# Driven oscillation with negligeable damping problem

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...

• 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? – Floris Aug 23 '16 at 11:47
• @Floris Oups sorry I said $1$ $mm$ but it is $0.1$ $mm$. I'm correcting that in my question. – Dory Aug 23 '16 at 11:49
• 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... – Floris Aug 23 '16 at 11:52