Asked by Eric
When a 65 kg person climbs into a 1000 kg car, the car's springs compress vertically by 3.1 cm. What will be the frequency of vibration when the car hits a bump? Ignore damping.
Answers
Answered by
Damon
65*9.8 = 637 Newtons weight
so
k = weight/deflection = 637/.031 = 20548 Newtons/meter
now the vibration problem
mass = 1068
k = 20548
You probably know about sqrt (k/m) but
F = m a = -kx
if x = A sin2pif t
v = 2pif Acos 2pif t
a = -(2pif)^2 A sin2pift
so
-m(2pif)^2 A sin 2pift = -kAsin 2pift
or
(2 pi f)^2 = k/m
f = (1/2pi) sqrt(k/m)
so
k = weight/deflection = 637/.031 = 20548 Newtons/meter
now the vibration problem
mass = 1068
k = 20548
You probably know about sqrt (k/m) but
F = m a = -kx
if x = A sin2pif t
v = 2pif Acos 2pif t
a = -(2pif)^2 A sin2pift
so
-m(2pif)^2 A sin 2pift = -kAsin 2pift
or
(2 pi f)^2 = k/m
f = (1/2pi) sqrt(k/m)
Answered by
Damon
blargh
Thoust claim is forsaken, please revise and realize. <p>Hi</p>
asd
Answered by
lol rawr xd
I can make a better physics response by just looking up this question
Answered by
sdasa
sdasfsaf
Answered by
sdasa
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