Natural and Resonance frequencies of a damped oscillator The damped oscillator equation is
\begin{equation}
m\ddot{x}+b\dot{x}+kx=0
\end{equation}
And its solution has natural frequency $\omega_0$
\begin{equation}
\omega_0=\sqrt{\frac{k}{m}-(\frac{b}{2m})^2}
\end{equation}
However, when one adds a driving force to the equation
\begin{equation}
m\ddot{x}+b\dot{x}+kx=D\cos(\Omega t + \phi)
\end{equation}
the resonance frequency $\Omega=\omega_R$ that maximizes amplitude is
\begin{equation}
\omega_R=\sqrt{\frac{k}{m}-2(\frac{b}{2m})^2}
\end{equation}
I'm wondering why the resonance frequency isn't the natural frequency. I've read this formulas in the wikipedia page of the harmonic oscillator.
 A: The difference is subtle - and only really matters with "somewhat damped" systems (where $\zeta$ is "not very small" compared to 1).
The key here is that the maximum AMPLITUDE is not reached at the same frequency as the maximum POWER DISSIPATED. For the former, you would like the frequency to be slightly lower (because you dissipate a certain amount of power per cycle). For the latter, you need the driving force to be exactly in quadrature with the velocity. But that gives higher power dissipation, and smaller amplitude (recall also that at higher frequencies, the velocity increases - and that you therefore have more dissipation at the same amplitude).
That's the intuitive explanation. Alternatively, one could just say "that's just how the math works"... 
A: See https://en.wikipedia.org/wiki/Resonance
It shows the resonance in damped system deviates from the point at which Forced w/Natural w = 1
Indeed the resonance occurs when Forced w/Damped w = 1
where e.g. for a base exited system Damped w = [1-square(damping factor)]^0.5 * Natural w 
