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

Answer:

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

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:

$\omega_0 =\sqrt{\frac{k}{m}}$(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:

$\omega_0=\frac{2\pi}{T}$(2)

We can equate (1) and (2):

$\sqrt{\frac{k}{m}}=\frac{2\pi}{T}$

solving for k:

$k=m(\frac{2\pi}{T})^2$ (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:

$m=\frac{1300+300}{4}=400kg$

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

$k=m(\frac{2\pi}{T})^2 =400(\frac{2\pi}{0.75})^2$

$k=28073.5 \frac{N}{m}$