The springs of a 1800 kg car compress 4.4 mm when its 72 kg driver gets into the driver's seat, If the car goes over a bump, what will be the frequency of oscillations? Ignore damping?

Question

2 SIGNIFICANT FIGUERS + UNIT

Answer 1

F = kx so k = F/x

The gravitational force of a 72 kg driver is (72)(9.8) = 705.6 N (because f = ma and a = the acceleration of gravity 9.8 m/s²)

So k = 705.6/(4.4/1000) = 705.6/0.0044 = 160363.6364 N/m

The frequency (ω) = √(k/m) = √[160363.6364/(1800+72)] = √85.664336 = 9.3 hz