- #1
LostTexan07
- 3
- 0
The springs of a 1100 kg car compress 6.0 mm when its 68 kg driver gets into the driver's seat. If the car goes over a bump, what will be the frequency of vibrations?
I tried to use the weight of the man (68 kg) and the compression distance (6 mm) to find the spring constant. I then tried to use the spring constant to find the period.
F = kx
68 = 6k
k = 11.33
T = 2(pi)(sq.rt 1100/11.33)
But I didn't get the correct answer.
I tried to use the weight of the man (68 kg) and the compression distance (6 mm) to find the spring constant. I then tried to use the spring constant to find the period.
F = kx
68 = 6k
k = 11.33
T = 2(pi)(sq.rt 1100/11.33)
But I didn't get the correct answer.