1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ This question appears to be off-topic because it shows insufficient research effort. $\endgroup$
    – BMS
    Jul 29, 2014 at 15:26

2 Answers 2

1
$\begingroup$

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 $c\neq 0$ ($\Rightarrow |A| = \infty$).

$\endgroup$
0
$\begingroup$

For a nonlinear system, the resonance frequency may depend on the amplitude of excitation.

Have a look in this particular nonlinear system with cubic stiffness (Duffing)

https://en.wikipedia.org/wiki/Duffing_equation

$\endgroup$
1
  • 1
    $\begingroup$ Hi Paolo, thanks for your post. However, if you read the question, the answers, and the comments carefully, you realize that the person asking the question was already aware of the fact that it is a nonlinear problem. $\endgroup$
    – flaudemus
    Feb 23, 2019 at 22:25

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.