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If we have a driven damped harmonic oscillator: $$ \frac{d^2x}{dt^2}+\gamma\frac{dx}{dt}+\omega^2_0x=\frac{F}{m} e^{i\omega t} $$

amplitude resonant frequency occurs at: $ \omega_R^2 = \omega^2_0x-\frac{\gamma^2}{2} $ As energy of a spring is proportional to displacement squared, the maximum energy of the system is here.

But, velocity resonance occurs at: $\omega=\omega_0$ as kinetic energy is proportional to velocity squared, the maximum energy of the system is here.

There is clearly a paradox here, I cannot understand how it can resonate at 2 different frequencies.

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  • $\begingroup$ For a linear, second order system with complex poles there is by definition only one resonant frequency. Amplitude of displacement is not frequency but you do have a maximum amplitude of oscillation (displacement or velocity) at the resonant frequency. $\endgroup$ – docscience Dec 30 '16 at 17:31
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For simple harmonic motion the maximum speed $v_{\rm max}$ is related to the frequency $f$ and the amplitude $A$ as follows $v_{\rm max}=2\pi f A$.

So your two resonance graphs are showing how two different (but related) quantities, amplitude and maximum speed vary with the frequency of the driver.
For one you are looking for maximum $A$ and for the other maximum $(2\pi ) fA$.


Similarly when you compare the torque $\tau$ and the power $\tau 2 \pi f$ ($f$ is the frequency or engine speed) delivered by a motor car engine the two related quantities have different variations with regard to engine speed.

The peak torque which dictates the acceleration of a car occurs at a different frequency (engine speed, rps) from the peak power which determines the maximum speed of a car.

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