4

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.

P. C. Spaniel
  • 4,649
  • Probably you've calculated something wrong. The driving frequency which maximizes amplitude is the natural frequency. – Physicist137 Aug 21 '17 at 22:09
  • 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. – Samà Aug 21 '17 at 22:29
  • @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. – Physicist137 Aug 21 '17 at 23:15
  • Related: https://physics.stackexchange.com/q/153197/2451 , https://physics.stackexchange.com/q/228279/2451 and links therein. – Qmechanic Aug 22 '17 at 04:40
  • See also answers and links here: https://physics.stackexchange.com/q/749963/226902 – Quillo Feb 22 '23 at 10:52

2 Answers2

5

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"...

Floris
  • 118,905
  • 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. – Physicist137 Aug 21 '17 at 22:20
  • 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. – Floris Aug 21 '17 at 22:22
  • Then something is wrong. The frequency in which maximizes amplitude is the natural frequency.... Or not? – Physicist137 Aug 21 '17 at 22:25
  • @Physicist137 it's not. See my comment on OP. – Samà Aug 21 '17 at 22:36
  • 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. – Physicist137 Aug 21 '17 at 23:19
0

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

MSJ
  • 1