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.
From the step response plot, the peak overshoot, defined as
$$M_p = \frac{y_{peak}-y_{steady-state}}{y_{steady-state}}\approx\frac{1.25-0.92}{0.92}=0.3587$$
Also, the relationship between \$M_p\$ and damping ratio \$\zeta\$ (\$0\leq\zeta<1\$) is given by:
$$M_p=e^\frac{-\pi\zeta}{\sqrt{1-\zeta^2}}$$
Or, in terms of \$\zeta\$:
$$\zeta=\sqrt{\frac{\ln^2M_p}{\ln^2M_p+\pi^2}}$$
So, replacing that estimated \$M_p\$:
$$\zeta\approx0.31$$
Also, from the step response plot, the damped natural frequency is aprox. 0.5 Hz or \$\pi\$ rad/s. The relationship with the undamped natural frequency is:
$$\omega_n=\frac{\omega_d}{\sqrt{1-\zeta^2}}\approx3.3 rad/s$$
Finally, the gain \$G_{DC}=y_{steady-state}\approx0.92\$
A standard second order transfer function has the form:
$$H(s)=G_{DC}\frac{\omega_n^2}{s^2+2\zeta\omega_ns+\omega_n^2}$$
Putting the obtained values:
$$H(s)\approx\frac{10}{s^2+2s+11}$$
Compare the step response below with that supplied by you:
Best Answer
From a state-space representation, the \$H_2\$ norm can be computed as \$\sqrt{\text{Trace}\left(b q b^T\right)}\$ or \$\sqrt{\text{Trace}\left(c p c^T\right)}\$, where \$b\$ and \$c\$ are the input and output matrices, and \$q\$ and \$p\$ are the observability and controllability gramians.
I have done the calculations below using Mathematica and the result is \$\frac{1}{2}\sqrt{\frac{\omega _n}{\zeta }}\$.