Question

(I) The springs of a 1700-kg car compress 5.0 mm when its 66-kg driver gets into the driver's seat If the car goes over abump, what will be the frequency of oscillations? Ignore damping.

Answer 1

Answer:

$f=7 \ Hz$

Explanation:

Simple Harmonic Motion

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

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

And the frequency is

${\displaystyle f=\frac {w}{2\pi}$

Thus

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

The total mass of the car and the driver is

$m=1700+66=1766\ kg$

They both weigh

$W=m.g=1766\ kg*9.8\ m/s^2$

$W=17306.8\ N$

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

$F=k.x$

We know the spring stretches 5 mm (0.005 m) when holding the total weight of the car and the driver. Solving for k

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

$\displaystyle k=\frac{17306.8}{0.005}$

$k=3,461,360\ N/m$

Thus, the frequency of oscillations is

${\displaystyle f =\frac {1}{2\pi}{\sqrt {\frac {3,461,360}{1,766}}}}$

$\boxed{f=7 \ Hz}$