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.

  • $\begingroup$ 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, 2016 at 17:31

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.