Question

The body of a 1300 Kg car is supported by four (4) springs. The people in the car have a combined mass of 300 Kg. When driven over a pothole in the road, the car vibrates with a period of 0.75 seconds in a manner that approximates simple harmonic motion. Find the spring constant of a single (1) spring.

Answer 1

The spring constant for a single spring is 28073.5 N/m

Step-by-step explanation:

First if we assume the vibration of the approximates simple harmonic motion, the to each of the springs we can associate a natural frequency of vibration that is defined for a simple spring as:

`\omega_0 =\sqrt{(k)/(m)}`(1)

with k the spring constant and m the mass attached to the spring

Natural frequency is related with the period (T) of the oscillation as follows:

`\omega_0=(2\pi)/(T)`(2)

We can equate (1) and (2):

`\sqrt{(k)/(m)}=(2\pi)/(T)`

solving for k:

`k=m((2\pi)/(T))^2` (3)

k is the spring constant for a single spring, if we assume the weight of the car and the people is uniformly distributed on the four springs, then the mass m is:

`m=(1300+300)/(4)=400kg`

Now we have all the values to put on (3):

`k=m((2\pi)/(T))^2 =400((2\pi)/(0.75))^2`

`k=28073.5 (N)/(m)`