# Resonance of van driving over speed bumps

An old van with an undamped suspension system drives over three speed bumps 10m apart at a speed of $2.5\text{m/s}$. The front of the van begins to resonate. State the natural frequency of the suspension and explain why driving over the bumps at a different speed would reduce the ampiltude of the oscillations.

I thought to calculate the time between the bumps as $10/2.5 = T = 4.0$ seconds. The solution states that this is the period of the oscillation.

I progressed to calculating the natural frequency of the suspension as $\frac{1}{4.0} = 0.25$ Hz. However, I don't understand why the time between the bumps is the period of the oscillation. I understand the second step, which involves calculating the natural frequency from $f = \frac{1}{T}$ but I don't understand why the time period T is 4.0 seconds from the calculation $\frac{10}{2.5}$.

Does that mean that one oscillation has a time period $T$ equal to 4 seconds? If so, why is this true?

The trick to this question is that you're really having to solve it backwards, and that can get confusing when you go back and check your results.

The natural frequency of the suspension is what it is. That frequency is based on the mass of the vehicle, the spring coefficient of the shocks, etc. You aren't given any information to calculate it directly. Instead, you are expected to calculate it indirectly.

The problem states that the van resonated when going over speedbumps 10m apart at 2.5m/s. Thus we know that the speedbump gave the van a "kick" at the same point in the van's natural oscillating cycle. That's how we know the rate at which the van goes over the speedbumps equals its resonant frequency.

What is probably making it hard is that the data you're given is not causal. If we kept the van the same, but spaced the speedbumps 8m apart, the van's natural frequency would not change. The speedbump spacing is not causing the natural frequency of the shocks. Instead, the van simply wouldn't resonate over those speed bumps.