Dynamics cheat sheet

and
leading to the solution where tan θ = 2ξr
1−r 2
p 2 = − c
√ ( c
) 2
2m − k −
2m m
= −ωξ − ω √ ξ 2 − 1
and
is
p 1 = − c
√ ( c
2m + 2m
p 2 = − c
2m − √ ( c
2m
) 2 k −
m = −ωξ + ω √
n ξ 2 − 1
) 2 k −
m = −ωξ − ω √
n ξ 2 − 1
(
−ξ+ √ (
ξ
u (t) = Ae
2 −1
)ωt −ξ− √ )
ξ + Be 2 −1 ωt F + β sin (ϖt − θ)
k
u ′ (0) + u (0) ωξ + u (0) ω √ ξ 2 − 1 + F k
((ξ β + √ )
ξ 2 − 1
A =
2ω √ ξ 2 − 1
u ′ (0) + u (0) ωξ − u (0) ω √ ξ 2 − 1 + F k
((ξ β − √ )
ξ 2 − 1
B = −
2ω √ ξ 2 − 1
1
β = √
(1 − r 2 ) 2 + (2ξr) 2
)
ω sin θ − ϖ cos θ
)
ω sin θ − ϖ cos θ
For t > t 1 . From Eq(1) and at t = t 1
Taking derivative of Eq (1)
(
−ξ+ √ )
(
ξ
u (t 1 ) = Ae
2 −1 ωt 1 −ξ− √ )
ξ
+ Be
2 −1 ωt 1
+ D sin (ϖt 1 − θ) (2)
(
−ξ+ √ (
ξ
˙u (t) = ωAe
2 −1
)ωt −ξ− √ )
ξ + ωBe 2 −1 ωt + ϖD cos (ϖt − θ)
At t = t 1
−ξ+
˙u (t 1 ) = ωAe( √ )
(
ξ 2 −1 ωt 1 −ξ− √ )
ξ
+ ωBe
2 −1 ωt 1
+ ϖD cos (ϖt 1 − θ) (3)
Equation of motion now is
which has solution for over-damped given by
ü + 2ξω ˙u + ω 2 u = 0
(
−ξ+ √ (
ξ
u (t) = Ae
2 −1
)ω nt −ξ− √ )
ξ + Be 2 −1 ω nt
where
˙u (t 1 ) + u (t 1 ) ω n
(ξ − √ )
ξ 2 − 1
A = −
√
2ω n ξ 2 − 1
˙u (t 1 ) + u (t 1 ) ξω n
(ξ + √ )
ξ 2 − 1
B =
√
2ω n ξ 2 − 1
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