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The Forced Damped Vibrations (Harmonic) Chapter of The AE2135 II Vibrations course taught at the University of Technology Delft.
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Aerospace Structures & Computational Mechanics
Lecture NotesVersion 1.6
AE21
35-I I
-Vib
ratio
ns
Forced Damped Vibrations (Harmonic)
6 Forced damped vibrations (harmonic) 17
6 Forced damped vibrations (harmonic)
Let’s look at the harmonic forced vibrations of an underdamped system, this has the followingequation of motion:
x+ 2ζωnx+ ω2nx = f cos (ωxt)
6-1 Solution method
To solve for the displacement, xp = Re(Apeiωxt
)is used, because: eiωt = cos (ωt) + i sin (ωt),
so cos (ωt) = Re(eiωt
). Entering this into the EOM gives:
Re((−ω2
x + 2ζωniωx + ω2n
)Apeiωxt
)= Re
(f eiωxt
)−→ Ap = Re
(f
ω2n − ω2
x + 2ζωniωx
)The denominator can be rewritten as a complex number u = v + iw, with:
v = ω2n − ω2
x
w = 2ζωnωxNow, denoting u as the complex conjugate of u:
Ap = Re(f
u
)= Re
(fu
uu
)= Re
(fu
v2 + w2
)Also:
u = v − iw = |u| (cos(θ)− i sin(θ)) = |u| (cos(−θ) + i sin(−θ)) = |u| e−iθ
with θ = arctan(wv
). Hence:
Ap = Re(f
√v2 + w2e−iθ
v2 + w2
)= Re
(f e−iθ√v2 + w2
)= Re
f e−iθ√(ω2n − ω2
x)2 + (2ζωnωx)2
So the particular solution becomes:
xp = Re(Apeiωxt
)= Re
f ei(ωxt−θ)√(ω2n − ω2
x)2 + (2ζωnωx)2
= f cos (ωxt− θ)√(ω2n − ω2
x)2 + (2ζωnωx)2
6-2 Initial conditions
x = Ahe−ζωnt sin (ωdt+ ϕ) +Ap cos (ωxt− θ)
x = Ah(−ζω−ζωnt
n sin (ωdt+ ϕ) + e−ζωntωd cos (ωdt+ ϕ))−Apωx sin (ωxt− θ)
So when x(0) = 0 and x(0) = 0, then:{Ah sin (ϕ) +Ap cos (θ) = 0Ah (−ζωn sin (ϕ) + ωd cos (ϕ)) +Apωx sin (θ)
from which ϕ and Ah can be derived as a function of the known Ap and θ.
AE2135-II - Vibrations Lecture Notes
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