Fourier Series and Partial Differential Equations Lecture Notes

Chapter 4. The wave equation 39
Thus
an = 2
L
nπx
f(x)sin dx, (4.55)
L 0 L
L 2
nπx
bn = g(x)sin dx. (4.56)
nπc L
Example 4.3 (Guitar or lute) For the special case (4.42),
and
an = 2
L .2h
L
2
L
0
= 8h
π2 sin
n2 Hence the solution is
y(x,t) = 8h
π 2
= 8h
π 2
0
nπx
xsin dx+
L
2
L .2h
L
nπx
(L−x)sin dx, (4.57)
L L L
2
nπ
, (4.58)
2
bn = 2
L
nπx
0.sin dx = 0. (4.59)
nπc L
∞
n=1
1
sin
n2
1
sin
12 0
nπ
nπx
nπct
sin cos ,
L L L
(4.60)
πx
πct
cos −
L L
1
3πx 3πct
sin cos
32 L L
+
(4.61)
1
5πx 5πct
sin cos −... .
52 L L
(4.62)
Example 4.4 (Piano) The initial transverse displacement is zero and the section [l1,l2]
is given an initial transverse velocity v. Here f(x) = 0 for 0 ≤ x ≤ L, and
0 for 0 ≤ x < L1 and L2 < x ≤ L,
g(x) =
(4.63)
v for L1 ≤ x ≤ L2.
Thus an = 0 and
bn = 2
L2
vsin
nπc L1
The transverse displacement is
y(x,t) = 2vL
π 2 c
∞
n=1
nπx
dx =
L
2vL
n2π2
cos
c
1
n2
cos
nπL1
−cos
L
nπL1
−cos
L
nπL2
sin
L
4.5 Normal modes for a weighted string
nπx
sin
L
nπL2
. (4.64)
L
nπct
L
. (4.65)
A string of length 2L has fixed ends and a mass m is attached to the mid-point; find the
normal frequencies of vibration.
Figure 10