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First steps to understand small oscillations in CM +1 little problem

  1. Jul 18, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    I'm trying to teach myself Small Oscillations in Classical Mechanics. So far I've read in Landau, Golstein, Wikipedia and other internet sources but this subjet seems really tough to even understand to me.
    What I understand is that if we have a potential function that depends only on the position (to make it simple, let's keep it in 1 dimension) and that function has a minimum say in x0, then we can approximate U(x) close to U(x0) by using a quadratic. To be more exact, we use a Taylor's polynomial of order 2 centered in x0. So far so good.
    Now, how does this help in finding say the fundamental frequencies of a system?!
    Problem #1 in my university assignments for CM is "Find the frequencies for small oscillations if [itex]U(x)=V \cos (ax)-Fx[/itex]."
    So first I think I should find the minimum of this function to get x0. I found it to be [itex]x_0=\frac{\arcsin \left ( \frac{-F}{Va} \right) }{a}[/itex].
    I notice that the argument of the arcsine function has to be within -1 to 1 to even make sense, so that I get a condition on F and Va.
    But now what...? I'm totally stuck. Should I use a Taylor's polynomial and what does it have to see with the frequencies of oscillations?
     
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  3. Jul 19, 2011 #2

    vela

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    You can neglect the constant, so you end up with a potential that looks like[tex]U(x)\approx\frac{1}{2}kx^2[/tex]where k is U''(x0). Do you recognize that potential?
     
  4. Jul 19, 2011 #3

    fluidistic

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    That looks like the potential energy stored in a spring.
    When you say I can neglect the constant, which one do you mean exactly? I don't really see how to obtain a quadratic from the given equation.
     
  5. Jul 19, 2011 #4

    ehild

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    The reference point of the potential is arbitrary, you can choose U(x0)=0

    The Taylor expansion of the potential around x0 starts with a quadratic term 1/2 k (x-x0)2 as the potential is minimum at x0, so the linear term of the Taylor series is zero.


    ehild
     
  6. Jul 19, 2011 #5

    vela

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    Whoops! As ehild noted, I should have said [itex]U(x) \approx 1/2 k (x-x_0)^2[/itex]. Your Hamiltonian or Lagrangian will therefore look like that of a simple harmonic oscillator for small deviations from the equilibrium point.
     
    Last edited: Jul 19, 2011
  7. Jul 19, 2011 #6

    ehild

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    Usually the deviation from equilibrium is considered as a new variable, u=x-x0, and then the potential is of form U=1/2 k u2.

    ehild
     
  8. Jul 19, 2011 #7

    fluidistic

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    Ok thanks guys.
    Ah now I understand what is the constant we can neglect (I think it's the first term in the Taylor's expansion of U(x) around U(x0), namely U(x0)). I also understand that since U'(x0)=0, the Taylor expansion of order 2 is indeed [itex]\frac{U''(x_0)(x-x_0)^2}{2}[/itex].
    The Lagrangian is then [itex]\frac{m \dot u ^2}{2}- \frac{ku ^2}{2}[/itex]. I wrote down and solved Euler's equation. I get [itex]\ddot u + \omega ^2 u=0[/itex] where [itex]\omega = \pm \sqrt { \frac{k}{m}}[/itex].
    So the solution is [itex]u=A\cos (\omega t+ \phi )[/itex].
    I hope I didn't make any error so far.
    Now I have to determine A, omega and phi (which I know is arbitrary so I don't have to find it I guess).
    Starting with omega... I need to find k. [itex]U''(x_0)=-a^2 \cos \left ( \frac{\arcsin \left ( - \frac{F}{a} \right ) }{a} \right ) =k[/itex]. Which seems quite complicated (and ugly).

    So now I have to determine A, right? Hmm I'm not sure how to do it. Should I work with the Lagrangian containing the approximation of U or the true value of U?
    I know that [itex]A=u[/itex] when [itex]\dot u=0[/itex]...

    Edit: Hmm what?! My value for k is negative, it's impossible! I made an error somewhere!

    Edit 2: Nevermind, k could be negative if the cos value is negative.
     
    Last edited: Jul 19, 2011
  9. Jul 19, 2011 #8

    vela

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    Make sure you get the right extremum. The equation [tex]\sin ax = -\frac{F}{aV}[/tex] has multiple solutions, half of which correspond to maxima.
     
  10. Jul 19, 2011 #9

    fluidistic

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    Oh right. Well I think I had it right.
    [itex]ax=\arcsin \left ( - \frac{F}{Va} \right ) +2 \pi n[/itex] with [itex]n \in \mathbb{Z}[/itex].
    If I take n=0 I get what I had. If I take any other n too... so I guess I'm missing something. Hmm.
     
  11. Jul 19, 2011 #10

    ehild

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    You need to discuss at what range of the parameter values are small oscillations possible: there has to be a point where U'=0 and U">0.
    Is equilibrium possible if |F/Vα|>1?

    I was in error in my previous post when I wrote that the constant term of the potential is 0, as it is Vcos(ax0)-Fx0, but it does not influence the equation of motion you get from the Lagrangian, [itex]\ddot u + k u^2=0[/itex]

    The solution is Acos(ωt+φ) if k positive. What is the solution in case k<0?

    ehild
     
  12. Jul 21, 2011 #11

    fluidistic

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    No, the equilibrium point x0 I obtained in my first post makes sense only for |F/Vα| lesser or equal to 1. So equilibrium isn't possible for this modulus to be greater than 1.
    Ok thanks a lot, I'll come back to this tomorrow (have to sleep soon).
     
  13. Jul 22, 2011 #12
    As you said, [itex]U(x)[/itex] can be written as:[tex]U(x) = U(x_0) + \frac{1}{2} k (x-x_0)^2 + \cdots[/tex]
    for some positive number [itex]k[/itex].

    Small oscillation means deviation from equilibrium is small, so [itex]x-x_0[/itex] is small. This justify dropping all higher order term abbreviated above. Then the potential becomes that of a simple harmonic oscillator up to an unimportant additive constant. So fundamental frequencies corresponds to the frequencies of the SHO.
     
  14. Jul 22, 2011 #13

    fluidistic

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    Hmm I'm not sure. [itex]k=-a^2 \cos \left ( \frac{\arcsin \left ( - \frac{F}{a} \right ) }{a} \right )[/itex]. For it to be negative implies that the cosine is positive.
    So that [itex]-\frac{\pi a}{2}<\arcsin \left ( -\frac{F}{a} \right )<a\frac{\pi}{2}[/itex].
    I feel like I'm missing the point.
     
  15. Jul 23, 2011 #14

    ehild

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    As nothing is said about F and a in the original problem, and arcsin is multi-valued, k can be both positive and negative. Try F=0.5 and a=1, for example and see what values are possible for k.


    ehild
     
    Last edited: Jul 23, 2011
  16. Jul 23, 2011 #15

    fluidistic

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    Hey ehild, now I just think about it... since [itex]k=U''(x_0)[/itex] and if it's negative, it means that U(x_0) is a maximum for U, not a minimum as one would expect in case of a stable equilibrium point. Thus discussing a possible negative value for k is out of discussion or I'm wrong?
    Anyway, assuming I want to see what happens if k is negative, I still do not see the point.
    If you take a=1 and F=0.5 (these are possible values since |-F/a|=0.5<1) then I'd get the motion equation [itex]m \ddot u -v \cos \left [ \arcsin \left ( - \frac{1}{2} \right ) \right ]u=0 \Rightarrow m \ddot u -\omega ^2 u=0[/itex]. Where u=x-x0.
    I think the solution is [itex]u=e^ \frac{u \omega }{\sqrt m}[/itex] if that is what you asked for (I hope I didn't make any error).
     
  17. Jul 23, 2011 #16

    ehild

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    Yes, and the same with negative exponent. A small deviation, both positive and negative, will increase. There are small oscillations only around the minimum point. Edit: It is a stable equilibrium, as there is a restoring force on the displaced object, although it will not be stationary any more.

    So the conditions that small oscillations occur around a point x0 are that

    sin(ax0)=-F/(Va) and Va2cos(ax0)<0

    There are infinite number of such points. Supposing both V and F are positive, ax0 must be in the third quadrant of the unit circle.
    If F <0 and V>0 sin(ax0)>0, ax0 is in the second quadrant.

    The function arcsin gives a value between -pi/2 and pi/2, so ax0 = pi-arcsin(-F/(Va)) + (2pi n).

    ehild
     
    Last edited: Jul 24, 2011
  18. Jul 23, 2011 #17

    fluidistic

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    Ok thanks a lot for all your help. I'll take some more time to fully understand the latest part of your post. Though for the first part, I think that taking an hypothetical negative value for k and finding+solving the equation of motion is an overkill.
    From analysis or calculus, we know that if f'(a)=0 and f''(a)<0 then f has a maximum at a.
    In our case we have [itex]U''(x_0)=k[/itex]. We also have that [itex]U'(x_0)[/itex]. So that if k<0 we do not get a stable equilibrium point ([itex]U(x_0)[/itex] is a maximum rather than a minimum). We do not have to show this explicitly by evaluating the equation of motion (though a confirmation is always welcome)
    However if k>0, [itex]U(x_0)[/itex] is a minimum and [itex]x_0[/itex] is a stable equilibrium point. I notice that k>0 implies that, as you said, Va2cos(ax0)<0.
    Now feel free to correct me. For instance I assumed that [itex]x_0[/itex] cannot be an equilibrium point if [itex]U(x_0)[/itex] is a local maximum. I hope I am right by saying this. Could you confirm it?
     
  19. Jul 24, 2011 #18

    ehild

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    It is an equilibrium point, but an unstable one. If you place an object there, it stays stationary, but it leaves x0 in case of the slightest displacement.

    The minimum point is also an equilibrium point, a stable one. If you slightly displace the object from there, it will experience a restoring force.

    I suggest to read http://en.wikipedia.org/wiki/Mechanical_equilibrium

    ehild
     
  20. Jul 24, 2011 #19

    fluidistic

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    Ok thank you very much! I just read wikipedia's article and I understood it. Thanks for all your help guys.
     
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