## Spring suspension on car

Ok, here is the problem:

A 1200 kg car carrying four 82 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 15 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?
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I know I'm supposed to find the spring constant of the suspension here and first I need to find the angular frequency...but how do I go about finding the angular frequency?

If anyone can help as soon as possible...thanks a lot
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 ω = 2Πf (f you can get from the speed and the space between the bumps.) Does that help?
 How do you get f from the speed and the bumps? I must have a really bad book...but it tells me that to get the frequency I need the angular frequency. My book says f= w/2pi...which gives you w right back. Thanks for the help

## Spring suspension on car

&omega; is the angular frequency (i.e. radians per second)

f is the linear frequency: cycles per second, bumps per second, bumps per minute, bumps per hour...

You have kilometers per hour and bumps per meter...

Got it?

Then, use
&omega; = 2&Pi;f
to get the angular frequency.

Then use what you know about simple harmonic motion to get the spring constant.
 Thanks a lot.