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Lecture notes AE2135 II Vibrations - Forced Undamped Vibrations (Harmonic)

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The Forced Undamped Vibrations (Harmonic) Chapter for The AE2135 II Vibrations course taught at the University of Technology Delft.

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Page 1: Lecture notes AE2135 II Vibrations - Forced Undamped Vibrations (Harmonic)

Aerospace Structures & Computational Mechanics

Lecture NotesVersion 1.6

AE21

35-I I

-Vib

ratio

ns

Forced Undamped Vibrations (Harmonic)

Page 2: Lecture notes AE2135 II Vibrations - Forced Undamped Vibrations (Harmonic)

14 5 Forced undamped vibrations (harmonic)

5 Forced undamped vibrations (harmonic)

m

x

Fe

Figure 18: A forced mass-spring system

5-1 General solution

EOM:mx+ kx = F e(t) = F cos(ωxt)

The displacement x of the forced mass-spring system in figure 18 is composed of a transientpart, xh, and a steady-state part, xp, i.e.:

x = xh + xp

The following is assumed for x:

x = A sin (ωnt+ ϕ)︸ ︷︷ ︸xh

+Ap cos(ωxt)︸ ︷︷ ︸xp

Solving the steady-state, or particular part of the displacement:

−mApω2x cos(ωxt) +Apk cos(ωxt) = F cos(ωxt)

so:Ap = f

ω2n − ω2

x

with:f = F

m

The amplitude and phase shift are displayed in figure 19.

Lecture Notes AE2135-II - Vibrations

Page 3: Lecture notes AE2135 II Vibrations - Forced Undamped Vibrations (Harmonic)

5 Forced undamped vibrations (harmonic) 15

θ

Figure 19: Amplitude of undamped forced motion and phase difference θ between loading andresponse. After ωx = ωn, the load and response are in opposite phase

Suppose: {x0 = 0x0 = 0

then:

A sin(ϕ) +Ap = 0

Aωn cos(ϕ)−Apωx · 0 = 0 −→ ϕ = π

2 + 2kπ

−→ A = −Ap

−→ x = Ap (cos(ωxt)− cos(ωnt))

5-2 Solution at resonance

x+ ω2nx = f cos (ωnt)

The normal assumption to reach a solution would be xp = Ap cos (ωnt), but in this case this

AE2135-II - Vibrations Lecture Notes

Page 4: Lecture notes AE2135 II Vibrations - Forced Undamped Vibrations (Harmonic)

16 5 Forced undamped vibrations (harmonic)

is not possible as it would lead to Ap →∞. Hence, in this case, we try:

xp = Apt sin (ωnt+ ϕ)xp = ωnA

pt cos (ωnt+ ϕ) +Ap sin (ωnt+ ϕ)xp = −ω2

nApt sin (ωnt+ ϕ) + ωnA

p cos (ωnt+ ϕ) + ωnAp cos (ωnt+ ϕ)

= 2ωnAp cos (ωnt+ ϕ)− ω2nA

pt sin (ωnt+ ϕ)

Substituting this in the equation above yields:

2ωnAp cos (ωnt+ ϕ)−((((((((

((ω2nA

pt sin (ωnt+ ϕ) +(((((((((

(ω2nA

pt sin (ωnt+ ϕ) = f cos (ωnt)−→ 2ωnAp cos (ωnt+ ϕ) = f cos (ωnt)

Hence:

ϕ = 0 + 2kπ

Ap = f

2ωn

So:−→ xp = f

2ωnt sin (ωnt+ ϕ) = f

2ωnt sin (ωnt)

which can be seen in figure 20.

t

Figure 20: Sinusoidal response, with an amplitude linearly increasing by f t2ωn

Lecture Notes AE2135-II - Vibrations