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SOLUTIONS: ASSIGNMENT 13
FALL 2009
Section 19.2: 5. (a) The wire loop is treated as massless, so any force on it would cause
infinite acceleration: there can be no (net) force on the loop. A nonzero
u
x
(
L, t
) would
lead to a nonzero vertical force, due to the string tension. Of course, the horizontal tension
force is balanced by a force exerted on the loop by the wire.
(b)–(d) The solution for (b) is given in the text; (c) and (d) are similar, using respectively
halfrange cosine and quarterrange cosine expansions:
(
c
) :
y
(
x, t
) =
V t
(
d
) :
y
(
x, t
) =
∞
summationdisplay
n
=1
A
n
cos
nπx
2
L
sin
nπct
2
L
,
A
n
=
4
nπc
integraldisplay
L
0
g
(
x
) cos
nπx
2
L
dx.
Note that there could be additional terms involving cos(
nπct/L
) or cos(
nπct/
2
L
), but
these vanish due to the condition that the initial displacement vanish.
6. Substituting
y
(
x, t
) =
X
(
x
)
T
(
t
) leads to
X
′′
/X
= (
T
′′
+
aT
′
)
/
(
c
2
T
); as usual both sides
must be constant, leading to the SturmLiouville problem
X
′′
+
λX
= 0
,
X
(0) =
X
(
L
) = 0
,
with known solutions
λ
n
= (
nπ/L
)
2
,
X
n
(
x
) = sin(
nπx/L
) for
n
= 1
,
2
, . . .
. Then
T
′′
n
+
aT
′
n
+
parenleftBig
nπc
L
parenrightBig
T
N
= 0
.
This equation has characteristic polynomial
r
2
+
ar
+ (
nπc/L
) = 0 with (given the as
sumption 0
< a <
2
πc/L
) complex roots

μ
n
±
iν
n
, where
μ
n
=
a
2
,
ν
n
=
radicalBigg
parenleftbigg
2
nπ
L
parenrightbigg
2

a
2
,
and with solution
T
n
(
t
) =
e
−
μ
n
t
(
A
n
cos
ν
n
t
+
B
n
sin
ν
n
t
). Thus the soluiton is
y
(
x, t
) =
∞
summationdisplay
n
=1
sin(
nπx
L
e
−
μ
n
t
(
A
n
cos
ν
n
t
+
B
n
sin
ν
n
t
)
.
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 Fall '08
 Staff
 Math, Sin, Boundary value problem, initial condition, Boundary conditions

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