Question

The springs of a 1300 kg car compress 4.5 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

T = 2*pi*sqrt(m/k) = 2*pi*sqrt(1368/248089) = 0.46657

F = 1/T = 1/0.46657 = 2.143 Hz

Answer 2

The spring constant k can be found using Hooke's Law

F = k*x

So k = F/x= mg/x

= 68 * 9.8 /0.0045 = 1.48*10⁵N/m

From SHM we get $\omega = \sqrt{\frac{k}{m}}$ $= \sqrt{\frac{1.48*10^{5}}{1300}} = 11.3846 rad /s$

So f = $\frac{\omega}{2\pi} =\frac{11.348}{2\pi}$