small?

    Thanks
     
  5. Oct 30, 2016 #4

    PeroK

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    You need to look at how ##F## varies about the equilibrium point, ##x_0##. Finding a Taylor series expansion often helps. Does that make any sense?
     
  6. Oct 30, 2016 #5
    I was thinking of using a taylor series expansion, but i'm unsure on how I would go about doing this? I"m not very proficient with taylor series, could you help me set it up?
     
  7. Oct 30, 2016 #6

    PeroK

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    Let me explain the idea. The Taylor series For ##F## about ##x_0## is:

    ##F(x) = F(x_0) + F'(x_0)(x-x_0) + \frac12 F''(x_0) (x-x_0)^2 \dots ##

    Now, if ##(x-x_0)## is small, then you can neglect the terms in ##(x-x_0)^2## and higher. Further, if ##F(x_0) = 0## (an equilibrium point), then:

    ##F(x) \approx F'(x_0)(x-x_0)##

    And, if ##x_0## is a local minimum for ##V##, then ##V''(x_0) = -F'(x_0) > 0##

    So, we have an equation for simple harmonic motion:

    ##m\ddot{x} = -k(x-x_0)##, for some ##k = -F'(x_0) > 0##
     
  8. Oct 30, 2016 #7
    Okay, thanks. Here is what we have so far, but we are getting stuck when trying to solve for the frequency, since we get it as an implicit function? For reference, our local min ##x_0=(\frac{k_2}{k_1})^{\frac13}##

    upload_2016-10-30_13-8-43.png
     
  9. Oct 30, 2016 #8

    PeroK

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    I can't read that. The frequency is always the same, based on ##k/m##
     
  10. Oct 30, 2016 #9
    So, we solved for k, and got it to be equal to ##2k_1##. So the corresponding period then would just be ##2 \pi \sqrt{\frac{2k_1}{m}}## and the frequency is just ##\frac{1}{2\pi}\sqrt{\frac{m}{2k_1}}##?
     
  11. Oct 30, 2016 #10

    PeroK

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    Do you think a larger mass would have a higher frequency? You might want to check ##2k_1## as well.
     
  12. Oct 30, 2016 #11
    Sorry, I got the inside of of roots mixed up.

    Based on the taylor series expansion, we get ##F(x_0) = -k(x-x_0)##, where ##k=-F(x_0)##. Then for an initial point ##x_0=(\frac{k_2}{k_1})^{\frac 13}## ##k=-(k_1+\frac{k_2}{x^3})=-(k_1+k_2\frac{k_2}{(\frac{k_2}{k_1}^3)^{\frac 13}}=-2k_1##. The problem that we run in to here, is that to solve for the angular frequency ##\omega_0## in the equation ##f=\frac{\omega_0}{2\pi}## using equations that we have derived in other parts of our assignment and in class, ##x(t)=Asin(\omega t)## where A is the amplitude, we get an implicit function which appears to be unsolvable due to it being ##-k(x-x_0)## rather than ##-kx## which is the typical function for a spring. If we do sub k into the standard equations for a spring, we end up getting a negative within a square root, which gives an imaginary frequency.

    Could you possibly point us in the right direction from here?
     
  13. Oct 30, 2016 #12
    Sorry, our ##k## should be ##3k_1##, not ##-2k_1##. That would give us a real number for the frequency, but I'm still not sure if it makes sense based on the derivation for the angular frequency that we did in class
     
  14. Oct 30, 2016 #13

    PeroK

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    You need to check ##F'(x)##, especially the derivative of ##1/x^2##.

    Second, there's no real difference having ##x-x_0##, as that just adds a nonzero equilibrium point. You could change variables to ##u = x - x_0## if you're really stuck.

    For positive ##k##, you get ##\sin## and ##\cos## solutions to your differential equation.
     
  15. Oct 30, 2016 #14

    PeroK

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    Yes, ##3k_1## is correct. It's just a positive constant, so why would it not make sense. The period and frequency are as you had them before, although you has them inverted.
     
  16. Oct 30, 2016 #15
    Thank you, I fixed my errors and everything worked out
     
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