100
3 Forced Vibration of Single Degree of Freedom System
=
2 A
ω 0 τ
[ − cos ω 0 t]
τ/2
0 =
2 A
π
b n = 2/τ
τ
0
F (t) sin nω 0 t dt
= 2/τ
τ
0
A | sin ω 0 t | sin nω 0 t dt
= 0
a n =
2
τ
τ
0
A | sin ω 0 t | cos nω 0 t dt
=
4
τ
τ/2
0
A sin ω 0 t cos nω 0 t dt
=
2
τ
A
τ/2
0
[sin (n + 1) ω 0 t − sin (n − 1) ω 0 t ] dt
=
−2 A
τ
cos (n + 1) π
(n + 1) ω 0
−
cos (n − 1) π
(n − 1) ω 0
−
1
(n + 1) ω 0
+
1
(n − 1) ω 0
For odd values of n, n +1 and n–1, the above expressions are zero. For even values
of n, the value of a n is given by
a n = −
2 A
τ ω 0
−
1
n + 1
+
1
n − 1
−
1
n + 1
+
1
n − 1
= −
2 A
2π
4
n 2 − 1
=
− 4 A
π
1
n 2 − 1
Therefore,
F (t) = a 0 +
∞
n=2, 4 ···
a n cos nω 0 t
(3.97)
or
F (t) =
2 A
π
+
∞
n =2,4
−
4 A
π
1
n 2 − 1
cos nω 0 t