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Harmonic Oscillator questions

  1. Feb 26, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    1)Consider a particle subject to the following force ##F = 4/x^2 - 1## for x>0.
    What is the particle's maximal velocity and where is it attained?

    2)A particle of unit mass moves along positive x axis under the force ##F=36/x^3 - 9/x^2##
    a)Given that E<0 find the turning point(s) as a function of E.
    b)Expand the potential about the equilibrium and obtain the period in the harmonic approximation.

    3. The attempt at a solution
    1) is a part of a bigger question, but only here do I need some advice. I said that the max velocity occurs at E=T => ##\frac{m}{2}\dot{x}^2 = E##, so ##\dot{x} = \pm \sqrt{2E/m}##. So this must occur when ##V(x) = 0 => \frac{4}{x} + x = 0##, which has no solutions. This does not seem right somehow?

    2)
    a) Just wondering: Why is the plural suggestive? For E>0, the particle is unbounded, but for E<0, then I think there would always exist two points x<x' such that oscillation would occur between these points.
    b) So ## V(x) \approx V(x_{equil}) + V''(x_{equil})(x-x_{equil})^2/2##. I found equilibrium point x=4 so I think what the question wants is V(x) ≈ -9/8 + k/2 (x-4)2. To find the period do I assume T = 2π √(m/k)?

    Many thanks
     
  2. jcsd
  3. Feb 26, 2013 #2

    NascentOxygen

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    For part (1), have you sketched F? Find the point where F changes from a push to a pull, and that will co-incide with maximum speed.

    You probably have a sign wrong in V(x).
     
  4. Feb 26, 2013 #3
    Total energy is conserved. When velocity is maximal, kinetic energy is maximal. What does that mean w.r.t. potential energy?

    This assumption is baseless. There can be any number of turning points, they don't have to come paired. For example, imagine there is some great impenetrable wall in space and there is nothing else. Then the wall is just one turning point, and motion can continue infinitely in both directions from the wall.
     
  5. Feb 27, 2013 #4

    CAF123

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    This occurs at x=2, so then max velocity is ##\dot{x} = \pm \sqrt{2/m} = \pm 1## for m=2.
    What was wrong with my way before? I worked out from a previous part that E = 5. So subbing this in does not give me the answer I just got now.

    Looking at my sketch of V(x) I see that V(x) is never zero. So it makes sense that 4/x + x never has any solns. Furthermore, if this is the case, then what I said about at max velocity E=T would be wrong since the particle would always have some potential energy.Yes?
     
  6. Feb 27, 2013 #5

    CAF123

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    Then potential energy is the smallest it can be and looking at my graph, this is when V(x)=4. So solving for 4/x + x = 4 => x = 2. Indeed, this is at max velocity so ##T + V = E => m/2 v_{max}^2 + 4 = 5##. Solve to get same answer as before. (I think this resolves my problem posted in response to NascentOxygen)

    So I think it is right to say that at x=2, potential energy is smallest but the particle has max velocity there.
    This is exactly the case here. For small x, the potential is tending to ∞, while for larger x V → 0. But for E < 0, I found two turning points. Is this right? EDIT: (I mean two turning points for each value of E<0)
     
  7. Feb 27, 2013 #6
    Not quite so. In my wall example, there is no force (constant potential) everywhere outside the wall, and a very strong repelling force in its immediate vicinity. In case 2), however, there is an attractive force as well, albeit waning with distance. Which is why for certain energy levels you have two turning point for x > 0.
     
  8. Feb 27, 2013 #7

    CAF123

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    Ok, I think maybe what I meant was that in my example when E > 0, the particle is bounded to the left (since V tends to infinity) but not so to the right (since V tends to 0). So here the particle would not undergo oscillations since there is no possible turning point for the particle at larger x. (and hence no confinement)

    Are my results good now? What about the period question in 2B)?
     
  9. Feb 27, 2013 #8
    I do not see that you have found the turning points as a function of E.

    2B seems good so far.
     
  10. Feb 27, 2013 #9

    CAF123

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    Turning points where E = V. So ##E = 18/x^2 - 9/x => Ex^2 + 9x -18 = 0##. Solve quadratic to give $$x_{1,2} = \frac{-9 \pm \sqrt{81 + 72E}}{2E}$$

    I found k to be 9/32 so T = ##2 \pi \sqrt{32/9} \approx 11.8##
     
  11. Feb 27, 2013 #10

    CAF123

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    I want to try to find this period without using the approximation and show that the results are consistent for small oscillations.
    I know $$T = \sqrt{2m} \int_{x_1}^{x_2} \frac{dx}{\sqrt{E - (18/x^2 - 9/x)}},$$ ##x_1, x_2## the two turning points.

    Rearranging the part inside the sqrt to ##Ex^2 + 9x -18## I made this equal to ##(x_2 - x)(x - x_1)##. So my intergal becomes $$T = \sqrt{2m} \int_{x_1}^{x_2} \frac{ x dx}{\sqrt{(x_2 - x)(x - x_1)}}$$

    Now let ##x = x_1 cos^2 \theta + x_2 sin^2 \theta##. Differentiating, subbing in and cancelling I end up with two apparantly simple looking integrals: $$2 \sqrt{2m} \int_{\theta_1}^{\theta_2} x_1 cos^2 \theta d \theta + 2 \sqrt{2m} \int_{\theta_1}^{\theta_2} x_2 sin^2 \theta d \theta.$$

    My question is: How do I solve for the limits?
     
  12. Feb 27, 2013 #11
    I recommend that you let z = 1/x and solve the original equation for z, and then find x. It will have a more readily understandable dependency on E so you can analyze it further.

    I have k = 9/64. How did you get yours?
     
  13. Feb 27, 2013 #12
    Obviously, ## x = x_1 ## corresponds to ## \theta = 0 ##, and ## x = x_2 ## corresponds to ## \theta = \pi / 2 ##.
     
  14. Feb 27, 2013 #13

    CAF123

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    Ok, I'll try this.

    Found V''(4), is it what you did?
     
  15. Feb 27, 2013 #14

    CAF123

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    I don't see how you got this. Putting x = x1 into the eqn defined for x gives x1 = x2 which is just nonsense. Thanks.
     
  16. Feb 27, 2013 #15
    ## V''(x) = 108 / x^4 - 18/ x^3 => V''(4) = 108/4^4 - 18/4^3 = 27/4^3 - 18/4^3 = 9/64 ##

    Let ## \theta = 0 ##. What do you get?
     
  17. Feb 27, 2013 #16

    CAF123

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    Yes, I made a silly error.

    Ok, I see. But there is no analytic solution to that eqn (it is transcendental, right?). So you basically solved it by inspection.

    Why if I just put x = x1 do I get nonsense?
     
  18. Feb 27, 2013 #17
    Yes, initially I just guessed. But then I made sure the guess made sense. In particular, I computed the derivative and determined that if x2 > x1, the derivative is positive when ## 0 \le \theta \le \pi/2 ##, so the function just grows monotonically. That means I can basically map any angular subrange into any [x1, x2] range. I just choose the angles and then solve for the coefficients at the sine and the cosine terms.

    If the coefficients are fixed like we have it here, then, due to the monotonicity, there is just one solution. If they are fixed in some other way, we may or may not have a solution, but that requires messy trigonometry to decide.
     
  19. Feb 27, 2013 #18

    CAF123

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    Thanks voko for your help.
    One more question though: Going back to question 1). At E=4,the energy is all potential (since at E=4, this corresponds to the minimum of potential V = 4). Does this mean if a particle had energy E=4 then it would be stationary? And only for E>4 does the particle exhibit oscillatory motion.
     
  20. Feb 27, 2013 #19
    Yes, when the total energy of a system is at a minimum of its potential energy, then the system can only be stationary. This follows directly from conservation of energy.
     
  21. Feb 27, 2013 #20

    CAF123

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    In the integral method to find the period (non approximation), I get that T = ##\sqrt{2 \pi} - \sqrt{2}##. However, the next part of the question asks to verify that this is consistent with what I did in the harmonic approximation.

    Can I write that for small oscillations: ##cos 2\theta \approx 1 - \frac{(2\theta)^2}{2}## and ##sin 2 \theta \approx 2 \theta##? When I do this, I don't get the solutions to match.

    EDIT: Using T = ##2\pi \sqrt{m/k}## gives a period of about 16s which is much larger than what I get when I do the integral properly.
     
    Last edited: Feb 27, 2013
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