Posted by **Sarah** on Wednesday, January 4, 2012 at 7:56pm.

A car with bad shocks has a mass of 1500 kg. Before you go for a drive with three of your friends you notice that the car sinks a distance of 6.0 cm when all four of you get in the car. You estimate that the four of you together have a mass of 271 kg. As you are driving down the highway at 65 mph you notice that the car is starting to bounce up and down with large amplitude. You realize that there is a periodic series of small bumps and dips in the road that is driving the bouncing. What is the distance (in meters, to two signi cant gures) between adjacent bumps on the road, assuming that damping by the shocks is negligible?

- Physics -
**drwls**, Wednesday, January 4, 2012 at 8:05pm
Compute the natural frequency of vibration of the shocks with the added mass of passengers. That frequency is

W = [1/(2*pi)]sqrt (k/M) Hz

k is the spring constant, which you can get from the additional deflection due to a known load.

M = 1771 kg

Set the natural frequency equal to the fre

- Physics -
**drwls**, Wednesday, January 4, 2012 at 8:07pm
Compute the natural frequency of vibration of the shocks with the added mass of passengers. That frequency is

W = [1/(2*pi)]sqrt (k/M) Hz

k is the spring constant, which you can get from the additional deflection due to a known load.

M = 1771 kg

Set the natural frequency equal to the frequency that bumps are encountered, V/d, and solve for the bump separqation, d.

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**Sarah**, Wednesday, January 4, 2012 at 8:43pm
I did that including making k=(1/2)mvsqared, and I changed 65mph to mps. I get 8.89 as my answer and the computer says I am wrong and should get 36.5 what am I missing

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**drwls**, Thursday, January 5, 2012 at 7:44am
k is the spring constant, NOT the kinetic energy.

k = 271*9.8 N/(0.06 m)= 4.43*10^4 N/m

- Physics -
**Sarah**, Thursday, January 5, 2012 at 11:32am
Thank you so much

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