Dynamics cheat sheet

The following table gives the solutions for initial conditions are u (0) and u ′ (0) under all damping conditions.
The roots shown are the roots of the quadratic characteristic equation λ 2 + 2ζωλ + ω 2 λ = 0. Special handling is
needed to obtain the solution of the differential equation for the case of ζ = 0 and ϖ = ω as described in the
detailed section below.
⎧
⎨ −iω
roots
ζ = 0
⎩
⎧
⎪⎨
u (t)
⎪⎩
⎧
⎨
roots
⎩
+iω
ϖ = ω → u (0) cos ϖt + u′ (0)
ϖ
sin ϖt − F K
ϖ ≠ ω → u (0) cos ωt +
−ξω + iω n
√
1 − ξ 2
−ξω − iω n
√
1 − ξ 2
(
u ′ (0)
ω
− F K
ϖt
2
cos (ϖt)
)
r
sin ωt + F 1
1−r 2 K
sin ϖt
1−r 2
ζ < 1
u (t) = e −ξωt (A cos ω d t + B sin ω d t) + F K
A = u 0 + F K
1 √(1−r 2 ) 2 +(2ξr) 2 sin θ
1
sin (ϖt − θ)
√(1−r 2 ) 2 2
+(2ξr)
B = v 0
ω d
+ u 0ξω
ω d
⎧
⎨ −ω
roots
⎩
−ω
+ F √ 1
K
ω d (1−r 2 ) 2 +(2ξr)
2
(ξω sin θ − ϖ cos θ)
ζ = 1
ζ > 1
u (t) = (A + Bt) e −ωt + F K
A = u 0 + F K
1 √(1−r 2 ) 2 +(2r) 2 sin θ
1
sin (ϖt − θ)
√(1−r 2 ) 2 2
+(2r)
F/k
B = v 0 + u 0 ω +
(ω sin θ − ϖ cos θ)
√(1−r 2 ) 2 2
+(2r)
⎧
√
⎨ −ω n ξ + ω n ξ 2 − 1
roots
⎩
√
−ω n ξ − ω n ξ 2 − 1
(
−ξ+ √ (
ξ
u (t) = Ae
2 −1
)ω nt −ξ− √ )
ξ + Be 2 −1 ω nt +
F
K
β sin (ϖt − θ)
A = v 0+u 0 ωξ+u 0 ω √ ((
ξ 2 −1+ F K β ξ+ √ )
)
ξ 2 −1 ω sin θ−ϖ cos θ
2ω √ ξ 2 −1
B = − v 0+u 0 ωξ−u 0 ω √ ((
ξ 2 −1+ F K β ξ− √ )
)
ξ 2 −1 ω sin θ−ϖ cos θ
2ω √ ξ 2 −1
23