Asked by Anonymous
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?
Answers
Answered by
Damon
6 mm = 6 * 10^-3 m
68 * 9.81 = 667 Newtons
k of spring = 667/(6*10^-3) = 111,167 N/m
w = sqrt(k/m) = sqrt (111,167/1100)
=10 radians/s
2 pi f = w = 10
f = 1.6 Hz
68 * 9.81 = 667 Newtons
k of spring = 667/(6*10^-3) = 111,167 N/m
w = sqrt(k/m) = sqrt (111,167/1100)
=10 radians/s
2 pi f = w = 10
f = 1.6 Hz
Answered by
Adam
Can someone please explain to me how 2pi f = w = 10
f = 1.6 HZ
because this is the only part I can't figure out for myself.I highly appreciate it,thanks
f = 1.6 HZ
because this is the only part I can't figure out for myself.I highly appreciate it,thanks
Answered by
Joe
since we know that w = 10 and that 2pi f = w we can divide w by 2pi so 10/(2pi) = 1.6
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