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Small Oscillations

  1. Sep 23, 2009 #1
    1. The problem statement, all variables and given/known data

    Consider an atom of mass m bonded to the surface of a much larger immobile body by
    electromagnetic forces. The force binding the atom to the surface has the expression

    F = eacosz + bsinz + dtanz

    where a, b and d are constants and z is positive upwards. The equilibrium point is defined
    to be the origin, so z=0 there. Ignore any motion except in the vertical direction. The
    whole assembly is subject to normal Earth gravity.

    a) For small oscillations, give an approximate expression for the binding force on the
    atom.
    b) What restrictions are there on the values of a, b and d so that the force on the atom
    actually is a restoring force and the atom can reach a stationary equilibrium?
    c) What is the angular frequency ω0 and frequency ν0 of oscillation of this undamped
    system?
    d) Would the oscillation frequency change if there were no gravity? Why?
    e) Subject to the considerations above, if the atom has a mass of 1 atomic mass unit,
    30 b = −1.658×10 and 4 d = −3.2361×10 N, what is the frequency ν0 of the oscillation?
    f) Suppose that a photon with this frequency that is incident on this atom would be
    absorbed. What wavelength does this correspond to? What part of the spectrum does it
    fall in?


    2. Relevant equations



    3. The attempt at a solution

    I have completed the problem, and think I'm correct. Any confirmation, thoughts would be greatly appreciated!

    a) in the z component, we can simplify the expression as a taylor series, disregarding the z terms with higher power terms than 2.

    F(z) = F0 + (dF/dz)0z

    F0 = 0 at equilibrium so F(z) is simply the first derivative of F evaluated at 0

    =(bea + d)z

    b) restoring force means that k (spring constant) must be negative.

    so bea + d < 0

    re-arranging yields that d < 0, b > 0 and a < ln(-d/b)

    c) w0/SUB] = [tex]\sqrt{k/m}[/tex] where k = bea + d

    and v0 = w0/2[tex]\pi[/tex]

    d) No, because with microscopic distances and particles, electromagnetic force dominates over gravity, so gravity has a negligible effect.

    e) Need to solve for a, so under equilibrium, force must balance gravity so F0 = mg. Fo from original eqn is ea so we can solve for a.

    a = ln(mg) = 64.96 N

    plugging in then yields vo as 6.42 x 1014 Hz, which is 467.3 nm, in the visible.

    Thoughts?
     
  2. jcsd
  3. Sep 23, 2009 #2

    gabbagabbahey

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    Careful, equilibrium means that the net force on the particle is zero....there is more than just the binding force acting on the particle...what do you really get when you plug the equilibrium point into your original expression for the binding force?
     
  4. Sep 23, 2009 #3
    I have taken that into account, in part e. under equilibrium, binding force is balancing with gravity. I am quite sure part a is correct.
     
  5. Sep 23, 2009 #4

    gabbagabbahey

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    I'm quite sure part (a) is incorrect...the question asks for an approximate expression for the binding force, not the net force....[itex]F_{net}(z=0)=0[/itex], but [itex]F_{binding}(z=0)=[/itex]____?
     
  6. Sep 23, 2009 #5
    my textbook (that the prof follows very closely) states that for small oscillations it is acceptable to expand as a taylor series, disregard the powers of z higher than 1, and that Fo vanishes from equilibrium, and you're left with the first derivative.

    If I'm wrong, please explain.
     
  7. Sep 23, 2009 #6

    gabbagabbahey

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    The problem is that you are asked to expand the binding force for small equilibriums...F0 only equals zero, if F0 is the net force at equilibrium...the binding force does not need to be zero at equilibrium.

    Plug z=0 into you original expression for the binding force...and into your answer for part (a)...do you get the same result?
     
  8. Sep 23, 2009 #7
    so you're saying F(z) = aexpe + bexpa + d ? the aexpe term being binding force at 0.
     
  9. Sep 23, 2009 #8
    sorry thats expa not aexpe
     
  10. Sep 23, 2009 #9

    gabbagabbahey

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    Close, Taylor expanding the binding force around the equilibrium point ([itex]z=0[/itex]) gives you:

    [tex]F_{binding}(z)\approx F_{binding}(0)+F'_{binding}(0)z=e^a+(be^a+d)z[/tex]

    ....make sense?

    Now redo part (b).
     
  11. Sep 23, 2009 #10
    doesn't b) remain the same? because only k needs to be negative yes? which is just the part attatched to the z?
     
  12. Sep 23, 2009 #11

    gabbagabbahey

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    Don't forget about the last part of this question...what is the net force? What must it be for the particle to be in equilibrium at [itex]z=0[/itex]? That should tell you the value of [itex]a[/itex].

    Right.

    No, this doesn't follow from your previous statement any more than [itex]2<0[/itex], [itex]-2>0[/itex] and [itex]0<\ln(-2/-2)[/itex] would follow from the equation [itex]-2e^0+2<0[/itex]
     
  13. Sep 23, 2009 #12
    Right, I solved for a in part (e), so I bring that down, and say that to reach an equilibrium, gravity must balance the electric force. Then I can solve for a.

    Fo = exp(a) = mg. so
    a = ln(mg) = 64.96 N

    As for restrictions on b and d, is it not correct that the ratio of -d/b needs to be positive?
     
  14. Sep 23, 2009 #13

    gabbagabbahey

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    Good, but I wouldn't use the numeric value of 64.96N until part (e)...just leave it as a=ln(mg) for now...

    Yes, but that doesn't mean it has to be greater than 'a'....plugging in a=ln(mg), you get [itex]mgb+d<0[/itex]...that's all you can really say about 'b' and 'd'.
     
  15. Sep 23, 2009 #14
    great thanks! do all the other answers look okay?
     
  16. Sep 23, 2009 #15

    gabbagabbahey

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    You should definately rethink part (d)...what is the net force when there is no gravity....will that net force actually result in oscillations at all?
     
  17. Sep 23, 2009 #16
    ah yes, I see now that because a is dependent on g, and a is in the expression for oscillation frequency, that it will be affected if gravity should be neglected.

    Is this correct?
     
  18. Sep 23, 2009 #17

    gabbagabbahey

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    Well yes, but how will it be affected?
     
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