Dynamics cheat sheet

8.3.2 Under-damped forced vibration c < c r , ξ < 1
u
K
C
M
Fsin t
mu ′′ + cu ′ + ku = F sin ϖt
u ′′ + 2ξωu ′ + ω 2 u = F sin ϖt
m
The solution is
where
and
u (t) = u h + u p
u h (t) = e −ξωt (A cos ω d t + B sin ω d t)
u p (t) =
F
√(k − mϖ) 2 + (cϖ) 2 sin (ϖt − θ)
where
tan θ =
cϖ
k − mϖ 2 =
2ξr
1 − r 2
Very important note here in the calculations of tan θ above, one should be careful on the sign of the
denominator. When the forcing frequency ϖ > ω the denominator will become negative (the case of ϖ = ω is
resonance and is handled separately). Therefore, one should use arctan that takes care of which quadrant the
angle is. For example, in Mathematica use
ArcTan[1 - r^2, 2 Zeta r]]
and in Matlab use
atan2(2 Zeta r,1 - r^2)
Otherwise, wrong solution will result when ϖ > ω The full solution is
u (t) = e −ξωt (A cos ω d t + B sin ω d t) + F k
1
√(1 − r 2 ) 2 + (2ξr) 2 sin (ϖt − θ) (1)
Applying initial conditions gives
A = u (0) + F k
B = u′ (0)
ω d
+
1
√(1 − r 2 ) 2 + (2ξr) 2 sin θ
u (0) ξω
ω d
+ F k
Another form of these equations is given as follows
1
ω d
√
(1 − r 2 ) 2 + (2ξr) 2 (ξω sin θ − ϖ cos θ)
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