AMATH 351
Assignment No. 2
Fall 2005
Due: Friday, September 30, 2005
1. Recall that in class, we studied the clamped string problem to arrive at the PDE,
∂
2
y
∂t
2
=
T
ρ
∂
2
y
∂x
2
.
(1)
Here
y
(
x, t
) denotes the displacement of the string from equilibrium. Using separation of variables
and the boundary conditions
y
(0
, t
) =
y
(
L, t
) = 0, we arrived at the following solutions
y
n
(
x, t
) =
u
n
(
x
)cos(
ω
n
t
)
,
n
= 1
,
2
,
3
, . . . ,
(2)
where
u
n
(
x
) = sin
p
nπx
L
P
,
ω
n
=
nπ
L
r
T
ρ
.
(3)
Show that a linear combination of the above solutions, i.e.,
N
s
n
=1
c
n
y
n
(
x, t
)
,
(4)
is also a solution to the PDE in Eq. (1).
2. Now consider the PDE model in Eq. (1), but for a string that is clamped only at one end, i.e.
y
(0
, t
) = 0. (Once again, assume that the tension
T
and lineal density
ρ
are constant throughout
the string.) Assume that the other boundary condition is given by
∂y
∂x
(
L, t
) = 0. (This would also
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 Spring '08
 SivabalSivaloganathan
 Boundary value problem, Ωn, Wronskian method

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