SHM - Worn out shocks on car, helpp

[Student3.41]

Homework Statement

A 94.0 kg man climbs onto a car with worn out shock absorbers and this causes the car to drop down 5.90 cm and comes to equilibrium. As he drives along, he hits a bump, and this starts the car oscillating at an angular frequency of 4.65 rad/s. What is the mass of the car?

Homework Equations

The Attempt at a Solution

I found the spring constant using Newtons 2nd law, k=mg/x =1563N/m

angular frequency = \sqrt{k/m}

m=1563/21.6 = 72.4kg

72.4kg x 4 = 289kg b/c its distributed by 4 shocks.. but not the right answer.. def missing something

 

[collinsmark]

I found the spring constant using Newtons 2nd law, k=mg/x =1563N/m

Try that calculation again. You're off by a factor of 10.

72.4kg x 4 = 289kg b/c its distributed by 4 shocks.. but not the right answer.. def missing something

I think you're supposed to assume that the spring constant that you solved for above is the combined spring constant of all the shocks on the car. In other words, you shouldn't have to multiply by 4 or anything like that.

 

[Student3.41]

Try that calculation again. You're off by a factor of 10.

I think you're supposed to assume that the spring constant that you solved for above is the combined spring constant of all the shocks on the car. In other words, you shouldn't have to multiply by 4 or anything like that.

Oops, haha...

Well, k=15 600N/m

When I substitute this value for k

ω=\sqrt{k/m}

m=k/ω^2

m=15600/21.6=722kg-94.0kg =628kg Thank you!, that was the correct answer

When i got the answer wrong, a note came up stating "The weight is distributed over the four shock absorbers"

 

[collinsmark]

Oops, haha...

Well, k=15 600N/m

When I substitute this value for k

ω=\sqrt{k/m}

m=k/ω^2

m=15600/21.6=722kg-94.0kg =628kg Thank you!, that was the correct answer

When i got the answer wrong, a note came up stating "The weight is distributed over the four shock absorbers"

Okay, I think I know what's going on.

My original assumption was that displacement of 5.90 cm was measured after the man settled into the car, and that it was averaged over the all sides of the car. If that were the case, it wouldn't matter if the car was supported by one big shock absorber, four [identical] shock absorbers, or even four thousand [identical] shock absorbers (as long as as the weight distribution was uniform). The answer would end up being the same.

But doesn't seem to be the way it was measured. I'm guessing that when the man "climbs" into his car (and after the shocks reach an equilibrium), the man is still completely to one side of the car such that only half of the shock absorbers are supporting his weight, and that's when the 5.90 cm displacement was measured. Later, the man continues to climb into the car all the way and centers himself (and the displacement changes accordingly). So the overall spring constant when the car+man hits the bump is twice what you originally calculated (since there are now twice as many shock absorbers in action).

Confusing? Yes. I think the problem statement should have been more specific about when and where the 5.90 cm displacement was measured in the first place. Specifying the details would have made the problem less ambiguous, and keep the student from doing pointless guesswork. But that's just my opinion. (A figure or diagram would have proved very useful to show the weight distribution when things were being measured.)

Good luck!

[Edit: Toned down my criticism of the problem statement just a tad.]