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. (obviouslly due to resonance)
(ii) Calculate the separation of the humps if it occurs at a speed of 15 ms-1.

the car has suspensions those are springs. now when the man enters the car goes down by 2cm. f=kx.

 

800=k 2/100  => k = 40000. the time period of ocillation of the spring = 2piroot (m/k).

 for max amplitude the time taken to reach the other bump = n time period of

 

ocillation of the spring. you can get the min distance by putting n=1.

 

i dont know this method is purely based on my instincts. i'am not sure if it is correct.

thanks...I thought exactly the same, just wanted to verify as the solution wasnt provided in the source.. And why do u think they have provided the mass of the car? The 'm' in time period is only the mass of the man..