# Quality Factor in damped oscillation

1. Apr 14, 2008

### Mattofix

Working through my lecture summaries, I have been given that $$Q (the quality factor) =\frac{2\pi}{(\Delta E/E)cycle}$$

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

The notes carry on to say

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

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

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

Knowing that $$E = E_{0}exp^{-bt/m}$$ 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. Apr 14, 2008

### Mattofix

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

3. Apr 14, 2008

### Mattofix

ok - i appreciate that $$\((\Delta E/E)cycle}$$ means energy loss per cycle divided by energy stored - where energy stored would be $$E = E_{0}exp^{-bt/m}$$

but i still cannot prove it

4. Apr 14, 2008

### rbj

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

$$\omega_{0}\sqrt{1 - \frac{1}{4Q^{2}}$$

if $\omega_0$ 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?