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Energy quantization of oscillator

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A simple pendulum has a length equal to 0.6 m and has a bob that has a mass equal to 0.5 kg. The energy of this oscillator is quantized, and the allowed values of energy are given by En = (n + 1/2)hf0, where n is an integer and f0 is the frequency of the pendulum. Find n if the angular amplitude is 1.0°.

    2. Relevant equations

    En = (n + 1/2)hf0
    f0=w(omega)/(2pi)

    3. The attempt at a solution

    I'm stuck at the beginning of finding out the total energy of the pendulum. I think I might have to use the equation U(x)=(1/2)m(w0)^2x^2 to find the energy but I get stuck with unknown values. My biggest question is what does the angular amplitude of 1 degree get used for. I don't think it can just be plugged into the equation as is for an amplitude. Any help on how to approach the problem would be much appreciated.
     
    Last edited: Sep 27, 2009
  2. jcsd
  3. Sep 27, 2009 #2

    kuruman

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    Express the potential energy in terms of θ then expand it for small values of θ. The total energy of the oscillator is the potential energy at maximum displacement. That's where the 1 degree comes in.
     
  4. Sep 27, 2009 #3
    Okay, so I'm using the equation U=(1/2)lw^2(theta)^2. I found this equation and I think the l in it stands for the length of the pendulum. So after doing the calculations I get the potential energy to be .0855 but my units are confusing me. Should theta be in radians or degrees and is this in fact what l stands for?
     
  5. Sep 27, 2009 #4

    kuruman

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    Yes, on both accounts. The angle must be expressed in radians and l is the length of the pendulum in meters. With these units, the energy should come out in Joules.
     
  6. Sep 27, 2009 #5
    When I do the calculations, I get U=.00149 m/s^2. I do not know why I am missing units.
     
  7. Sep 27, 2009 #6

    kuruman

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    Show me the numbers you put in and your answer. The correct expression is

    U = mglθ2/2.
     
  8. Sep 27, 2009 #7
    I'm confused where you got mgl from, but using your formula I was able to get the correct answer
     
    Last edited: Sep 27, 2009
  9. Sep 27, 2009 #8

    kuruman

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    Derive (or look up) the potential energy for a pendulum referenced to the lowest point of the motion (equilibrium position) in terms of the angle θ. Then use the small angle approximation for the cosine

    [tex]cos\theta\approx1-\frac{\theta^{2}}{2}.[/tex]
     
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