2.4 Time Domain Analysis
19
t 1
t 2
f (t)
q
t
0
1
2
3
4
5
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
t
x(t)
Fig. 2.6 Forcing function f (t) = q in (t 1 , t 2 ) and zero outside this time interval (left panel) and
displacement x(t) for x st = 1, ω = π , ζ = 0.05, t 1 = 1.5, and t 2 = 1.8 (right panel)
The oscillator remains at rest till t 1 since it has zero ICs and is subjected to no
action. It exhibits forced and free vibrations for times t in (t 1 , t 2 ) and times t ≥ t 2
larger than t 2 . The general solution in (t 1 , t 2 ) is the sum of the general homogeneous
solution and a particular solution, i.e.,
x(s) = e
−ζ ω s
A cos(ω d s) + B sin(ω d s)
+ x p (s), 0 < s < t 2 − t 1 ,
where s = t − t 1 > 0 denotes a local time which is used for convenience and
x p (s) = x st = q/k from Example 2.4. The ICs x(0) = 0 and ˙
x(0) = 0 imply
x(0) = 0 ⇒ A + x st = 0 and
˙
x(0) = 0 ⇒ −ζ ω A + B ω d = 0
so that A = −x st , B = −ζ ω x st /ω d = −ζ q/
k
1 − ζ 2
, and
x(s) = x st
1 − e
−ζ ω s
cos(ω d s) +
ζ
1 − ζ 2
sin(ω d s)
, 0 < s < t 2 − t 1 .
(2.26)
The oscillator displacement and velocity at the end of the time interval (t 1 , t 2 ), i.e.,
x((t) and ˙
x((t), result from the above equation with s = t = t 2 − t 1 . The free
vibration solution which occurs for times t ≥ t 2 results from Eq. 2.21 and has the
form
x(t) = e
−ζ ω u
x((t) cos(ω d u) +
˙
x((t) + ζ ω x((t)
ω d
sin(ω d u)
,
(2.27)
where u = t − t 2 ≥ 0 denotes the time measured from the end of the time interval
(t 1 , t 2 ). In summary, the oscillator displacement has the expression