Page 128 - Advanced engineering mathematics

108 CHAPTER 3 The Laplace Transform

By Newton’s second law,

m 1 x =−(k 1 + k 2 )x 1 + k 2 x 2 + f 1 (t)
1
and

m 2 x (t) = k 2 x 1 − (k 2 + k 3 )x 2 + f 2 (t),
2
where f 1 (t) and f 2 (t) are forcing functions. We have assumed here that damping is negligible.
As a specific example, let
m 1 = m 2 = 1,k 1 = k 3 = 4,k 2 = 5/2.
Also suppose f 2 (t) = 0 and f 1 (t) = 2(1 − H(t − 3)). This acts on the first mass with a force of
constant magnitude 2 for the first three seconds, then turns off. Now the system is
13 5

x =− x 1 + x 2 + 2[1 − H(t − 3)]
1
2 2
5 13

x = x 1 − x 2 .
2
2 2
Suppose the masses are initially at rest in the equilibrium position:

x 1 (0) = x 2 (0) = x (0) = x (0) = 0.

1
2
Apply the transform to the system to obtain
13 5 2(1 − e −3s )
2
s X 1 =− X 1 + X 2 +
2 2 s
5 13
2
s X 2 = X 1 − X 2 .
2 2
Solve these to obtain

2 13 1
2
X 1 (s) = s + (1 − e −3s )
2
2
(s + 9)(s + 4) 2 s
13 1 1 s
= −
36 s 4 s + 4
2
1 s 13 1
− − e −3s
2
9 s + 9 36 s
1 s 1 s
+ e −3s + e −3s
2
2
4 s + 4 9 s + 9
and
5 1 1 s
X 2 (s) = −
36 s 4 s + 4
2
1 s 5 1
+ − e −3s
2
9 s + 9 36 s
1 s −3s 1 s −3s
+ e − e .
2
2
4 s + 4 9 s + 9
Apply the inverse transform to obtain
13 1 1
x 1 (t) = − cos(2t) − cos(3t)
36 4 9
13 1 1

+ − + cos(2(t − 3)) − cos(3(t − 3)) H(t − 3)
36 4 9
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October 14, 2010 14:14 THM/NEIL Page-108 27410_03_ch03_p77-120