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Fractional energy loss per cycle in a heavy damped oscillator

  1. Apr 7, 2013 #1
    http://www1.gantep.edu.tr/~physics/media/kunena/attachments/382/chapter2.pdf [Broken]

    On page 9 and 10 of the above PDF the method for deriving the fractional energy loss per cycle in a lightly damped oscillator is described.

    I understand and follow this derivation.

    What would the derivation for a heavily damped oscillator be ? What approximations would I have to make ?
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 7, 2013 #2

    Philip Wood

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    The site seems to use 'lightly damped' to mean damped, but less than critically. So any damped response that is oscillatory is 'lightly damped'. 'Heavily damped' must. then, mean critically or more than critically damped, so there are no oscillations and it doesn't make sense to talk about fractional energy loss per cycle!
  4. Apr 7, 2013 #3


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    I would be happier if you had asked why the derivation in the PDF was wrong (or being charitable, it's an approximate result with no explanation of how it was derived).

    Judging by that sample chapter, this is a terrible book - don't buy it!

    Here's a website that does it right:
  5. Apr 7, 2013 #4


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    A text, prepared with Microsoft Word has to be treated with great suspicion. That's at least my experience. :devil:

    Fitzpatrick's manuscripts are always a pleasure to read and, no surprise, they are written in LaTeX :smile:
  6. Apr 7, 2013 #5
    I understand the derivation (sort of) due to a similar derivation in my lecture notes.

    To everyone else : The problem is I have a question where the Q factor is 3 and I am asked to calculate the fractional energy loss per cycle.

    I simply have no idea how to determine this....
  7. Apr 7, 2013 #6

    Philip Wood

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    Taking the standard form of the damped shm equation as [itex] \ddot{x} + 2\alpha \dot{x} + \omega_0^2 x = 0[/itex], then Q is defined as [itex]\frac{\omega_0}{2\alpha}[/itex].
    So, knowing Q, you can express [itex] \alpha[/itex] in terms of [itex]\omega_0[/itex].

    But you know that the solution of the damped shm equation is a sinusoid with an exponentially decaying amplitude, and you have equations for the amplitude, in terms of [itex] \alpha[/itex], and for the frequency of the sinusoid, in terms of [itex] \omega_0 [/itex] and [itex] \alpha[/itex]. From the second of these you can find the periodic time, T. Substitute this for t in the amplitude equation and the job is virtually done.

    I think that, for Q = 3, the amplitude after a cycle is 0.35 of its original value. I would say, "Check your answer against mine", but I'm prone to slips.
    Last edited: Apr 7, 2013
  8. Apr 7, 2013 #7
    My workings and answer are below. Could ou tell me if I've worked this out correctly ?

    Last edited: Apr 7, 2013
  9. Apr 8, 2013 #8

    Philip Wood

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    The structure of your solution is fine, but I've spotted one arithmetical mistake, and one place where you are making things more complicated than they need be...

    First the arithmetic. The expression of [itex]\omega_D^2[/itex] in terms of [itex]\gamma[/itex] should read
    [itex]\omega_D^2 = 9 \gamma^2 - \frac{\gamma^2 }{4} = \frac{35 \gamma^2 }{4}[/itex]

    Now the simplification I recommend. Consider the fall in amplitude from t= 0 to t = T, rather than from t= T to t = 2T. [Incidentally, it's easier to think about what's going on if you use the cosine rather than the sine form of the solution. Then the displacement at t = 0 is simply the initial amplitude, A.]
  10. Apr 8, 2013 #9
    After doing what you recommended I've found that ΔE/E = 0.65. Can you confirm to me whether that is correct ?

    Thank you for all your help so far :)
  11. Apr 8, 2013 #10

    Philip Wood

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    This agrees with my answer given in post 6 !

    Good luck with your studies. Physics is wonderful!
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