Physics 214 Spring 2008 Solutions to Assignment
]
7
Problem 1.
(
a
). Given that
ω
1

ω
2
=
0
.
1
2
(
ω
1
+
ω
2
), we have
ω
2
=
19
21
ω
1
. Therefore,
s
(0
,t
) =
s
m
cos(
ω
1
t
) +
s
m
cos(
ω
2
t
) =
s
m
cos(
ω
1
t
) +
s
m
cos(
19
21
ω
1
t
).
0.02
0.04
0.06
0.08
t
L
s
R
M
2
M
1
1
2
s
L
cm
R
Figure 1:
Plot of
s
(0
,t
)
versus
t
for
ω
1
= 1000
π rad/s
and
s
m
= 1
cm
.
(
b
). With
k
2
=
ω
2
v
=
19
21
ω
1
v
=
19
21
k
1
, we have
s
(
x,
0) =
s
m
cos(
k
1
t
) +
s
m
cos(
k
2
t
) =
s
m
cos(
k
1
t
) +
s
m
cos(
19
21
k
1
t
).
5
10
15
20
25
x
L
m
R
M
2
M
1
1
2
s
L
cm
R
Figure 2:
Plot of
s
(
x,
0)
versus
x
for
ω
1
= 1000
π rad/s
which gives
k
1
=
ω
1
v
≈
1000
π
343
m

1
for sound
wave in air with
v
≈
343
m/s
and
s
m
= 1
cm
.
(
c
). Yes. Because the wavelengths of the two sound waves are slightly diﬀerent,
there will be points where they are out of phase and destructively interfere as we
see in ﬁgure 2.
Problem 2.
Given
y
(
x,t
) =
A
cos(
kx

ωt
), we have
dy
(
x,t
)
dx
=

kA
sin(
kx

ωt
) and
dy
(
x,t
)
dt
=
ωA
sin(
kx

ωt
) which give
u
p
(
x,t
) =
1
2
τ
(
dy
(
x,t
)
dx
)
2
=
1
2
τk
2
A
2
sin
2
(
kx

ωt
) and
u
k
(
x,t
) =
1
2
μ
(
dy
(
x,t
)
dx
)
2
=
1
2
μω
2
A
2
sin
2
(
kx

ωt
). Therefore,
u
p
(
x,t
) =
u
k
(
x,t
),
since
τk
2
=
μω
2
or
v
=
q
τ
μ
=
ω
k
. The total energy per unit length at time
t
is
then
u
(
x,t
) = 2
u
p
(
x,t
) = 2
u
k
(
x,t
).
Problem 3.
No, the energies do not add up, since energy is not linear in
y
. The displacement
waves add up.
(
a
). The energy density of
y
= 2
f
1
(
x

vt
) is 4 times that of 2
f
1
(
x

vt
), since
dy
dx
= 2
df
1
dx
and
dy
dt
= 2
df
1
dt
and squaring these in calculating the potential and the
kinetic energy densities (and consequently the total energy density) gives a factor
of 4, and the total energy density in