$Q$-factor of oscillator

Question

Given the equation for a damped oscillator is $$\ddot{x}+\gamma\dot{x}+\omega_0^2x=0$$

Is the $Q$ factor of the system given by $\omega_0/\gamma$ or $\omega_0/2\gamma$?

I have seen both forms come up and so I do not know which one to use.

Answer 1

you start with:

$$m\ddot{x}+d\dot{x}+k\,x=0$$

or :

$$\ddot{x}+\frac{d}{m}\dot{x}+\frac{k}{m}\,x=0\tag 1$$

we want to write equation (1) with the parameters $\omega_0$ and $D$

$$\ddot{x}+2\,D\,\omega_0\,\dot{x}+\omega_0^2\,x=0\tag 2$$

with:

$$\omega_0=\sqrt{\frac{k}{m}}$$

$$D=\frac{1}{2}\frac{d}{\sqrt{k\,m}}$$

Q Factor is defined as $Q=\frac{1}{2 D}$