Posted by Anonymous on Wednesday, January 11, 2012 at 6:53pm.
The springs of a 1100 kg car compress 6.0 mm when its 68 kg driver gets into the driver's seat. If the car goes over a bump, what will be the frequency of vibrations?

Physics  Damon, Wednesday, January 11, 2012 at 7:11pm
6 mm = 6 * 10^3 m
68 * 9.81 = 667 Newtons
k of spring = 667/(6*10^3) = 111,167 N/m
w = sqrt(k/m) = sqrt (111,167/1100)
=10 radians/s
2 pi f = w = 10
f = 1.6 Hz