Appendix: All Solutions to the Problems
903
⎧
⎨
⎩
3
4
λl
2
−
1
2
π
2
l
a 1 −
8λl
2
9π 2 a 2 = 0
−
8λl
2
9π 2 a 1 +
3
4
λl
2
−
2π
2
l
a 2 = 0
Multiplying the various terms by 36π
2 l, we get
(27π
2
λl
3
− 18π
4
)a 1 − 32λl
3 a 2 = 0
− 32λl
3 a 1 + (27π
2
λl
3
− 72π
4
)a 2 = 0
The sufficient and necessary conditions that the homogeneous linear equations
have non-null solution are
27π
2
λl
3
− 18π
4
−32λl
3
−32λl
3
27π
2
λl
3
− 72π
4
= 0
Expanding the determinant and arranging, we get
(729π
4
− 1024)l
6
λ
2
− 2430π
6 l
3
λ + 1296π
8
= 0
This is a quadratic equation with one unknown, the solution is
λ =
1215π
6
± π
4
(1215π 2 ) 2 − 1296(729π 4 − 1024)
(729π 4 − 1024)l 3
Solving for two characteristic values
λ 1 =
6.548458
l 3
, λ 2 =
26.831570
l 3
Therefore the vibration frequency of the string can be obtained
ω 1 = a
λ 1 =
2.558995a
√
l
l 2
, ω 2 = a
λ 2 =
5.179920a
√
l
l 2
Substituting the two characteristic values into the homogeneous linear equations,
we get
a 1 =
32λ i l
3
27π 2 λ i l 3 − 18π 4 a 2 (i = 1, 2)
The corresponding characteristic function is
v 2 =
32λ i l
3
27π 2 λ i l 3 − 18π 4 sin
π
l
x + sin
2π
l
x
a 2 (i = 1, 2)