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?

Answer 1

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

Answer 2

Can someone please explain to me how 2pi f = w = 10
f = 1.6 HZ
because this is the only part I can't figure out for myself.I highly appreciate it,thanks

Answer 3

since we know that w = 10 and that 2pi f = w we can divide w by 2pi so 10/(2pi) = 1.6