# Physics

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The springs of a 1100 kg car compress 6.0 mm when its 68 kg driver gets into the driver's seat. If the car goes over a bump, what will be the frequency of vibrations?

• Physics -

6 mm = 6 * 10^-3 m

68 * 9.81 = 667 Newtons

k of spring = 667/(6*10^-3) = 111,167 N/m

w = sqrt(k/m) = sqrt (111,167/1100)

2 pi f = w = 10
f = 1.6 Hz

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