Physics
posted by Robert on .
A 1125 kg car carrying four 80 kg people travels over a rough "washboard" dirt road with corrugations 4.0 m apart which causes the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is 17 km/h. The car now stops, and the four people get out. By how much does the car body rise on its suspension owing to this decrease in weight?

Can't figure it out. Does it have something to do with finding w?
w=2pi/T = square root of k/m but I don't have K or T? 
Use the information provided to get the natural (and resonant) frequency of the suspension system with a total mass of M = 1205 kg. From that you can derive the spring constant, k.
f = [1/(2 pi)] sqrt(k/M)
The natural frequency will be the rate at which the car goes over washboard bumps at the resonant (maximum amplitude) condition.
When the people get out, the car body will rise by an amount
(80 kg)(9.8 m/s^2)/k 
Thank you!