Dynamics cheat sheet

When ϖ ≈ ω but less than ω, letting
where ∆ is very small positive quantity. And since ϖ ≈ ω let
Multiplying Eq (2) and (3) gives
Eq (1A) can now be written in terms of Eqs (2,3) as
( u ′ (0)
u (t) = u (0) cos ωt +
ω
(
v0
= u (0) cos ωt +
ω − F k
ω − ϖ = 2∆ (2)
ω + ϖ ≈ 2ϖ (3)
ω 2 − ϖ 2 = 4∆ϖ (4)
− F ωϖ
k 4∆ϖ
ω
4∆
Since ϖ ≈ ω the above becomes
( u ′ (0)
u (t) = u (0) cos ωt +
ω
− F k
= u (0) cos ωt + u′ (0)
ω
sin ωt + F k
)
sin ωt + F k
)
sin ωt + F k
)
ω
sin ωt + F 4∆ k
Using sin ϖt − sin ωt = 2 sin ( ϖ−ω
2
t ) cos ( ϖ+ω
2
t ) the above becomes
u (t) = u (0) cos ωt + u′ (0)
ω
From Eqs (2,3) the above can be written as
sin ωt + F k
u (t) = u (0) cos ωt + u′ (0)
ω
ω
2∆
sin ωt + F k
(
sin
ω 2
sin ϖt
4∆ϖ
ω 2
sin ϖt
4∆ϖ
ω
sin ϖt
4∆
ω
(sin ϖt − sin ωt)
4∆
( ϖ − ω
2
) ( )) ϖ + ω
t cos t
2
ω
(sin (−∆t) cos (ϖt))
2∆
Since lim ∆→0
sin(∆t)
∆
= t the above becomes
This is the solution to use for resonance.
u (t) = u (0) cos ωt + u′ (0)
ω
sin ωt − F k
ωt
2
cos (ωt)
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