2.4 Time Domain Analysis
21
∼ 1 −
1 − ζ ω ωt + O
ω ωt
2
1 + ζ ω ωt + O
ω d t
2
∼ O
ω d t
2 .
The displacement and velocity x((t) ∼ O
ω ωt
2 and ˙
x((t) ∼ (q qt)/m +
O
ω ωt
2 provide the input for the free vibration solution for t > t 2 under
the assumption ω ωt, ω d t 1. The reader is encouraged to construct these
approximations by using the above approximations of exp
− ζ ω ωt
, cos
ω d t
,
and sin
ω d t
.
The latter result has an important physical meaning. It represents the displacement of an oscillator following an impulse of duration t and intensity q. This
interpretation provides the ingredient for constructing particular solutions for SDOF
systems subjected to arbitrary forcing functions f (t). The construction involves the
following three steps.
– Step 1: Discretize the support of the forcing function f (t) in small time intervals
u such that ω ωu, ω d u 1 and represent f (t) by a piecewise constant
function equal to f (u) in the time interval [u, u + u), as illustrated in Fig. 2.7,
i.e., the forcing function f (t) for t in [u, u + u) is approximated by its value
f (u) at the left end of this time interval.
– Step 2: The free vibration solution of the oscillator following the forcing function
which is constant and equal to f (u) in the time interval (u, u + u) and zero
otherwise results from the last expression in Eq. 2.28 and has the form
x(t; u) ∼ e
−ζ ω (t−u−u)
O(ω d u)
2 cos
ω d (t − u − u)
+
f (u) )u/m + O(ω d u) 2
ω d
sin
ω d (t − u − u)
∼ e
−ζ ω (t−u−u) f (u) )u/m
ω d
sin
ω d (t − u − u)
+ O(ω d u)
2
under our assumption of small u.
– Step 3: Add the contributions to the oscillator displacement caused by the forcing
function prior to the time t, i.e.,
x(t)
u+u≤t
x(t; u) =
u+u≤t
e
−ζ ω (t−u−u) f (u)
m ω d
sin
ω d (t−u−u)
u,
and take the limit as u → 0, which gives
x(t) =
1
m ω d
t
0
e
−ζ ω (t−u) sin
ω d (t − u)
f (u) du.
(2.29)