Oscillation Frequency and Spring Constant

[katdesutter]

Homework Statement

An automobile is supported by four wheels. These wheels are connected to the automobile by four springs. When six 68Kg teenagers get into the automobile, it settles closer to the road by 3.2 centimeters. What is the spring constant of each of the springs? If the automobile has a mass of 1400 Kg when empty, what would be its oscillation frequency when carrying this load of people.

Homework Equations

F=-kx
F=ma
w(omega)=sqrt(k/m)

The Attempt at a Solution

I assumed the springs are identical, the mass is equally distributed over the four springs, and springs are unextended when there are no passengers
For the first question: I first converted cm to m. (0.032m)
I then set -kx=ma
After plugging in my known values:
x=0.032m
m=408kg
g=9.8 N/m^2
I came up with a k value of 124950 N/m and then divided this number by 4 to get one spring's k constant.
Thus k=31238 N/m

I think I did the first part right? For the second question I didn't know what values to use in my equation. I am pretty sure that I use the equation w(omega)=sqrt(k/m)
I don't know if I am supposed to use the k constant for only one spring or if I am supposed to use the total k constant. I know the mass will be 1808kg.

 

[eladtan]

The force the car feels is the total force from all of the four springs. If you write down Newton's second law:

m\ddot{x}=-4kx

From this equation it is obvious that the relevant frequency is \omega=\sqrt{4k/m}

 

[squarks]

Or divide the mass by four, since the mass is distributed equally to each of the spring (i.e. the same reasoning as your first part)