Aerospace Structures & Computational Mechanics

Lecture NotesVersion 1.6

AE21

35-I I

-Vib

ratio

ns

Free Damped Vibrations

10 4 Free damped vibrations

4 Free damped vibrations

k c

m

x

Figure 13: A damped mass-spring system

Consider the mass-damper-spring system as shown in figure 13. It has the following equationof motion:

mx+ cx+ kx = 0To solve for the motion x(t), substitute x = xeλt:

m�xλ2��eλt + c�xλ��e

λt + k�x��eλt = 0

−→ mλ2 + cλ+ k = 0

Solve this with the quadratic formula:

λ = −c±√c2 − 4mk

2mThe discriminant c2 − 4mk makes the difference whether λ is real or complex. The criticalfactor is when it is zero:

c2 = 4mk −→ ccrit = 2√mk

Define the damping ratio ζ:ζ = c

ccrit= c

2√mk

= c

2mωnThen the equation of motion can be rewritten:

x+ 2ζωnx+ ω2nx = 0

4-1 Case 1: Underdamped motion

In this case: 0 < ζ < 1:

λ1,2 = −2ζωn ±√

4ζ2ω2n − 4ω2

n

2= −ζωn ± ωn

√ζ2 − 1

= −ζωn ± i ωn√

1− ζ2︸ ︷︷ ︸ωd

so:

λ1 = −ζωn − iωdλ2 = −ζωn + iωd

Lecture Notes AE2135-II - Vibrations

4 Free damped vibrations 11

which leads to the following solution (see also figure:

x = e−ζωnt(a1e

iωdt + a2e−iωdt

)= Ae−ζωnt︸ ︷︷ ︸

decay

sin (ωdt+ ϕ)︸ ︷︷ ︸frequency content

(4.1)

t

x

Figure 14: Decaying sinusoidal response (the dashed lines indicate ±e−ζωnt)

The amplitude A and phaseshift ϕ can be found from the initial conditions x(0) = x0 andx(0) = x0:

x(0) = A sinϕ = x0 −→ A = x0sin(ϕ) (4.2)

From equation 4.1 the velocity can be deduced:

x = −ζωnAe−ζωnt sin(ωdt+ ϕ) +Ae−ζωntωd cos(ωdt+ ϕ)

So the initial velocity condition can be written as:

x0 = −ζωnA sin(ϕ) +Aωd cos(ϕ) (4.3)

Substituting equation 4.2 in this equation yields:

x0 = −ζωnx0 + x0ωdtan(ϕ)

Hence:−→ ϕ = arctan

(x0ωd

x0 + ζωnx0

)Also, from equation 4.2 (see figure 15):

sin(ϕ) = x0A

= x0ωdP

−→ P = Aωd

and using the Pythagorean theorem:

−→ A =

√x2

0ω2d + (x0 + ζωnx0)2

ωd

AE2135-II - Vibrations Lecture Notes

12 4 Free damped vibrations

ϕ

Figure 15: Obtaining P through angle formulae

4-2 Case 2: Overdamped motion

Now: ζ > 1In this case the roots of the equation of motion are real:

λ1 = −ζωn − ωn√ζ2 − 1

λ2 = −ζωn + ωn

√ζ2 − 1

and the displacement becomes non-oscillatory (see also figure 16):

t

x

Figure 16: An overdamped motion with x(0) = 1 and x(0) = 0

4-3 Case 3: Critically damped motion

For this type of motion: ζ = 1Now the roots are equal:

λ1 = λ2 = −ωnFor repeated roots, the solution takes the form:

x = (a1 + a2t)e−ωnt

where

a1 = x0

a2 = x0 + ωnx0

A critically damped motion can be seen in figure

Lecture Notes AE2135-II - Vibrations

4 Free damped vibrations 13

t

x

Figure 17: A critically damped motion with x(0) = 1 and x(0) = 0

AE2135-II - Vibrations Lecture Notes