posted by johnny .
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?
(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?
I will be happy to critique your thinking.
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
b) k =mw^2, k = 1400 * 18.84^2, k = 496924/4 = 124231 N/M
Christiaan your answer for the spring constant is wrong
not clear enough you have to demostrate all calculation
for a)you must first calculate the angular frequency,w, you are only given the frequency,f. to do that you use the equation:
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