Question

A car of mass 1000 kg had its springs compressed vertically by 2.6 cm when a driver of mass 62 kg climbed into it. If damping is ignored what is the frequency of vibration when the car hits a bump while driving? 

Answer 1

The spring-mass system is a typical case of a simple harmonic motion, since the distance traveled by the mass describes an oscillatory behaviour. The natural angular frequency of a spring-mass system is computed by

$\omega=\sqrt{\frac{k}{m}}$

And the frequency is

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

Thus

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

The total mass of the car and the driver is

$m = 1000 + 62 = 1062\ kg$

They both weight

$W = mg = 1062\times9.81 = 10418.2 N$

We need to know the constant of the spring. It can be found by using the formula of the Hook's law:

$F=kx$

We know the spring stretches 2.6 cm (0.026 m) when holding the total weight of the car and the driver. Solving for k

$k=\frac{F}{x}$

$k=\frac{10418.2}{0.026}=400700.77\ N/m$

Thus, the frequency of oscillations is

$f=\frac{1}{2\pi}\sqrt{\frac{400700.77}{1062}}=3.09\ Hz$

Answer: f = 3.09 Hz.