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CHAPTER 9 PROBLEM SOLUTIONS
S
OLUTION TO
P
ROBLEM
9.1.
If we can compute expressions for K and q that are real,
then these quantities exist by construction.
Consider that A, B, K and q must satisfy the
following relationship:
K
cos(
ϖ
t
+θ
)
=
K
cos(
θ
)
( 29
cos(
ϖ
t
)
+

K
sin(
θ
)
( 29
sin(
ϖ
t
)
≡
A
cos(
ϖ
t
)
+
B
sin(
ϖ
t
)
Therefore
K
cos(
θ
)
=
A
and

K
sin(
θ
)
=
B
.
Consequently,
K
cos(
θ
)
( 29
2
+

K
sin(
θ
)
( 29
2
=
K
2
=
A
2
+
B
2
in which case
K
=
A
2
+
B
2
.
Further,
K
sin(
θ
)
K
cos(
θ
)
=
tan(
θ
)
=

B
A
in which case
θ =
tan

1

B
A
with due regard to quadrant.
S
OLUTION TO
P
ROBLEM
9.2.
For the inductor,
W
L
(
t
)
=
1
2
L
⋅
i
L
2
(
t
)
=
1
2
L V
o
C
L
sin
1
LC
t
2
=
CV
0
2
2
sin
2
1
LC
t
and for the capacitor,
W
C
(
t
)
=
1
2
C
⋅
v
C
2
(
t
)
=
1
2
C V
o
cos
1
LC
t
2
=
CV
0
2
2
cos
2
1
LC
t
.
Hence,
W
C
+
W
L
=
1
2
C
⋅
v
C
2
(
t
)
+
1
2
L
⋅
i
L
2
(
t
)
=
CV
0
2
2
sin
2
1
LC
t
+
cos
2
1
LC
t
=
CV
0
2
2
S
OLUTION TO
P
ROBLEM
9.3.
Since
x
(
t
)
=
(
K
1
+
K
2
t
)
e
α
t
,
x
'(
t
)
=α⋅
K
1
e
α
t
+
K
2
e
α
t
α⋅
t
⋅
K
2
e
α
t
and
x
''(
t
)
=α
2
⋅
K
1
e
α
t
α⋅
K
2
e
α
t
α⋅
K
2
e
α
t
+α
2
⋅
t
⋅
K
2
e
α
t
Substituting into the differential equation, we have
α
2
⋅
K
1
e
α
t

2
α⋅
K
2
e
α
t
+α
2
⋅
t
⋅
K
2
e
α
t
+
2
αα⋅
K
1
e
α
t
+
K
2
e
α
t
α⋅
t
⋅
K
2
e
α
t
[ ]
+α
2
K
1
e
α
t
+
K
2
te
α
t
[ ]
=
0
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View Full DocumentThis means that the solution form satisfies the differential equation.
S
OLUTION TO
P
ROBLEM
9.4.
(a)
Suppose
x
(
T
)
=
0
at some T.
Then
K
1
e
s
1
T
=
K
2
e
s
2
T
.
Since
e
s
i
T
0
whenever s
i
is real and T is finite, K
1
2
must have opposite signs.
(b)
For
this we solve for T and show there can only be one solution.
Since
K
1
e
s
1
T
=
K
2
e
s
2
T
and
e
s
i
T
0
,
K
1

K
2
=
e
s
2
T
e
s
1
T
implies
l
n
K
1

K
2
=
l
n
e
s
2
T
e
s
1
T
=
s
2

s
1
( 29
T
Hence the unique solution is given by
T
=
l
n
K
1

K
2
s
2

s
1
( 29
provided s
2
≠
s
1
which is the case for distinct roots.
S
OLUTION TO
P
ROBLEM
9.5.
Suppose
x
(
T
)
=
0
at some T >0.
This is true if and only if
K
1
e
s
1
T
=
K
2
Te
s
1
T
(*)
Since
e
s
1
T
0
and T > 0, (*) is true if and only if
K
1
= 
K
2
T
which is true if and only if
K
1
2
have opposite signs.
S
OLUTION
9.6
.
(a)
Denote one period of oscillation by T.
Then by definition
9950
T
=
2
π
.
Hence,
T
=
0.63148
ms.
The time constant of decay is 1 ms.
Therefore,
NT
=
N
×
0.63148=
.
Hence
N
=
1.5836
cycles.
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