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Homework Help: Coupled oscillation: time interval between maxima

  1. Mar 30, 2010 #1
    1. The problem statement, all variables and given/known data

    I calculated [tex]T_o[/tex] to be 1.27 seconds and [tex]"T_o"'[/tex] to be 1.23 seconds, each representing a normal mode of oscillation. These are correct according to the text.

    Here is the question: what is the time interval between successive maximum possible amplitudes of one pendulum after one pendulum is drawn aside and released?

    (answer: 40 seconds)

    3. The attempt at a solution

    I'm not even quite sure how to begin this problem, or what it is asking precisely. Is this suggesting that both modes of oscillation are superimposed? If this is the case, the superimposed period would be 1.25s.... where am I going wrong here?
     
    Last edited: Mar 30, 2010
  2. jcsd
  3. Mar 30, 2010 #2

    vela

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    What are the two modes of oscillation, and what initial conditions would you need to excite each one?
     
  4. Mar 30, 2010 #3
    The first mode starts with equal displacement for both pendulums, both oscillate with natural frequency [tex]\sqrt(\frac{g}{l})[/tex]. The second mode starts with equal displacement but opposite signs for the pendulums, both oscillating with frequency [tex]\sqrt(w_0^2 + w_c^2)[/tex].
     
  5. Mar 30, 2010 #4

    vela

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    Right. The initial condition given doesn't excite just one mode, so the motion will be the superposition of the two modes, which have different frequencies. What happens when you combine two oscillations with different frequencies?
     
  6. Mar 30, 2010 #5
    You get a beat with a frequency of each pendulum equal to the average of the two normal frequencies = ~ 0.80 hz = ~ 5.02 rad/s.
     
  7. Mar 30, 2010 #6

    vela

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    Yes, you get a beat. But there's two frequencies associated with it. You mentioned one, but what's the frequency of the envelope, i.e. the beat frequency?
     
  8. Mar 30, 2010 #7
    It's half the difference of the two, so roughly 0.0256 hz or 0.1609 rad/s. I can see where the answer comes from now, but I don't really understand the roles of these two respective frequencies.
     
  9. Mar 30, 2010 #8

    vela

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  10. Mar 30, 2010 #9
    Thanks!
     
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