Matakuliah
Tahun
: K0124 / Matematika Teknik II
: 2006/2007
PERTEMUAN 17 & 18
VIBRATING STRING EQUATION
1
To solve this boundy-value problem , let y = XT as
usual.
Then XT "  a 2 X "T or T " a 2T  X " X
Calling the separation constant
 2
, we have
T "  2 a 2T  0, X "  2 X  0
2
and
T  A1 sin at  B1 cos at,  A2 sin ax  B2 cos ax A1 s
A solution is thus given by
y( x, t )  XT   A2 sin x  B2 cos x A1 sin t  B1 cos t 
From y(0, t )  0, A2  0 ,
3
then
y( x, t )  B2 sin x A1 sin t  B1 cos t   sin   Asin t  B cos t 
From y ( L, t )  0 , we have sin L Asin t  B cos t   0
4
So that sin L  0, L  m or  m L since the
second factor must not be equal to zero. Now
yt ( x, t )  sin x Aa cos t  Ba sin at 
And yt ( x,0)  sin x a   0 from which
A0
.
Thus
m x
m at
y( x, t )  B sin
cos
L
L
5
PERTEMUAN 18
VIBRATING STRING EQUATION
(LANJUTAN)
6
To solve this boundy-value problem , let y = XT as
usual.
Then XT "  a 2 X "T or T " a 2T  X " X
To satisfy the condition y ( x,0)  f ( x) , it will be
necessary to superimpose solutions.
7
This yields

y ( x, t )   Bm sin
m 1
m x
m at
cos 
L
L
Then

m x
y ( x,0)  f ( x)   Bm sin
L
m 1
8
And from the theory of Fourier series,
m x
2 L
Bm   f ( x) sin
dx
L 0
L
The result is

m x  m x
m at
2 L
y ( x, t )     f ( x) sin
dx  sin
cos
0
L
L
L

m 1  L
which can be verified as the solution.
9
TERIMA KASIH
10