How Is the Spring Constant Calculated for Car Vibrations?

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EroAlchemist
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Homework Statement



The springs of a 1500kg car compress 5mm when its 68kg driver gets into the drivers seat. If the car goes over a bump what will be the frequency of the vibrations?


Homework Equations



Spring force = F = -kx
Time = (4pi^2)(m/k)
Freq = 1/T


The Attempt at a Solution



Change of mass = 68kg
(68kg * 9.8m/s2)/.005m = -k = 1.33 * 10^5 Nm
k = -1.33 * 10^5 Nm

Time = (4pi^2)(68kg/-1.33 * 10^5 Nm) = 0.02s (in neg y direction)

Freq = 1/T = 1/0.02 = 50Hz

Book gives 1.5 Hz as correct answer.
Should I be using a different mass value? Different equations? Thanks for taking a look!
 
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EroAlchemist said:
If the car goes over a bump what will be the frequency of the vibrations?

Spring force = F = -kx
Time = (4pi^2)(m/k)

k = -1.33 * 10^5 Nm

Time = (4pi^2)(68kg/-1.33 * 10^5 Nm) = 0.02s (in neg y direction)

Freq = 1/T = 1/0.02 = 50Hz

Book gives 1.5 Hz as correct answer.
Should I be using a different mass value? Different equations? Thanks for taking a look!


The book is correct.
Your equation should be (double check your notes or text):
Time = sqrt[(4pi^2)(m/k)]
The mass should be the mass of the system (car + driver), not just the driver.
 
Thanks tvavanasd -

Got it now!

EA