Physics 315:
Oscillations and Waves
Homework 6: Solutions
1. Making use of the identity cos(
α
−
β
)
≡
cos
α
cos
β
+ sin
α
sin
β
, we can
write the traveling wave solution
ψ
(
x,t
) =
A
cos(
kx
−
ωt
)
(1)
in the form
ψ
(
x,t
) =
A
cos(
kx
) cos(
ωt
) +
A
sin(
kx
) sin(
ωt
)
.
(2)
But, this is just the sum of two standing wave solutions.
Likewise, making use of the identity cos
α
cos
β
≡
(1
/
2) [cos(
α
−
β
)+cos(
α
+
β
)], we can write the standing wave solution
ψ
(
x,t
) =
A
cos(
kx
) cos(
ωt
)
(3)
in the form
ψ
(
x,t
) = (
A/
2) cos(
kx
−
ωt
) + (
A/
2) cos(
kx
+
ωt
)
.
(4)
But, this is just the sum of two counterpropagating traveling wave solu
tions.
Finally, making use of the identities cos(
α
−
β
)
≡
cos
α
cos
β
+ sin
α
sin
β
and cos(
α
+
β
)
≡
cos
α
cos
β
−
sin
α
sin
β
, we can write the following super
position of traveling waves,
ψ
(
x,t
) =
A
cos(
kx
−
ωt
) +
AR
cos(
kx
+
ωt
)
,
(5)
in the form
ψ
(
x,t
)
=
A
cos(
kx
) cos(
ωt
) +
A
sin(
kx
) sin(
ωt
)
+
AR
cos(
kx
) cos(
ωt
)
−
AR
sin(
kx
) sin(
ωt
)
(6)
=
A
(1 +
R
) cos(
kx
) cos(
ωt
) +
A
(1
−
R
) sin(
kx
) sin(
ωt
)
.
Of course, this is just a superposition of two standing waves.
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2. If a transmission line of characteristic impedance
Z
carries a current
I
(
x,t
) =
I
i
cos(
kx
−
ωt
) +
I
r
cos(
kx
+
ωt
)
,
(7)
then the corresponding voltage is [see Eq. (7.104) in lecture notes]
V
(
x,t
) =
I
i
Z
cos(
kx
−
ωt
)
−
ZI
r
cos(
kx
+
ωt
)
.
(8)
If the line is open circuited at
x
= 0 then this implies that
I
(0
,t
) = 0 (since
current cannot flow through an infinite resistance). Hence, it follows from
(7) that
I
(0
,t
) =
I
i
cos(
ωt
) +
I
r
cos(
ωt
) = 0
,
(9)
which implies that
I
r
=
−
I
i
.
(10)
Hence, the current takes the form
I
(
x,t
) =
I
i
[cos(
kx
−
ωt
)
−
cos(
kx
+
ωt
)] = 2
I
i
sin(
kx
) sin(
ωt
)
,
(11)
where use has been made of a trigonometric identity. Likewise, the voltage
is written
V
(
x,t
) =
I
i
Z
[cos(
kx
−
ωt
) + cos(
kx
+
ωt
)] = 2
I
i
Z
cos(
kx
) cos(
ωt
)
.
(12)
Note that the current at a given point on the line oscillates as sin(
ωt
)
whereas the corresponding voltage oscillates as cos(
ωt
) = sin(
ωt
+
π/
2).
In other words, the current and the voltage oscillations are everywhere
π/
2
radians out of phase. The energy flux down the line is
I
=
IV
. Hence, the
mean energy flux is
(I)
= 4
I
2
i
Z
sin(
kx
) cos(
kx
)
(
sin(
ωt
) cos(
ωt
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 Spring '10
 Fitzpatrick
 Work, Wavelength, Cos, Standing wave, wave solutions

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