Physics 214 PS #5 Solutions
by Stephen Poprocki
2008–02–26
Problem 1.
a) A standing wave has the form
y
(
x, t
)=
A f
(
kx
)
cos
(
ωt
+
ϕ
)
,
where
f
is either cosine or sine. Since the string is fixed at both ends, then
y
(0
, t
)=0
and
y
(
L, t
)=
0
, at all times
t
. Thus
f
(0) =
f
(
kL
) = 0
. This cannot be satisfied if
f
=
cos, so we must have
f
=
sin and hence sin
(
kL
) = 0
kL
=
nπ
, for
n
= 1
,
2
,
3
,
. Since the possible
k
’s depend on
n
, we
write
k
n
=
nπ
/
L
, so
y
(
x, t
)=
A
sin
parenleftBig
nπ
L
x
parenrightBig
cos
(
ωt
+
ϕ
)
.
i. The nodes occur when
sin
parenleftBig
nπ
L
x
parenrightBig
=0
nπ
L
x
=
mπ,
m
=0
,
1
,
2
,
, n
Hence the first node (besides at
x
=0
(
m
=0
)) corresponds to
m
=1
and occurs at
x
=
L
n
,
n
=1
,
2
,
3
,
.
ii. The antinodes occur when
sin
parenleftBig
nπ
L
x
parenrightBig
=
±
1
nπ
L
x
=
mπ
2
,
m
=1
,
3
,
5
,
,
2
n
−
1
.
The first antinode corresponds to
m
=1
and occurs at
x
=
L
2
n
,
n
=1
,
2
,
3
,
.
b) Take the ideal free end to be at
x
=
L
. The node at
x
= 0
requires
f
(0) = 0
so again
f
=
sin, but
now the antinode at
x
=
L
requires
sin
(
kL
)=
±
1
k
n
=
nπ
2
L
,
n
=1
,
3
,
5
,
,
so
y
(
x, t
)=
A
sin
parenleftBig
nπ
2
L
x
parenrightBig
cos
(
ωt
+
ϕ
)
.
i. The first node (excluding the wall at
x
=0
) occurs when
nπ
2
L
x
=
π
x
=
2
L
n
,
n
=1
,
3
,
5
,
.
1