I cannot find a simple explanation of the damping ratio formula.
The natural frequency for a spring mass system seems pretty simple:
position, velocity and acceleration are given by: $$x(t)=Acos(\omega t)$$ $$v=x'(t)=-A\omega sin(\omega t)$$ $$a=x''(t)=-A\omega^2 cos(\omega t)$$
Replacing a and x in $ma = -kx$ with the formulas above, we have: $$-mA\omega^2 cos(\omega t)=-kAcos(\omega t)$$
And therefore $m\omega^2=k$, from which we get $m\omega^2=k$ and hence:
$$\omega=\sqrt{\frac{k}{m}}$$
How do we get to the following formula for the damping ratio?
$$\frac{b}{m}=2\zeta\omega$$
We have $2 \zeta \omega= \frac{b}{m}$ by definition, $\zeta$ is just defined to be this quantity.