4
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • $\begingroup$ Probably you've calculated something wrong. The driving frequency which maximizes amplitude is the natural frequency. $\endgroup$
    – Physicist137
    Commented Aug 21, 2017 at 22:09
  • $\begingroup$ You have a problem with notation, your $\omega_0$ is actually $\omega_D$, in the literature $\omega_0=\sqrt{\frac{k}{m}}$. That said, the maximum amplitude is indeed attained at a frequency $\omega_R$, somewhat less than $\omega_0$, this is the result of calculations, I don't know if there is an inherent physical reason for that.In most cases of any practical interest, however, the difference between $\omega_R$ and $\omega_0$ is (as you see) negligibly small. $\endgroup$
    – Samà
    Commented Aug 21, 2017 at 22:29
  • $\begingroup$ @Samà I don't think so. The natural frequency of that oscillator is OP's $\omega_0$. It is written that... Sure, he used notation different from usual, but that doesn't matter. However, what he claims to be the resonance frequency is clearly in error. It should be $\omega_R = \omega_0$, and not what he stated. $\endgroup$
    – Physicist137
    Commented Aug 21, 2017 at 23:15
  • $\begingroup$ Related: physics.stackexchange.com/q/153197/2451 , physics.stackexchange.com/q/228279/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Aug 22, 2017 at 4:40
  • $\begingroup$ See also answers and links here: physics.stackexchange.com/q/749963/226902 $\endgroup$
    – Quillo
    Commented Feb 22, 2023 at 10:52

2 Answers 2

Reset to default
5
$\begingroup$

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"...

Improve this answer
$\endgroup$
5
  • $\begingroup$ Still, the frequency in which maximizes kinetic energy is the natural frequency of the oscillator without damping. Which is quite different from the frequency the OP inserted there. $\endgroup$
    – Physicist137
    Commented Aug 21, 2017 at 22:20
  • $\begingroup$ The question was not "maximize kinetic energy" but "maximize amplitude". Because kinetic energy depends on both frequency and amplitude, these two won't be the same. $\endgroup$
    – Floris
    Commented Aug 21, 2017 at 22:22
  • $\begingroup$ Then something is wrong. The frequency in which maximizes amplitude is the natural frequency.... Or not? $\endgroup$
    – Physicist137
    Commented Aug 21, 2017 at 22:25
  • $\begingroup$ @Physicist137 it's not. See my comment on OP. $\endgroup$
    – Samà
    Commented Aug 21, 2017 at 22:36
  • $\begingroup$ Ah. I see. Thanks @Samà. However, the resonance frequency for amplitude maximization is the natural frequency. I think this answer would be correct if it stated it is explaining the difference of the resonance frequency from the natural frequency without damping. But then, that's not what the OP asked. $\endgroup$
    – Physicist137
    Commented Aug 21, 2017 at 23:19
0
$\begingroup$

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

Improve this answer
$\endgroup$
1

Your Answer

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

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