Chapter 4
The maximum displacement of the mass is found by determining the first time the velocity
of the mass becomes zero. Therefore, we have
y
(
t
) = 0 =
2 + 0
.
2
√
5
e
−
2
√
5
t
−
2
√
5 0
.
1 +
2 + 0
.
2
√
5
t e
−
2
√
5
t
,
which gives
t
=
2
2
√
5(2 + 0
.
2
√
5)
=
1
√
5(2 + 0
.
2
√
5)
.
Thus the maximum displacement is
y
1
√
5(2 + 0
.
2
√
5)
= 0
.
1 +
2 + 0
.
2
√
5
1
√
5(2 + 0
.
2
√
5)
e
−
2
√
5
/
[
√
5(2+0
.
2
√
5)]
≈
0
.
242 (m)
.
11.
The equation of the motion of this massspring system is
y
+ 0
.
2
y
+ 100
y
= 0
,
y
(0) = 0
,
y
(0) = 1
.
Clearly, this is an underdamped motion because
b
2
−
4
mk
= (0
.
2)
2
−
4(1)(100) =
−
399
.
96
<
0
.
So, we use use equation (16) on page 213 in the text for a general solution. With
α
=
−
b
2
m
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 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations, Velocity, c1 cos, 5T, maximum displacement

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