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

The required frequency = 0.442 Hz

Step-by-step explanation:

Frequency
`f = ( (1)/(2 \pi)) \omega`

where;

`\omega = \sqrt{(k)/(m) }`

Then;

`f = \Bigg ( (1)/(2 \pi) \Bigg ) \Bigg( \sqrt{(k)/(m) } \Bigg )`

However;

`k = (F)/(x)` and;

mass
`m = m_(car ) + m_(person)`

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

`f = \Bigg ( (1)/(2 \pi) \Bigg ) \Bigg( \sqrt{\frac{{F}}{x(m_(car)+m_(person))} } \Bigg )`

where;

`F = m_(person)g`

Then;

`f = \Bigg ( (1)/(2 \pi) \Bigg ) \Bigg( \sqrt{\frac{ {m_(person)g }}{x(m_(car)+m_(person))} } \Bigg )`

replacing the values;

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

`\mathbf{f = 0.442 \ Hz}`