Answer to Question #171540 in Physics for Jerick

Question

 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 a bump, what will be the frequency of oscillations? Ignore damping.

Answer 1

Find spring compression without the driver:

$$x_1=\frac{Mg}{k}.$$

With the driver:

$$x_2=x_1+\Delta x=\frac{(M+m)g}{k},\\\space\\ \frac{Mg}{k}+\Delta x=\frac{(M+m)g}{k},\\\space\\ \Delta x=\frac{mg}{k},\\\space\\ k=\frac{mg}{\Delta x}.$$

Now remember the formula for the period of oscillations of the spring-mass system:

$$T=2\pi\sqrt{\frac{m_\text{total}}k}=2\pi\sqrt{\frac{(m+M)}k}.$$

And frequency of oscillations is... inverse of period:

$$f=\frac 1T=\frac{1}{2\pi}\sqrt{\frac{k}{m+M}}=\frac{1}{2\pi}\sqrt{\frac{mg}{\Delta x(m+M)}}.\\\space\\ f=\frac{1}{2\pi}\sqrt{\frac{66·9.8}{0.005(66+1700)}}=1.36\text{ Hz}.$$