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Exercises 4.8
and so
y
(
t
)=
±
−
3
4
cos 8
t
−
sin 8
t
²
e
−
8
t
=
5
4
e
−
8
t
sin(8
t
+
φ
)
,
where tan
φ
=(
−
3
/
4)
/
(
−
1) = 3
/
4andcos
φ
=
−
1
<
0. Thus,
φ
=
π
+ arctan(3
/
4)
≈
3
.
785
.
The damping factor is (5
/
4)
e
−
8
t
, the quasiperiod is
P
=2
π/
8=
4, and the quasifrequency
is 1
/P
=4
/π
.
9.
Substituting the values
m
,
k
= 40, and
b
=8
√
5 into equation (12) on page 213 in the
text and using the initial conditions, we obtain the initial value problem
2
d
2
y
dt
2
+8
√
5
dy
dt
+40
y
=0
,y
(0) = 0
.
1(m)
0
(0) = 2 (m
/
sec)
.
The initial conditions are positive to reﬂect the fact that we have taken down to be positive
in our coordinate system. The auxiliary equation for this system is
2
r
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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 Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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