Quality Factor in damped oscillation

Mattofix

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.

Mattofix

any help would be much appreciated so i can get rid of this irritating missing link.

Mattofix

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}
[/tex]

but i still cannot prove it

rbj

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?