15.79:
a)
.
2
1
)
2
1
(
2
2
k
∂
∂
=
∆
∆
=
∆
∆
=
t
y
μ
μ
m
mv
x
K
u
y
b)
so
and
)
sin(
ωt
kx
ωA
t
y

=
∂
∂
).
(
sin
2
1
2
2
2
ωt
kx
A
μω
u
k

=
c)
The piece has width
,
height
and
y
x
x
x
∂
∂
∆
∆
and so the length of the piece is
2
1
2
2
1
2
2
1
)
(
∂
∂
+
∆
=
∂
∂
∆
+
∆
x
y
x
x
y
x
x
,
2
1
1
2
∂
∂
+
∆
≈
x
y
x
where the relation given in the hint has been used.
d)
[
]
.
2
1
)
(
1
2
2
2
1
p
∂
∂
=
∆
∆

+
∆
=
∂
∂
x
y
F
x
x
x
F
u
x
y
e)
),
sin(
ωt
kx
kA
x
y


=
∂
∂
and so
)
(
sin
2
1
2
2
2
p
ωt
kx
A
Fk
u

=
and f) comparison with the result of part (c)with
,
2
2
2
2
F
μ
ω
v
ω
k
=
=
shows that for a
sinusoidal wave
.
p
k
uv
u
=
g) In this graph,
p
k
and
u
u
coincide, as shown in part
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