# 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?