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Chapter 4
The maximum displacement of the mass is found by determining the Frst time the velocity
of the mass becomes zero. Therefore, we have
y
0
(
t
)=0=
±
2+0
.
2
√
5
²
e
−
2
√
5
t
−
2
√
5
h
0
.
1+
±
.
2
√
5
²
t
i
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
³
1
√
5(2 + 0
.
2
√
5)
´
=
³
0
.
±
.
2
√
5
²
µ
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
+0
.
2
y
0
+ 100
y
=0
,y
(0) = 0
0
(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
=
−
0
.
2
2
=
−
0
.
1a
n
d
β
=
1
2
m
√
4
mk
−
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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