Physics 315:
Oscillations and Waves
Homework 4: Solutions
1. Consider the
i
th loop from the left in the circuit, which consists of the
i
th
capacitor, the
i
th inductor, and the
i
+ 1th capacitor. The current flowing
through the
i
th inductor is
I
i
. Thus, the potential drop across the inductor
is
L
˙
I
i
.
From Kirchoff’s first circuital law, the current flowing through
the
i
th capacitor (with the same sense of circulation about the
i
th loop
as the current flowing through the
i
th inductor) is
I
i

I
i
−
1
.
Hence, the
potential drop across the capacitor is
integraltext
t
0
[
I
i
(
t
′
)

I
i
−
1
(
t
′
)]
dt
′
/C
. Likewise,
the current flowing through the
i
+ 1th capacitor (with the same sense of
circulation) is
I
i

I
i
+1
.
Thus, the potential drop across the capacitor is
integraltext
t
0
[
I
i
(
t
′
)

I
i
+1
(
t
′
)]
dt
′
/C
. According to Kirchoff’s second circuital law, the
sum of the potential drops around any closed loop in a circuit is zero. Hence,
applying this law to the
i
th loop, we get
L
dI
i
(
t
)
dt
+
integraldisplay
t
0
[
I
i
(
t
′
)

I
i
−
1
(
t
′
)]
dt
′
/C
+
integraldisplay
t
0
[
I
i
(
t
′
)

I
i
+1
(
t
′
)]
dt
′
/C
= 0
.
(1)
Differentiating with respect to
t
, and rearranging, we obtain
d
2
I
i
dt
2
=
ω
2
0
(
I
i
−
1

2
I
i
+
I
i
+1
)
,
(2)
where
ω
0
= 1
/
√
L C
. Thus, we have a set of
N
coupled ordinary differential
equations in which the loop index
i
ranges from 1 to
N
.
However, the
first and last equations are special cases, since they govern the behavior
of the leftmost and rightmost circuit loops, respectively, and these loops
are different from the interior loops.
Consider the leftmost circuit loop.
In this loop, the current flowing through the 1st capacitor is
I
1
, rather
than
I
1

I
0
. We can incorporate this additional information into the 1st
differential equation by setting
I
0
to zero. Likewise, in the rightmost circuit
loop, the current flowing through the
N
+ 1th capacitor is
I
N
, rather than
I
N

I
N
+1
. Hence, it is necessary to set
I
N
+1
to zero in the
N
th differential
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equation. In other words, we need to solve Eqs. (2), for
i
= 1
, N
, subject
to the constraints
I
0
=
I
N
+1
= 0.
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 Spring '10
 Fitzpatrick
 Current, Work, Sin, The Circuit, Normal mode, Mode shape, potential drop

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