()
2
2
500 kg 9.8 m/s
=
4.9 10 N/cm.
10cm
k
=×
(b) The amplitude decreasing by 50% during one period of the motion implies
eT
bT
m
−
==
′
2
1
2
2
where
π
ω
.
Since the problem asks us to estimate, we let
′
≈=
km
/
. That is, we let
′
≈≈
49000
500
,
N/m
kg
9.9 rad / s
so that
T
≈
0.63 s. Taking the (natural) log of both sides of the above equation, and
rearranging, we find
3
2 500 kg
2
ln2
0.69
1.1 10 kg/s.
0.63 s
m
b
T
=≈
=
×
Note: if one worries about the
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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