Dynamics cheat sheet

at t = t 1
˙u (t 1 ) = −ξωe −ξωt 1
(A cos ω d t 1 + B sin ω d t 1 ) + e −ξωt 1
(−Aω d sin ω d t 1 + ω d B cos ω d t 1 )
u st
+ ϖ
cos (ϖt 1 − θ) (3)
√(1 − r 2 ) 2 + (2ξr) 2
Now for t > t 1 the equation becomes
which has the solution
where A = u (t 1 ) and B = ˙u(t 1)+u(t 1 )ξω
ω d
mü + c ˙u + ku = 0
u = e −ξωt (A cos ω d t + B sin ω d t)
critically damped with sin impulse ξ = c
c r
= 1 For t ≤ t 1 Initial conditions are u (0) = u 0 and ˙u (0) = v 0
then the solution is from above
u (t) = (A + Bt) e −ωt u st
+
sin (ϖt − θ) (1)
√(1 − r 2 ) 2 + (2r) 2
Where tan θ =
cϖ
k−mϖ 2
= 2ξr
1−r 2 . A, B are found from initial conditions
A = u 0 +
u st
√(1 − r 2 ) 2 + (2r) 2 sin θ
B = v 0 + u 0 ω +
u st
√(1 − r 2 ) 2 + (2r) 2 (ω sin θ − ϖ cos θ)
For t > t 1 the solution is
To find u (t 1 ) , from Eq(1)
u (t) = (u (t 1 ) + ( ˙u (t 1 ) + u (t 1 ) ω) t) e −ωt (2)
u (t 1 ) = (A + Bt) e −ωt 1
+
u st
√(1 − r 2 ) 2 + (2r) 2 sin (ϖt 1 − θ)
taking derivative of (1) gives
˙u (t) = −ω (A + Bt) e −ωt + Be −ωt + ϖ
sin (ϖt − θ) (3)
√(1 − r 2 ) 2 + (2r) 2
u st
at t = t 1
˙u (t 1 ) = −ω (A + Bt 1 ) e −ωt 1
+ Be −ωt 1
+ ϖ
sin (ϖt 1 − θ) (4)
√(1 − r 2 ) 2 + (2r) 2
u st
Hence Eq (2) can now be evaluated using Eq(3,4)
over-damped with sin impulse ξ = c
c r
> 1 For t ≤ t 1 Initial conditions are u (0) = u 0 and ˙u (0) = v 0 then
the solution is
u = Ae p1t + Be p2t u st
+
sin (ϖt − θ)
√(1 − r 2 ) 2 + (2ξr) 2
where tan θ = 2ξr
1−r 2 (make sure you use correct quadrant, see not above on arctan) and
p 1 = − c
√ ( c
) 2
2m + k −
2m m
= −ωξ + ω √ ξ 2 − 1
47