Dynamics cheat sheet

where r = ϖ ω = π/t 1
ω
= T
2t 1
where T is the natural period of the system. u st = F 0
k
, hence the above becomes
⎛
⎜
u (t) = u 0 cos ωt + ⎝ v 0
ω − F 0
k
When u 0 = 0 and v 0 = 0 then
u (t) = − F 0
k
u (t) = F 0
k
( )
π/t1
ω
1 −
(
π/t1
ω
1
1 −
(
π/t1
ω
( )
π/t1
ω
1 −
(
π/t1
ω
) 2
sin ωt + F 0
k
) 2
(sin
) 2
⎞
⎟
⎠ sin ωt + F 0
k
1
1 −
(
π/t1
ω
1
1 −
(
π/t1
ω
(
π t
t 1
)
− π/t 1
ω sin ωt )
(
) 2
sin π t )
t 1
The above Eq (1) gives solution during the time 0 ≤ t ≤ t 1
Now after t = t 1 the force will disappear, the differential equation becomes
mü + ku = 0 t > t 1
(
) 2
sin π t )
t 1
but with the initial conditions evaluate at t = t 1 . From (1)
( )
v0
u (t 1 ) = u 0 cos ωt 1 +
ω − u r
1
st
1 − r 2 sin ωt 1 + u st
1 − r 2 sin ϖt 1
( )
v0
= u 0 cos ωt 1 +
ω − u r r
st
1 − r 2 u st
1 − r 2 sin ωt 1 (2)
since sin ϖt 1 = 0. taking derivative of Eq (1)
˙u (t) = −ωu 0 sin ωt + ω
and at t = t 1 the above becomes
˙u (t 1 ) = −ωu 0 sin ωt 1 + ω
= −ωu 0 sin ωt 1 + ω
(
v0
)
ω − u r
st
1 − r 2 cos ωt + ϖ 1 cos ϖt
1 − r2 (
v0
)
ω − u r
st
1 − r 2 )
ω − u r
st
1 − r 2
(
v0
cos ωt 1 + ϖ 1
1 − r 2 cos ϖt 1
(1)
cos ωt 1 − ϖ 1
1 − r 2 (3)
since cos ϖt 1 = −1. Now (2) and (3) are used as initial conditions to solve mü + ku = 0 . The solution for
t > t 1 is
u (t) = u (t 1 ) cos ωt + ˙u (t 1)
ω
sin ωt
Resonance with undamped sin impulse When ϖ ≈ ω and t ≤ t 1 we obtain resonance since r → 1 in
the solution shown up and as written the solution can’t be used for analysis in this case. To obtain a solution
for resonance some calculus is needed. Eq (1) is written as
(
)
ϖ
v 0
u (t) = u 0 cos ωt +
ω − u ω
st
1 − ( 1
)
ϖ 2
sin ωt + u st
ω
1 − ( )
ϖ 2
sin ϖt
ω
( )
v0
= u 0 cos ωt +
ω − u ωϖ
ω 2
st
ω 2 − ϖ 2 sin ωt + u st
ω 2 sin ϖt
− ϖ2 (1A)
Now looking at case when ϖ ≈ ω but less than ω, hence let
ω − ϖ = 2∆ (2)
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