Oscillation Frequency and Spring Constant

In summary, the automobile is supported by four springs and when six 68kg teenagers get into it, the car settles closer to the road by 3.2 centimeters. The spring constant for each spring was calculated to be 31238 N/m, assuming the springs are identical and the mass is equally distributed. To find the oscillation frequency of the car when carrying this load, the equation w(omega)=sqrt(k/m) can be used, with the total mass of the car and passengers (1808kg) and the total spring constant (124950 N/m).
  • #1
katdesutter
1
0

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.
 
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  • #2
The force the car feels is the total force from all of the four springs. If you write down Newton's second law:

[tex] m\ddot{x}=-4kx[/tex]

From this equation it is obvious that the relevant frequency is [tex] \omega=\sqrt{4k/m}[/tex]
 
  • #3
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)
 

Related to Oscillation Frequency and Spring Constant

1. What is the relationship between oscillation frequency and spring constant?

The oscillation frequency and spring constant have a direct relationship. As the spring constant increases, the oscillation frequency also increases. This means that a stiffer spring will have a higher oscillation frequency compared to a more flexible spring.

2. How does mass affect the oscillation frequency of a spring?

The mass of an object attached to a spring affects the oscillation frequency. As the mass increases, the oscillation frequency decreases. This is because more massive objects require more force to oscillate, resulting in a lower frequency.

3. Can the length of a spring affect its oscillation frequency?

Yes, the length of a spring can affect its oscillation frequency. Longer springs will have a lower oscillation frequency compared to shorter springs. This is because longer springs require more time to complete one oscillation compared to shorter springs.

4. How do you calculate the spring constant?

The spring constant can be calculated using Hooke's Law, which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position. The formula for spring constant is k = F/x, where k is the spring constant, F is the force applied to the spring, and x is the displacement of the spring.

5. What factors can affect the spring constant?

The spring constant can be affected by the material and shape of the spring, as well as the temperature and strain on the spring. Additionally, the number of coils, the diameter of the wire, and the tension on the spring can also affect the spring constant.

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