I tried doing this problem but couldn't get the right answer. I don't get how the answer is 2.19cm. Is the distance referring to the vertical or horizontal distance

Four people, each with a mass of 73.5 kg, are in a car with a mass of 1190 kg. An earthquake strikes. The driver manages to pull of the road and stop, as the vertical oscillations of the ground surface make the car bounce up and down on its suspension springs. When the frequency of the shaking is 1.50 Hz, the car exhibits a maximum amplitude of vibration. The earthquake ends and the four people leave the car as fast as they can. By what distance does the car's undamaged suspension lift the car's body as the people get out?

2 answers

Wo = sqrt(k/m)
2 pi * 1.5 = Wo = 9.42 radians/s
so
9.42 = sqrt (k/(1190+4*73.5) )

88.8 = k/1484
k = 131,780 Newtons/meter

newtons lost = 4 * 73.5 * 9.81 = weight (not mass) of people

newtons lost = 2884 Newtons

elongation = F/k = 2884/131,780 = .022 meters = 2.2 centimeters, about an inch
why are we using the newton lost but not the newton remaining
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