Solutions to Practice Midterm 2
Problem 1:
We have the two waves
y
1
(
x, t
) =
A
sin(
kx

ωt
)
y
2
(
x, t
) =

A
sin(
kx
+
ωt
) =
A
sin(
kx
+
ωt
+
π
)
(1) Wave 1 is traveling in the direction of
positive
x
, due to the
kx

ωt
term. Wave 2 is
traveling in the direction of
negative
x
, due to the
kx
+
ωt
term.
(2) The phase difference between the two waves is Δ
φ
=
φ
+
π
. We make the definitions
θ
1
=
kx

ωt
θ
2
=
kx
+
ωt
+
φ
+
π
(3) The resultant wave is a standing wave, given by
y
(
x, t
) =
y
1
(
x, t
) +
y
2
(
x, t
)
=
A
sin
θ
1
+
A
sin
θ
2
= 2
A
sin
θ
2
+
θ
1
2
¶
cos
θ
2

θ
1
2
¶
= 2
A
sin
kx
+
φ
2
+
π
2
¶
cos
ωt
+
φ
2
+
π
2
¶
=

2
A
cos
kx
+
φ
2
¶
sin
ωt
+
φ
2
¶
(4) The resultant wave has amplitude 2
A
.
(5) The maximum transverse position of an element of the string is the maximum value of
y
(
x, t
) =

2
A
cos
kx
+
φ
2
¶
sin
ωt
+
φ
2
¶
This value is
y
max
= 2
A
. This value is attained at (
x
n
, t
m
) for which
cos
kx
n
+
φ
2
¶
= 1
,
x
n
=
1
k
2
nπ

φ
2
¶
,
n
= 0
,
1
,
2
,
.....
sin
ωt
m
+
φ
2
¶
=

1
,
t
m
=
(4
m
+ 3)
π

φ
2
ω
,
m
= 0
,
1
,
2
,
.....
or at (
x
n
, t
m
) for which
cos
kx
n
+
φ
2
¶
=

1
,
x
n
=
1
k
•
(2
n
+ 1)
π

φ
2
‚
,
n
= 0
,
1
,
2
,
.....
sin
ωt
m
+
φ
2
¶
= 1
,
t
m
=
(4
m
+ 1)
π

φ
2
ω
,
m
= 0
,
1
,
2
,
.....
1