# Oscillations Car carrying four people find how much body rises

1. Nov 23, 2004

A 1175 kg car carrying four 80 kg people travels over a rough "washboard" dirt road with corrugations 4.0 m apart which causes the car to bounce on its spring suspension. The car bounces with maximum amplitude when its speed is 17 km/h. The car now stops, and the four people get out. By how much does the car body rise on its suspension owing to this decrease in weight?

I first tried figuring the natural frequency of the car:

$$\omega_o=(2\pi\upsilon)/(\Delta(x))=(2\pi(17km/h)(1hr/60s)/(.004km)$$ This gives 445.059 rad/s.

I then tried to figure out k:

$$\kappa=(m1+m2)\omega_o^2=((1175 kg+(4*180kg))(445.059rad/s)^2$$ This gives 3.75e8.

Finally, I tried finding $$\Delta(x)=\Delta(F)/\kappa=(m_2*g)/(\kappa)=((180 kg*4)(9.8m/s^2))/(3.75*10^8)$$This gives 1.8e-5 m, or .00188 cm, this is wrong, where did I go wrong? Help!

PS This is the first time I use Latex so if it looks odd I'm sorry!

2. Nov 23, 2004

Help...Anyone lol...

3. Nov 27, 2004

### vtech

your conversion for the first part seems to be off... recheck it... and make sure you convert to meter seconds... but you're on the right track

4. Nov 30, 2011

### mmayorga

i just solved my homework problem with your equations and it worked. the only problem with your method is your unit conversions: to go from km/h to m/s you multiply by 1000/(60^2)