Rather than the frequency domain, let's look at this in the time domain and particularly, the characteristic equation associated with a linear homogeneous 2nd order differential equation for some system:
\$r^2 + 2 \zeta \omega_n r + \omega^2_n = 0\$.
If the roots of the characteristic equation are real (which is the case if \$\zeta \ge 1\$), the general solution is the sum of real exponentials:
\$Ae^{\sigma_1 t} + Be^{\sigma_2t} \$
where
\$\sigma_1 = -\zeta \omega_n + \sqrt{(\zeta ^2 - 1)\omega^2_n} \$
\$\sigma_2 = -\zeta \omega_n - \sqrt{(\zeta ^2 - 1)\omega^2_n} \$
Since these are real exponentials, there is no oscillation in these solutions.
If the roots are complex conjugates (which is the case if \$\zeta < 1\$), the general solution is the sum of complex exponentials:
\$e^{\sigma t}(Ae^{j\omega t} + Be^{-j\omega t})\$
where
\$\sigma = -\zeta \omega_n\$
\$\omega = \sqrt{(1 - \zeta ^2)\omega^2_n}\$
This solution is a sinusoid with angular frequency \$\omega\$ multiplied by a real exponential. We say the system has a "natural frequency" of \$\omega\$ for a reason that I think is obvious.
Finally, setting \$\zeta = 0\$ (an undamped system) , this solution becomes:
\$Ae^{j\omega_n t} + Be^{-j\omega_n t}\$
which is just a sinusoid of angular frequency \$\omega_n\$.
In summary, a system may or may not have an associated natural frequency. Only systems with \$\zeta < 1\$ have a natural frequency \$\omega\$ and only in the case that \$\zeta = 0\$ will the natural frequency \$\omega = \omega_n\$, the undamped natural frequency.
What you have initially described is a 2nd order low pass filter then you've made a bit of a complication of things because the bandwidth is \$\omega_n\$ i.e. \$\omega_n\$ is the 3dB point when \$\zeta\$ is not causing the filter to peak i.e. has a value of \$\frac{1}{\sqrt2}\$: -
OK you may be trying to derive a general expression but that doesn't help much visualize the problem so for now I'm assuming \$\zeta\$ = \$\frac{1}{\sqrt2}\$.
Anyway, moving on and ignoring your expression for \$\omega_0\$, the bandwidth of the low-pass filter is from dc to \$\omega_n\$. Then you've modified the 2nd order low-pass filter expression with a 1st order high pass filter (\$s +z\$).
When s is very low, the low pass filter is unaffected other than having a "gain factor" of z. If z is unity then the low pass filter response is unaffected until jw approaches z in magnitude - this then marks a lower 3dB point and the net response climbs with increasing frequency until the original \$\omega_n\$ is reached then, because the low pass filter is a second order, the response starts to fall again.
What you have proposed is a band pass filter with finite gain at dc.
I'm not sure about the bandwidth
If the high pass filter (\$s +z\$) comes into play at significantly lower frequencies than \$\omega_n\$ then the net bandwidth reduces from \$\omega_n\$ to \$\omega_n - z\$.
That's how I see it anyway.
Best Answer
The first thing I recommend is to rewrite the equation in a low-entropy format in which the numerator and the denominator are unitless. In your case, as underlined by a concerned citizen, the presence of the zero at the origin does not change the exercise.
A second-order polynomial can be put under the following normalized form: \$D(s)=1+b_1s+b_2s^2\$. Then, you can equate this expression with the normalized form where the quality factor \$Q\$ appears and find the correspondence between the terms: \$D(s)=1+\frac{s}{\omega_oQ}+(\frac{s}{\omega_o})^2\$. From there, if you do the maths ok, the you find \$\omega_o=\frac{1}{\sqrt{b_2}}\$ and \$Q=\frac{\sqrt{b_2}}{b_1}\$.
So, in your case, you have \$H(s)=\frac{\omega_n^2 s}{s^2+2\zeta\omega_n s+\omega_n^2}\$. Factor \$\omega_n^2\$ in the numerator and the denominator and you will have the above form for the identification of your coefficients.