28. (a) In Eq. 3325, we set
q
=0and
t
= 0 to obtain 0 =
Q
cos
φ
.T
h
i
sg
i
v
e
s
φ
=
±
π/
2 (assuming
Q
6
= 0). It should be noted that other roots are possible (for instance, cos(3
π/
2) = 0) but the
±
π/
2
choices for the phase constant are in some sense the “simplest.” We choose
φ
=
−
π/
2tomakethe
manipulation of signs in the expressions below easier to follow. To simplify the work in part (b),
we note that cos(
ω
0
t
−
π/
2) = sin(
ω
0
t
).
(b) First, we calculate the timedependent current
i
(
t
) from Eq. 3325:
i
(
t
)=
dq
dt
=
d
dt
³
Qe
−
Rt/
2
L
sin(
ω
0
t
)
´
=
−
QR
2
L
e
−
Rt/
2
L
sin(
ω
0
t
)+
Qω
0
e
−
Rt/
2
L
cos(
ω
0
t
)
=
Qe
−
Rt/
2
L
µ
−
R
sin(
ω
0
t
)
2
L
+
ω
0
cos(
ω
0
t
)
¶
,
which we evaluate at
t
=0
:
i
(0) =
Qω
0
.I
fw
ed
en
o
t
e
i
(0) =
I
as suggested in the problem, then
Q
=
I/ω
0
. Returning this to Eq. 3325 leads to
q
=
Qe
−
Rt/
2
L
cos(
ω
0
t
+
φ
)=
µ
I
ω
0
¶
e
−
Rt/
2
L
cos
³
ω
0
t
−
π
2
´
=
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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