Dynamics cheat sheet

where ∆ is very small positive quantity. and we also have
Multiplying Eq (2) and (3) with each others gives
Going back to Eq (1A) and rewriting it as
ω + ϖ ≈ 2ϖ (3)
ω 2 − ϖ 2 = 4∆ϖ (4)
( v0
u (t) = u 0 cos ωt +
ω − u ωϖ
)
ω 2
st sin ωt + u st sin ϖt
4∆ϖ
4∆ϖ
( v0
= u 0 cos ωt +
ω − u ω
)
ω 2
st sin ωt + u st sin ϖt
4∆
4∆ϖ
Since ϖ ≈ ω the above becomes
( v0
u (t) = u 0 cos ωt +
ω − u ω
)
ω
st sin ωt + u st sin ϖt
4∆
4∆
= u 0 cos ωt + v 0
ω sin ωt + u ω
st (sin ϖt − sin ωt)
4∆
now 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 + v ( ( )
0
ω sin ωt + u ω ϖ − ω
st sin t cos
2∆ 2
From Eq(2) ϖ − ω = −2∆ and ω + ϖ ≈ 2ϖ hence the above becomes
or since ϖ ≈ ω
Now lim ∆→0
sin(∆t)
∆
This can also be written as
( ϖ + ω
u (t) = u 0 cos ωt + v 0
ω sin ωt + u ω
st (sin (−∆t) cos (ϖt))
2∆
u (t) = u 0 cos ωt + v 0
ω sin ωt − u ω
st (sin (∆t) cos (ωt))
2∆
= t hence the above becomes
u (t) = u 0 cos ωt + v 0
ω sin ωt − u ωt
st cos (ωt)
2
2
))
t
u (t) = u 0 cos ϖt + v 0
ϖ sin ϖt − u ϖt
st cos (ϖt) (1)
( )
2
π
= u 0 cos t + v ( ) ( ) ( )
0 π π π
t 1 ϖ sin t − u st t cos t
t 1 2t 1 t 1
since ϖ ≈ ω in this case. This is the solution to use for resonance and for t ≤ t 1
Hence for t > t 1 , the above equations is used to determine initial conditions at t = t 1
u (t 1 ) = u 0 cos ϖt 1 + v 0
ϖ sin ϖt 1 − u st
ϖt 1
2 cos (ϖt 1)
but cos ϖt 1 = cos π t 1
t 1 = −1 and sin ϖt 1 = 0 and ϖt 1
2
= π 2
, hence the above becomes
Taking derivative of Eq (1) gives
u (t 1 ) = −u 0 + u st
π
2
ϖ 2 t
˙u (t) = −ϖu 0 sin ϖt + v 0 cos ϖt + u st
2 sin (ϖt) − u ϖ
st cos (ϖt)
2
45