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Simple Harmonic Oscillator and Damping

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data

    After four cycles the amplitude of a damped harmonic oscillator has dropped to 1/e of it's initial value. Find the ratio of the frequency of this oscillator to that of it's natural frequency (undamped value)

    2. Relevant equations

    x'' +(√k/m) = 0

    x'' = d/dt(dx/dt)

    x'' + 2βx' + Wo^2= 0

    Here we are assuming a damping force linear in v,

    f = -bv

    2β is defined as b/m where b is a constant.

    Wo^2 = k/m

    3. The attempt at a solution

    Since the problem statement makes no mention of whether this is underdamped (β then than Wo), overdamped (β greater than Wo), or critically damped (β= Wo), I can only conclude that it does not matter what the case is.

    Two of the solutions can be expressed as sines and cosines and would be periodic with 2∏, but one of the solutions is simply two exponential terms, which does not make sense to me.

    That's kind of where I'm stuck, any thoughts would be helpful. Thanks.
     
  2. jcsd
  3. Nov 16, 2013 #2
    Well, it tells you that it oscillates 4 times before amplitude decreases by a factor of e.
    Could this be over-damped oscillator?

    I don't think you need to find the solution from scratch here. Don't you have it in your notes or textbook? It contains a product of a cosine (or sine) and an exponentially decreasing term.
     
  4. Nov 16, 2013 #3
    Yeah mate I have all the solutions, that isn't my problem. I am thinking it must be the underdamped oscillator, because the overdamping case doesn't oscillate harmonically, it is real, so the solution is just two exponentials. With critical damping, the solutions are imaginary so I am pretty sure it is underdamping..

    Excuse me with underdamping the solution is imaginary and can be wrote as

    x(t) = Ae^-ßt cos(w1t -ø)

    Where W1 = (Wo^2+ß^2)^1/2

    Since it says 4 cycles, I am thinking 8 pie, but it's just tough to put together the mathematics here.
     
  5. Nov 17, 2013 #4
    Isn't there a minus in that last formula?
    The damping decreases the frequency.

    It's good to have all the formulas. Now you need to understand them as well.
    Looking at the formula for x(t), after how long the amplitude decreases to A/e?
    You can take the initial phase (∅) equal to zero. You don't have to but it may help to understand what is going on.
     
  6. Nov 17, 2013 #5
    Yes that is a minus, sorry. I was thinking about putting the phase equal to zero. I have the solution, i got the problem wrong, I just thought it would be helpful to talk about it and try to solve it on my own before looking at it. Right now I'm caught up in the Lagrangian but I will return to that problem at some point in the night. I will try to set the phase equal to zero and reanalyze. Thanks.
     
  7. Nov 20, 2013 #6

    ehild

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    Your equations for the oscillator are not correct. They should be

    x'' +(√k/m)x = 0

    x'' + 2βx' + Wo^2x= 0

    As the oscillator makes oscillations (after four has its amplitude dropped to 1/e of its initial value), it can not be critically damped or overdamped.

    You have the equation x(t) = Ae^-ßt cos(w1t -ø). Ae^-ßt is the amplitude of the cosine factor.
    As you know that the amplitude Ae^-ßt decreases to[STRIKE] A/4[/STRIKE] A/e in 4 cycles, you can determine β, by replacing t=4T. You do not need to worry about the cosine factor. How is the period T related to the angular frequency w1? If you know β you can get w1 in terms wo.

    ehild
     
    Last edited: Nov 20, 2013
  8. Nov 20, 2013 #7
    It decreases to A/e and not A/4.

    But anyway, he is busy with other problems now. :smile:
     
  9. Nov 20, 2013 #8

    ehild

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    Yes, stupid of me :redface:

    ehild
     
  10. Nov 23, 2013 #9
    Yes sorry I was onto other problems, it was just my teacher solved it very strangely by taking ratios, so I just let it go. Thank you for your help though.
     
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