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 $f_o$?
 A: 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.
Thus, somewhat paradoxically, maximum power transfer occurs when the forcing is at the natural frequency, even though the vibrational amplitude is not maximum.
(Minor note: I am using the two-argument arctangent function $\tan^{-1}(a,b)$ as it is defined in Mathematica, ArcTan[a,b]).
A: Your question has been answered by following standard mathematical procedures. Nonetheless, your concern about intuitive sense is quite fair. On one hand, the idea that energy transfer should be maximum at a frequency below the natural frequency $\omega_0$ is not wrong at all. The work done by the external force over one period of the stationary oscillation is maximum at $$\omega''=\omega_0 \sqrt{\frac{\lambda ^2+2 \sqrt{\lambda ^4-\lambda
   ^2+1}-2}{3\lambda ^2}},$$ where $\lambda=\omega_0 \tau$, and $\tau$ is the characteristic decay time in the factor $\exp(-t/\tau)$ of the underdamped transient. The frequency $\omega''$ is below $\omega_0$ and above the frequency of the periodic part of the underdamped oscillation, namely $\omega'=\sqrt{\omega_0^2-1/\tau^2}$. On the other hand, the maximum averaged power, which is the work done over one period divided by the period $T=2\pi/\omega$, will have its maximum to the right of $\omega''$. The standard procedure shows it is at $\omega_0$. The point here is whether our intuition sense deals with the work done over one period or with the average power.
