Page 115 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

102 Transient Vibration Chap. 4

-Fr Figure 4.4-3.
The response to the second step function started at i is
^ = - [1 - coso)„(i - t^)] (4.4-5)

and by adding, the response in the second interval t > becomes
kx
^ = {[1 - cos«„i] - [1 - cos W„(i - i,)]}
= -cos (Oj + cos - ti) t > ti (4.4-6)
Half-sine pulse. For a pulse of time duration ij, the excitation is
Tit
F{t) = Fqsin — f ort <

h
= 0 for t > (4.4-7)
and the differential equation of motion is

X + o)tx sin TTt/t^ t < t, (4.4-8)
m
ution
The general solution is the sum of the free vibration and the particular solution
, . ^ sin pi
x(t) = Asmco^t -h B cos H------- =------j (4.4-9)
-
^ 0)^- p
where p = ir/t^. To satisfy the initial conditions x(0) = i(0) = 0, we find
P_
Fn
B = 0 and A = - ( ^ r
1

and the previous solution reduces to
_ £_

sm + sin pt
( f ) =
\^n I I "n
In t / \ . 77/
r 2ij T \ T ] il i < it (4.4-10)
2ij T