Dynamics cheat sheet

Since t 1 is very small, it can be assumed that u changes is negligible as well as the change in velocity, hence the
above reduces to the same result as in the case of undamped. Therefore, the system is solved as free system, but
with initial velocity u ′ (0) = F 0 /m and zero initial position.
Initial conditions are u (0) = 0 and u ′ (0) = 0 then the solution is
applying initial conditions gives A = 0 and B =
u impulse = e −ξωt (A cos ω d t + B sin ω d t)
( F0
)
m
ω d
, hence
u impulse (t) = e −ξωt (
F0
mω d
sin ω d t
If the initial conditions were not zero, then ( the solution for these are added ) to the above. From earlier, it
was found that the solution is u (t) = e −ξωt u(0) cos ω d t + u′ (0)+u(0)ξω
ω d
sin ω d t , therefore, the full solution is
due to IC only
{ (
}} ){
{ ( }} ){
u (t) = e −ξωt u(0) cos ω d t + u′ (0) + u(0)ξω
sin ω d t + e −ξωt F0
sin ω d t
ω d
mω d
)
due to impulse
critically damped with impulse input ξ = c
c r
= 1 with initial conditions u (0) = 0 and u ′ (0) = 0 then the
solution is
u (t) = (A + Bt) e −( cr
2m
)
t
= (A + Bt) e −ωt
where A, B are found from initial conditions A = u (0) = 0 and B = u ′ (0) + u (0) ω = F 0
m
, hence the solution
is
u impulse (t) = F 0t
m e−ωt
If the initial conditions were not zero, then the solution for these are added to the above. From earlier, it
was found that the solution is u (t) = (u 0 (1 + ωt) + u ′ (0) t) e −ωt , therefore, the full solution is
u (t) =
due to IC only
{ }} {
(
u (0) (1 + ωt) + u ′ (0) t ) e −ωt +
due to impulse
{ }} {
F 0 t
m e−ωt
over-damped with impulse input ξ = c
c r
> 1 With initial conditions are u (0) = 0 and u ′ (0) = 0 the
solution is
u impulse (t) = Ae λ 1ωt + Be λ 2ωt
where A, B are found from initial conditions and
λ 1 = −ωξ + ω √ ξ 2 − 1
λ 2 = −ωξ − ω √ ξ 2 − 1
A = u′ (0) − u (0) λ 2
2ω √ ξ 2 − 1
B = −u′ (0) + u (0) λ 1
2ω √ ξ 2 − 1
Hence the solution is
(
−ξ+ √ (
ξ
u impulse (t) = Ae
2 −1
)ωt −ξ− √ )
ξ + Be 2 −1 ωt
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