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Quality Factor in damped oscillation

  1. Apr 14, 2008 #1
    Working through my lecture summaries, I have been given that [tex] Q (the quality factor) =\frac{2\pi}{(\Delta E/E)cycle}[/tex]

    and accepted this as a statement, taking [tex]\((\Delta E/E)cycle}[/tex] to mean the 'energy loss per cycle'.

    The notes carry on to say

    'The frequency [tex]\widetilde{\omega}[/tex] of under(damped) oscillator as function of the frequency [tex]\omega_{0}[/tex] and the Q factor:

    [tex]\widetilde{\omega} = \omega_{0}\sqrt{1 - (\frac{b}{2m\omega_{0})^{2}}} = \omega_{0}\sqrt{1 - \frac{1}{4Q^{2}}[/tex]

    My problem being that I cannot prove that [tex]\frac{b}{2m\omega_{0}} = \frac{1}{4Q^{2}} [/tex]

    Knowing that [tex]E = E_{0}exp^{-bt/m}[/tex] i tried finding the energy loss per cycle by finding the difference between the energy at time t and the energy at time t + T (where T is the time period) but just ened up with an unhelpfull equation.
    Last edited: Apr 14, 2008
  2. jcsd
  3. Apr 14, 2008 #2
    any help would be much appreciated so i can get rid of this irritating missing link.
  4. Apr 14, 2008 #3
    ok - i appreciate that [tex]
    \((\Delta E/E)cycle}
    [/tex] means energy loss per cycle divided by energy stored - where energy stored would be [tex]
    E = E_{0}exp^{-bt/m}

    but i still cannot prove it
  5. Apr 14, 2008 #4


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    i can tell you why, if [itex]Q>\sqrt{1/2}[/itex] that the peak resonant frequency is

    [tex]\omega_{0}\sqrt{1 - \frac{1}{4Q^{2}}[/tex]

    if [itex]\omega_0[/itex] the "natural" resonant frequency (i dunno what to call it) of the system. but i do not know what b and m are and can't tell from the context. is this a second order mechanical system or an electrical system?
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