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PHYS 2214  Fall 2008
PS7 Solutions
PHYS 2214: Waves, Optics, and Particles — Fall 2008
Problem Set 7 – Solutions
October 22, 2008
c
±
Joe P. Chen
1. (a) Recall that for a sound wave travelling in the
²
x
direction,
s
(
x, t
) =
f
∓
(
x
∓
vt
), the ﬁrst partial
derivatives of
s
(
x, t
) are related via
∂s
(
x, t
)
∂t
=
∓
v
∂s
(
x, t
)
∂x
We’ll refer to this as the “pulse equation.” So for a rightmoving wave, we can rewrite the
instantaneous power per unit crosssectional area along +
x
as
I
inst
(
x, t
) =

B
∂s
∂x
∂s
∂t
=

1
2
B
∂s
∂x
∂s
∂t

1
2
B
∂s
∂x
∂s
∂t
=

1
2
B
±

1
v
∂s
∂t
²
∂s
∂t

1
2
B
∂s
∂x
±

v
∂s
∂x
²
=
1
2
B
v
±
∂s
∂t
²
2
+
1
2
Bv
±
∂s
∂x
²
2
=
v
"
1
2
B
v
2
±
∂s
∂t
²
2
+
1
2
B
±
∂s
∂x
²
2
#
=
v
"
1
2
ρ
±
∂s
∂t
²
2
+
1
2
B
±
∂s
∂x
²
2
#
=
v
[
u
k
(
x, t
) +
u
p
(
x, t
)] =
vu
(
x, t
)
In the penultimate line we used the fact that the sound speed
v
=
p
B/ρ
. Similarly we can show
that for a leftmoving wave
s
(
x, t
) =
s
(
x
+
vt
), the instantaneous power per crosssectional area
along +
x
satisﬁes
I
inst
(
x, t
) =

B
∂s
∂x
∂s
∂t
=
···
=

vu
(
x, t
)
Figure 1: Setup of the continuity equation for sound.
We can interpret this result by considering a small volume of the ﬂuid which ranges from
x
to
x
+ Δ
x
and spans a small crosssectional area Δ
A
(Fig. 1). Without loss of generality, assume
that sound is travelling in the +
x
direction. First, we repeat the derivation of the continuity
equation done in Lecture 14 (10/16/08): At time
t
, the total energy
E
(
t
) contained in this small
volume is
E
(
t
) =
X
x
≤
x
0
≤
x
+Δ
x
u
(
x
0
, t
)Δ
A
Δ
x
At a slightly later time
t
+ Δ
t
, the total energy contained in this volume is
E
(
t
+ Δ
t
) =
X
x
≤
x
0
≤
x
+Δ
x
u
(
x
0
, t
+ Δ
t
)Δ
A
Δ
x
1
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View Full DocumentPHYS 2214  Fall 2008
PS7 Solutions
By local energy conservation, the change in energy contained in this volume
E
(
t
+ Δ
t
)

E
(
t
)
must be equal to the energy ﬂow into the volume between times
t
and
t
+ Δ
t
. There are two
contributions to the energy ﬂow: 1) Net energy ﬂow through Δ
A
at
x
into
the volume, which
is just the (instantaneous power per unit crosssectional area along +
x
) times Δ
A
times Δ
t
, or
I
inst
(
x, t
)Δ
A
Δ
t
; 2) Net energy ﬂow through Δ
A
at
x
+ Δ
x
out of
the volume, which is equal to

I
inst
(
x
+ Δ
x, t
)Δ
A
Δ
t
. Putting it altogether we have
E
(
t
+ Δ
t
)

E
(
t
) =
I
inst
(
x, t
)Δ
A
Δ
t

I
inst
(
x
+ Δ
x, t
)Δ
A
Δ
t
or
X
x
≤
x
0
≤
x
+Δ
x
[
u
(
x
0
, t
+ Δ
t
)

u
(
x
0
, t
)] Δ
A
Δ
x
= [
I
inst
(
x, t
)

I
inst
(
x
+ Δ
x, t
)] Δ
A
Δ
t
If we divide both sides of the equation by Δ
x
Δ
t
Δ
A
:
X
x
≤
x
0
≤
x
+Δ
x
[
u
(
x
0
, t
+ Δ
t
)

u
(
x
0
, t
)]
Δ
t
=
[
I
inst
(
x, t
)

I
inst
(
x
+ Δ
x, t
)]
Δ
x
and take the limit Δ
x,
Δ
t
→
0, we obtain the continuity equation:
∂u
(
x, t
)
∂t
=

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