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GingerBread27
Nov23-04, 11:18 AM
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!

GingerBread27
Nov23-04, 07:28 PM
Help...Anyone lol...

vtech
Nov27-04, 08:20 PM
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

mmayorga
Nov30-11, 02:42 PM
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)