Question

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of 3 Hz.
(a) What is the spring constant of each spring if the mass of the car is 1400 kg and the weight is evenly distributed over the springs?
N/m

(b) What will be the vibration frequency if five passengers, averaging 71.0 kg each, ride in the car with an even distribution of mass?
Hz

Answer 1

I will be happy to critique your thinking.

Answer 2

f = sqrt(k/m)

where f is the frequency, k is the spring constant of the four springs, and m is the mass of the car.

=> k = f²*m = (3 Hz)²*1400 kg =12600 N/m

This results is the results for the total of the four springs, so for one spring we divide the answer by four:

k/4 = 3150 N/m

The extra weight on the car will amount to 355 kg, bringing the total of car and passengers to 1750 kg.

=> f = sqrt(k/m) = sqrt(12600 (N/m)/17150 kg) = 2.68 Hz

Answer 3

b) k =mw^2, k = 1400 * 18.84^2, k = 496924/4 = 124231 N/M

Answer 4

Christiaan your answer for the spring constant is wrong

Answer 5

not clear enough you have to demostrate all calculation

Answer 6

for a)you must first calculate the angular frequency,w, you are only given the frequency,f. to do that you use the equation:
w=2*pi*f
=2*pi*3Hz
next you take that and put it into the equation with k being the spring constant,w being the angular velocity, and m being the mass:
w= sqrt( k / m )
6*pi= sqrt( k / 1400)
...but since there are 4 springs you need to modify the equation to be:
6*pi= sqrt( 4k / 1400)
so the answer should be 124357.0155