Question

An 88 kg person steps into a car of mass 2002 kg, causing it to sink 5.36 cm on itssprings. Assuming no damping, with what fre-quency will the car and passenger vibrate onthe springs? Answer in units of Hz. The acceleration of gravity is 9.81 m/s^2.

Answer 1

Answer:

The required frequency = 0.442 Hz

Explanation:

Frequency $f = ( \dfrac{1}{2 \pi}) \omega$

where;

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

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{k}{m} } \Bigg )$

However;

$k = \dfrac{F}{x}$ and;

mass $m = m_{car } + m_{person}$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{\dfrac{F}{x}}{m_{car}+m_{person}} } \Bigg )$

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{{F}}{x(m_{car}+m_{person})} } \Bigg )$

where;

$F = m_{person}g$

Then;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {m_{person}g }}{x(m_{car}+m_{person})} } \Bigg )$

replacing the values;

$f = \Bigg ( \dfrac{1}{2 \pi} \Bigg ) \Bigg( \sqrt{\dfrac{ {(88 \ kg)* (9.81 \ m/s^2) }}{(5.36 \times 10^{-2} \ m) (2002 \ kg +88 \ kg)} } \Bigg )$

$\mathbf{f = 0.442 \ Hz}$