Spring constant of a car on springs (1 Viewer)

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1. The problem statement, all variables and given/known data
When four people with a combined mass of 325 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.55 cm.

A) What is the effective force constant of the springs?

B) The four people get out of the car and bounce it up and down. What is the frequency of the car's vibration?

2. Relevant equations

ΔForce = -k(Δx)

frequency = (1/(2pi))*Sqrt(k/m)

3. The attempt at a solution

A)

ΔForce = -k(Δx)

k = (2325*9.81)/(0.0055m)

k = 4146955 N/m (Wrong answer apparently)

B)

frequency = (1/2pi)*sqrt(4146955/2000)

frequency = 7.247 Hz
 
Last edited:

scottdave

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So the car has 4 wheels, each wheel has its own spring. OK so they want the effective force constant, so you can probably forget about the factor of 4.

Also the Force = -kx would probably make more sense expressed as:
ΔForce = -k(Δx). You are looking for the change in force to affect a change in displacement.
 
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So the car has 4 wheels, each wheel has its own spring. OK so they want the effective force constant, so you can probably forget about the factor of 4.

Also the Force = -kx would probably make more sense expressed as:
ΔForce = -k(Δx). You are looking for the change in force to affect a change in displacement.
Ok, fixed. But that wouldn't change the answer I got, so what have I done wrong? Thanks
 

scottdave

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Ok, fixed. But that wouldn't change the answer I got, so what have I done wrong? Thanks
The empty car weight compresses the spring a certain amount (you don't know this number). Adding 325 kg additional mass causes the weight on the springs to increase by an amount, causing the springs compresssion to change by 0.55 cm

So I can't resist asking "What is the delta force?" :biggrin:
 
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The empty car weight compresses the spring a certain amount (you don't know this number). Adding 325 kg additional mass causes the weight on the springs to increase by an amount, causing the springs compresssion to change by 0.55 cm

So I can't resist asking "What is the delta force?" :biggrin:
Oh, I see, so:

Fi = 9.81*2000 = 19600 N

Ff = 9.81*2325 = 3208.25 N

The change in force is 3208.25 N.

So 3208 = kx

k = 3208/0.0055

k = 583272.7273 N/m

Is this correct? If so, then I would just have to plug the numbers in for part B?

Thanks a lot
 

scottdave

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You're almost there.
Check your Ffinal calculation again. Then do the subtraction to get the proper change in force.
 
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You're almost there.
Check your Ffinal calculation again. Then do the subtraction to get the proper change in force.
Ff is 22808.25, my mistake. Change in force is 22808.25-19600 = 3208N.
 
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You're almost there.
Check your Ffinal calculation again. Then do the subtraction to get the proper change in force.
For b, I have:

frequency = (1/(2pi))*sqrt(583272.7/2000)

frequency = 2.718 Hz
 

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