Dynamics cheat sheet

Hence the full solution is
u p = p 0 1 ((
1 − r
2 )
k (1 − r 2 ) 2 + (2ζr) 2 sin ϖt − 2ζr cos ϖt )
u (t) = e −ξωnt (A cos ω d t + B sin ω d t) + F k
Applying initial conditions now gives
1
(1 − r 2 ) 2 + (2ζr) 2 ((
1 − r
2 ) sin ϖt − 2ζr cos ϖt ) (1)
A = u (0) +
B = u′ (0)
ω d
2F rξ
k
1
(1 − r 2 ) 2 + (2ξr) 2
+ u (0) ξω n
− F ( 1 − r 2) ϖ 2F rζ
ω d kω d (1 − r 2 ) 2 +
2
+ (2ζr) kω d
ω n
(1 − r 2 ) 2 + (2ζr) 2
The above 2 sets of equations are equivalent. One uses the phase angle explicitly and the second ones do not.
Also, the above assume the force is F sin ϖt and not F cos ϖt. If the force is F cos ϖt then in Eq 1 above, the
term reverse places as in
u (t) = e −ξωnt (A cos ω d t + B sin ω d t) + F k
Applying initial conditions now gives
(
1 − r
2 )
A = u (0) + F k (1 − r 2 ) 2 + (2ξr) 2
B = u′ (0)
ω d
+ u (0) ξω n
ω d
+
2F rζ
kω d
1 ((
1 − r
2 )
(1 − r 2 ) 2 + (2ζr) 2 cos ϖt − 2ζr sin ϖt )
ϖ
(1 − r 2 ) 2 + (2ζr) 2 − F ( 1 − r 2)
kω d
ω n
(1 − r 2 ) 2 + (2ζr) 2
When a system is damped, the problem with the divide by zero when r = 1 does not occur here as was the
case with undamped system, since when when ϖ ≈ ω or r = 1, the solution in Eq (1) becomes
u (t) = e −ξωt ((
u (0) + F k
) (
1
u ′
2ξ sin θ (0) u (0) ξω
cos ω d t + + + F ω d ω d k
) )
1
2ω d ξ (ξω sin θ − ϖ cos θ) sin ω d t
+ F k
1
sin (ϖt − θ)
2
and the problem with the denominator going to zero does not show up here. The amplitude when steady state
response is maximum can be found as follows. The amplitude of steady state motion is F 1
k
√(1−r . This
2 ) 2 +(2ξr)
√
2
1
is maximum when the magnification factor β = √
is maximum or when (1 − r 2 ) 2 + (2ξr) 2 or
(1−r 2 ) 2 +(2ξr)
√ 2
(
1 − ( ) )
ϖ 2 2 ( )
ω + 2ξ
ϖ 2
ω is minimum. Taking derivative w.r.t. ϖ and equating the result to zero and solving
for ϖ gives
ϖ = ω √ 1 − 2ξ 2
We are looking for positive ϖ, hence when ϖ = ω √ 1 − 2ξ 2 the under-damped response is maximum.
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