Dynamics cheat sheet

and at t = t 1
ϖ 2 t 1
ϖ
˙u (t 1 ) = −ϖu 0 sin ϖt 1 + v 0 cos ϖt 1 + u st sin (ϖt 1 ) − u st
2
2 cos (ϖt 1)
ϖ
= −v 0 + u st
2
Now the solution for t > t 1 is
u (t) = u (t 1 ) cos ωt + ˙u (t 1)
ω
(
π
= −u (0) + u st
2
under-damped with sin impulse c < c r , ξ < 1
sin ωt
)
cos ωt + −u′ (0) + u st
π
2t 1
ω
⎧
⎨ F 0 sin (ϖ) 0 ≤ t ≤ t 1
mü + c ˙u + ku =
⎩
0 t > t 1
sin ωt
or
⎧
⎨
ü + 2ξω ˙u + ω 2 F 0 sin (ϖ) 0 ≤ t ≤ t 1
u =
⎩
0 t > t 1
mü + c ˙u + ku = F sin ϖt
ü + 2ξω ˙u + ω 2 u = F sin ϖt
m
For t ≤ t 1 Initial conditions are u (0) = u 0 and ˙u (0) = v 0 and u st = F k
u (t) = e −ξωt (A cos ω d t + B sin ω d t) +
then the solution from above is
u st
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ) (1)
Applying initial conditions gives
For t > t 1 . From (1)
A = u 0 +
B = v 0
ω d
+ u 0ξω
ω d
+
u st
√(1 − r 2 ) 2 + (2ξr) 2 sin θ
u st
u (t 1 ) = e −ξωt 1
(A cos ω d t 1 + B sin ω d t 1 ) +
ω d
√
(1 − r 2 ) 2 + (2ξr) 2 (ξω sin θ − ϖ cos θ)
u st
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt 1 − θ) (2)
Taking derivative of (1) gives
˙u (t) = −ξωe −ξωt (A cos ω d t + B sin ω d t) + e −ξωt (−Aω d sin ω d t + ω d B cos ω d t)
u st
+ ϖ
cos (ϖt − θ)
√(1 − r 2 ) 2 + (2ξr) 2
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