The problem : A car's suspension system rolling on the pavement can be viewed as a driven oscillation system. The car's natural frequency is $100$ $ rad/s $. It has also a mass of $800$ $kg$. We know that when the car is going such that the driving frequency is equal to $150$ $rad/s$, the amplitude of the vibrations is equal to $0.1$ $mm$. Damping can be neglected.

The question : What is the amplitude of the vibrations at resonance?

The answer is $0.8$ $mm$ but I have no idea how to get there. If I use the formula for the amplitude of a driven oscillation $A = \frac{F_0/m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (\frac{\omega \beta}{m}})^2}$, the natural frequency $\omega_0$ and driving frequency $\omega$ are equal at resonance and $\beta$ approches 0 and so I divide by zero...

  • $\begingroup$ I hope you get your answer - and perhaps that you will update this posting when you do. The question makes no sense to me: "ignore damping" means that at resonance a system will end up with "infinite" amplitude (and infinite Q); and off-resonance, the response amplitude will be equal to the driving amplitude. If you consider the system a "mass on a spring", you can compute the spring constant from the resonant frequency, and the driving amplitude from the off resonance response. But without information about damping, you don't know Q. And why would the amplitude be less at resonance? $\endgroup$
    – Floris
    Commented Aug 23, 2016 at 11:47
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    $\begingroup$ @Floris Oups sorry I said $1$ $mm$ but it is $0.1$ $mm$. I'm correcting that in my question. $\endgroup$
    – Dory
    Commented Aug 23, 2016 at 11:49
  • $\begingroup$ That makes a little bit more sense. I still don't see how you can answer this without some information about damping. As you noted - there is a singularity... $\endgroup$
    – Floris
    Commented Aug 23, 2016 at 11:52

1 Answer 1


Your corrected data probably makes more sense. You can calculate the force for 0.1 mm and 150 rad/s: 1000 N. The answer (0.8mm) corresponds to a force that is 8 times larger, i.e. 8000N, which is roughly the weight of the car. I am not sure the force can grow above that, even at resonance and without damping, as the car will probably lose contact with the road.

  • $\begingroup$ How to you find that the force at resonance is 8 times larger? Is it because it cannot go higher? $\endgroup$
    – Dory
    Commented Aug 23, 2016 at 14:36
  • $\begingroup$ @Dory: because the answer (0.8 mm) is 8 times greater than the amplitude (0.1 mm) for 150 rad/s, for which the force can be calculated to be 1000 N (I use the Hook law for the suspension system - the force is proportional to displacement). I hypothesize that the force cannot be greater than the weight of the car as it will not grow if the car loses contact with the road (or maybe its movement downwards is also limited by the ground). $\endgroup$
    – akhmeteli
    Commented Aug 23, 2016 at 14:46

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