# Why maximum energy transfer at natural frequency even if max amplitude occurs below $f_0$

This is a paragraph from my book:

"For a damped system, the resonant frequency at which the amplitude is a maximum is lower than the natural frequency.However, maximum transfer of energy, or energy resonance always occurs when applied frequency is equal to natural frequency"

This doesn't make intuitive sense to me. I understand that if there is damping, maximum amplitude occurs below $f_0$... so, shouldn't maximum energy be transferred at this driving frequency as amplitude maximum instead of at $f_o$?

Interesting question. Note that the power transfer is $$P=\frac{1}{\tau}\int_0^\tau F(t)v(t)\,\mathrm dt$$ where $F(t)$ is the applied force, $v(t)$ is the velocity of the oscillation, and $\tau$ is the length of the period of the oscillation (this is just a simple extension of the usual rule for work, $w=Fd$, to time-varying systems).
The usual forced oscillator $$m x''(t)=-kx(t)-\gamma x'(t)+A\sin(\omega t)$$ has steady-state solution $$x(t)=\frac{A }{\sqrt{\left(k-m\omega^2\right)^2+\gamma ^2 \omega ^2}}\sin \left(\omega t+\tan ^{-1}\left(k-m \omega ^2,-\gamma \omega \right) \right).$$
Since the user-applied force is $F(t)=A\sin(\omega t)$, the velocity is $v(t)=x'(t)$ and $\tau=2\pi/\omega$, we obtain \begin{align}P&=\frac{\omega}{2\pi}\int_0^{2\pi/\omega} A\sin(\omega t)\frac{A \omega \cos \left(\omega t+\tan ^{-1}\left(k-m \omega ^2,-\gamma \omega \right)\right)}{\sqrt{\left(k-m\omega^2\right)^2+\gamma ^2 \omega ^2}}\,\mathrm dt\\&=\frac{\frac{1}{2}A^2 \gamma \omega ^2}{\left(k-m \omega ^2\right)^2+\gamma ^2 \omega ^2}\,.\end{align} Solving $$\frac{\mathrm dP}{\mathrm d\omega}=0$$ for $\omega$ yields $$\omega=\sqrt{\frac{k}{m}}$$ which is exactly the natural frequency.
(Minor note: I am using the two-argument arctangent function $\tan^{-1}(a,b)$ as it is defined in Mathematica, ArcTan[a,b]).