Simple harmonic motion of a car

[danielle36]

A 1000 kg car carrying two 100 kg football players travels over a bumpy "washboard" road with the bumps spaced 3.0 m apart. The driver finds that the car bounces with a max amplitude when he drives at a speed of 5.0 m/s. The car then stops and picks up three more 100 kg passengers. By how much does the car body sag on its suspension when these three additional passengers get in?

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$$\lambda = 3.0 m$$
$$f = v/ \lambda$$
$$f = 1.67 Hz$$
$$\omega = 2 \pi f = \sqrt{k/m}$$
$$k = 132300 N/m$$
$$\Delta L = mg/k$$
$$\Delta L = 0.11 m$$

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Answer is 2.23 cm

I think my basic problem here is I just don't know what I'm doing.

Thanks for taking the time to read :)

 

[CompuChip]

If you don't know what you are doing, then how did you get this far? Is this the way the answer tells you to do it? Or did the question hint at this and did you find it out yourself? Or did you put it together all by yourself?

As for the answer, I can say you are almost there. Just note that you are calculating the change in L in the last two lines, not L itself. This change is induced by the changing of another quantity, if you want the $\Delta L$ you will also have to plug in the change in this quantity, and not the value itself. Or, alternatively, you can first calculate L for the case before and after, and subtract them to find the difference.

 

[danielle36]

Ohhh I thought I was just way off... I had tried subtracting the change in L when m = 1000 kg, but I see now that I had to use m = 1200 kg and that gives me a 2.22 cm difference


Thank you :D

 

[CompuChip]

Great

I think my basic problem here is I just don't know what I'm doing.

Has that problem been solved too, or do you want someone to explain it?

 

[danielle36]

Actually, I think I got it now.. What I was doing made sense to me but since I was so far off I thought I was applying the equations to the wrong situation, but it turns out I just hadn't thought the question over enough