3010
Chapter 30
(c)
SET UP:
At critical damping,
4/
R
LC
=
.
EXECUTE:
()
3
9
422
.0 10 H
2420
15.0 10 F
R
−
−
×
==
Ω
×
EVALUATE:
The frequency with damping is almost the same as the resonance frequency of this circuit (
1/
LC
),
which is plausible because the 75
Ω
resistance is considerably less than the 2420
Ω
required for critical damping.
30.39.
IDENTIFY:
Follow the procedure specified in the problem.
SET UP:
Make the substitutions
1
,,
,
xq
mL
bR
k
C
→→→→
.
EXECUTE:
(a)
Eq. (13.41):
2
2
0
dx bd
x k
x
dt
m dt
m
++
=
. This becomes
2
2
0
dq Rdq
q
dt
L dt
LC
+
+=
, which is Eq.(30.27).
(b)
Eq. (13.43):
2
2
4
kb
ω
mm
′
=−
. This becomes
2
2
1
4
R
ω
L
CL
′
, which is Eq.(30.29).
(c)
Eq. (13.42):
(/
2 )
cos(
)
bm
t
xA
e
ω
t
φ
−
′
=+
. This becomes
2)
cos(
)
RL
t
qA
e
ω
t
−
′
=
+
, which is Eq.(30.28).
EVALUATE:
Equations for the
L

R

C
circuit and for a damped harmonic oscillator have the same form.
30.40.
IDENTIFY:
For part (a), evaluate the derivatives as specified in the problem. For part (b) set
qQ
=
in Eq.(30.28)
and set
/0
dq dt
=
in the expression for
/
dq dt
.
SET UP:
In terms of
ω
′
, Eq.(30.28) is
cos
(
)
t
qt
Ae
t
−
′
=
+
.
EXECUTE:
(a)
cos(
)
t
e
ω
t
−
′
.
cos(
)
sin(
).
2
t
t
dq
R
ω
t
ω
Ae
ω
t
dt
L
−−
′′
′
+
−
+
2
2
2
2
cos(
)
2
sin(
)
cos(
)
22
t
t
t
dq
R
R
ω
t
ω
ω
t
ω
ω
t
dt
L
L
φφ
−
⎛⎞
′
′
′
+
+
−
+
⎜⎟
⎝⎠
2
2
1
0
q
R
R
q
dt
L dt
LC
L
L
LC
′
=
−
−
, so
2
2
2
1
4
R
ω
L
′
.
(b)
At
0,
,
0
dq
tq
Q
i
dt
===
=
, so
cos
Q
and
cos
sin
0
2
dq
R
A
ω
A
dt
L
′
=
=
. This gives
cos
Q
A
=
and
tan
2
21
/
/
4
RR
L
ω
L
LC
R
L
′
−
.
EVALUATE:
If
0
R
=
, then
AQ
=
and
0
=
.
30.41.
IDENTIFY:
Evaluate Eq.(30.29).
SET UP:
The angular frequency of the circuit is
′
.
EXECUTE:
(a)
When
0
5
11
0,
298 rad s.
(0.450 H) (2.50 10
F)
R
ω
LC
−
=
=
×
(b)
We want
0
0.95
ω
ω
=
, so
2
2
(1
4
)
1(
0
.
9
5
)
14
LC
R
L
R C
LC
L
−
=
. This gives
2
5
4
4(0.450 H)(0.0975)
(1 (0.95) )
83.8
.
(2.50 10
F)
L
R
C
−
=
=
Ω
×
EVALUATE:
When
R
increases, the angular frequency decreases and approaches zero as
2/
R
→
.
30.42.
IDENTIFY:
L
has units of H and
C
has units of F.
SET UP:
1 H
1
s
=Ω
⋅
.
/
CqV
=
says 1 F 1 C/V.
=
VI
R
=
says 1 V/A
1
=
Ω
.
EXECUTE:
The units of
/
L C
are
2
HsV
.
FC
V A
Ω⋅
=
Ω
Therefore, the unit of
/
L C
is
.
Ω
EVALUATE:
For Eq.(30.28) to be valid,
1
LC
and
2
2
4
R
L
must have the same units, so
R
and
/
L C
must have the
same units, and we have shown that this is indeed the case.
30.43.
IDENTIFY:
The emf
2
E
in solenoid 2 produced by changing current
1
i
in solenoid 1 is given by
1
2
.
i
M
t
Δ
=
Δ
E
The
mutual inductance of two solenoids is derived in Example 30.1. For the two solenoids in this problem
01
2
,
AN N
M
l
μ
=
where
A
is the crosssectional area of the inner solenoid and
l
is the length of the outer solenoid.