car and hump?

Question

When the driver of mass 80 kg, steps into the car of mass 920 kg, the vertical height of the car above the road decreases by 2.0 cm. If the car is driven over a series of equally spaced humps, the amplitude of the vibration becomes much larger at one particular speed.
(i) Explain why this occurs.
(ii) Calculate the separation of the humps if it occurs at a speed of 15 ms-1.

Answer 1

You can model the suspension of the car as a spring. The spring constant can be computed as
80*9.81=k*2/100

where k is in N*m

Springs have a resonant frequency which can be related to the hump spacing.

Knowing k, the resonant period, T, is 1/f the frequency, the angular velocity is w, which is
f=w/(2*pi)
and w=sqrt(k/m)

relating the equations to the problem at hand:
T=2*pi*sqrt(m/k)
mass is 1000 kg
k=50*80*9.81
The period will be in seconds, so T*15 will give you the spacing

I computed 15 meters using pi=3.14

j