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Let's consider a spring which is subjected to forced vibrations: $$ F = F_0 \cos(\omega t) $$ Is the resonance frequancy $\omega_0$ of the spring dependent on the amplitude $F_0$?

I ask this because I am currently conducting tests with a plate which is forced to vibrate in the Z-direction orthogonally to its plan, thanks to a shaker, and it turns out that the resonance frequency of the plate is different for different values of the shaker amplitude (a higher amplitude gives a higher resonance frequency)

Thank you.

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Theoretically it is not. It is dependant on the damping and mass of the spring mostly. I suspect that as you drive it harder and harder however, and it is compressed more, that you leave the linear region of spring constants and damping parameters. The equation for an undamped spring's resonant frequency is sqrt(k/m) where k is the spring constant and m the spring mass –  aPhysicist Jul 29 at 14:36
Thank you for your answers but I know what you are telling me, I asked it naively to go further. So Jim thank you but I don't need your patronizing ton. –  user56288 Jul 29 at 14:48
This question appears to be off-topic because it shows insufficient research effort. –  BMS Jul 29 at 15:26
BMS, you can begin by saying you don't know how to deal with the problem. You're talking about research effort and indeed you don't really think beyond the question as good thinkers do. FYI it's a problem of non linear resonance here due to the stress in the plate. Avoiding writing useless answers would be really great, thank you. –  user56288 Jul 31 at 5:44

1 Answer 1

No, the resonance frequancy is just dependent on your forcing frequency $\omega$ and the attenuation $c$.

If you start with

$$m\ddot{x}+c\dot{x}+k x = F_0 \cos{(\omega t)}$$

you will get something like

$$|A| = \frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2 + 4r^2\omega^2}}$$

for your amplitude $A$, where $r=\frac{c}{2m}$. With $F_0 = \text{const.}$ just look at the denominator:

$$\frac{\partial}{\partial \omega} \sqrt{(\omega_0^2-\omega^2)^2 + 4r^2\omega^2}\stackrel{!}{=} 0$$

which results in

$$\omega = \omega_{\mathrm{res}} = \sqrt{\omega_0^2 - 2r^2}$$

Therefore your resonance frequency $\omega_{\mathrm{res}}$ only depends on $r$ but the amplitude $|A|$ linearly depends on $F_0$, as long as your attenuation $c\neq 0$ ($\Rightarrow |A| = \infty$).

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