physics

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

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asked by johnny
  1. I will be happy to critique your thinking.

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  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

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

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    posted by johnny
  4. Christiaan your answer for the spring constant is wrong

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  5. not clear enough you have to demostrate all calculation

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  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

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    posted by Kara

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