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Contents

1 Introduction 3

1.1 Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . 31.2 Definition and Scope of Vibration . . . . . . . . . . . . . . . . 31.3 Importance of the study of vibrations . . . . . . . . . . . . . . 41.4 Examples of vibrations . . . . . . . . . . . . . . . . . . . . . . 41.5 Degrees of freedom . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Free Undamped Vibration 52.1 Simple harmonic motion . . . . . . . . . . . . . . . . . . . . . 52.2 Equation of motion of simple harmonic motions: spring-mass

system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Forced Undamped Vibration 83.1 Types of forcing function . . . . . . . . . . . . . . . . . . . . . 83.2 General response . . . . . . . . . . . . . . . . . . . . . . . . . 83.3 Amplitude ratio (Magnification factor) . . . . . . . . . . . . . 10

4 Free Damped Vibration 114.1 Damping Effect . . . . . . . . . . . . . . . . . . . . . . . . . . 114.2 The EOM and Its Solution . . . . . . . . . . . . . . . . . . . . 114.3 Damping ratio . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

4.4 Logarithmic decrement . . . . . . . . . . . . . . . . . . . . . 144.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

5 Forced Damped Vibration 165.1 The EOM and Its Solution . . . . . . . . . . . . . . . . . . . . 165.2 Amplitude Ratio . . . . . . . . . . . . . . . . . . . . . . . . . 17

1

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2 CONTENTS

5.3 Vibration due to movement of base . . . . . . . . . . . . . . . 18

5.3.1 Force Transmissibility due to Base Excitation . . . . . 195.4 Vibration due to unbalanced mass . . . . . . . . . . . . . . . . 20

5.4.1 Force Transmissibility due to Unbalanced Mass . . . . 21

6 Vibrations of 2-DOF System 226.1 Torsional Vibration . . . . . . . . . . . . . . . . . . . . . . . . 226.2 Free Vibration of a 2 dof System . . . . . . . . . . . . . . . . 22

6.2.1 EOM . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226.3 Forced Vibration of a 2 dof System . . . . . . . . . . . . . . . 266.4 Design strategy for DVA . . . . . . . . . . . . . . . . . . . . . 29

Razi Abdul Rahman, Trans Krian, 22 January, 2009.

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Chapter 1

Introduction

1.1 Learning outcomes

As in the PEO, at the end of the course, the students are able to:

1. Determine vibration response for a 1 degree-of-freedom (dof) system.I.e., he/she should understand the response in terms of the displace-ment, velocity, and acceleration.

2. Calculate natural frequencies and mode shapes for a 2 dof system.

3. Design a tuned vibration absorber (dynamic vibration absorber) sys-tem.

1.2 Definition and Scope of Vibration

Vibration is

usually understood as oscillation between an equilibrium point.

in mechanical systems, can be viewed as the flow between kinetic

energy and potential energy

We limit to sufficiently small displacement, hence to linear vibration ,and also harmonic/single frequency motion.

We only consider vibration of a particle only. Vibration of rigid bodies isskipped.

3

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4 CHAPTER 1. INTRODUCTION

1.3 Importance of the study of vibrations

Vibrations cause damages to machines, equipments, buildings, etc. Vibrations create noise We need to find ways to limit the level of vibrations, and thus the

fundamentals of vibration must first be well understood

1.4 Examples of vibrations

Noise from computer fans

Noise from automobile tires. Vibration isolation in Shanghai tower as the application of dynamic

vibration absorber (DVA)

1.5 Degrees of freedom

definition: direction an object or particle is free to move

must be describable with independent coordinates independent coordinates by definition imply the minimum number of

coordinates

Example 1: mass-spring

Example 2: Pendulum

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Chapter 2

Free Undamped Vibration

2.1 Simple harmonic motion

Some definitions:

periodic motion a motion that is repeated after equal intervals of time.

harmonic motion periodic motion with one frequency

simple harmonic motion a harmonic motion where the acceleration isproportional to the displacement and directed toward the mean po-sition.

A simple harmonic motion can be described with a sinosoidal function

x = A sin = A sin t (2.1)

The 1st time derivative of x gives the velocity

x = A cos t (2.2)

The 2nd time derivative of x gives the acceleration

x = 2A sin t (2.3)

Refer to Figure 1.41 of Ref. [3].

5

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6 CHAPTER 2. FREE UNDAMPED VIBRATION

2.2 Equation of motion of simple harmonic

motions: spring-mass system

Undamped free vibration:

free: no input over time (no external force) undamped: no energy transfer out of the system (no damping)See Figure 8/1 (a) in Reference [2]: mass-spring system (horizontal)The mass displaces to the right from the neutral positionFirst step: Draw free body diagram (FBD), then analyze the forces

acting on the mass.Second step: Use Netwons Second Law of Motion:

F = ma = mx

kx = mxBring x and all its derivatives to the left hand side to write the equation ofmotion:

mx + kx = 0 (2.4)

We now want to solve this equation to have x as a function of time x(t)...Eq. 2.4 is a second order ordinary differential equation. The principle in

calculus tells us the solution to Eq. 2.4 has the form of

x(t) = Aet (2.5)

By substitution,(m2 + k)et = 0 (2.6)

The equation is zero if the parenthesis is zero.Looking inside the parenthesis, youll see

1 = +i

k

m

2 = i

k

m

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2.3. EXAMPLES 7

The form of Eq. 2.5 can now be expressed with two constants A1 and A2

x(t) = A1eik

mt + A2e

ikmt (2.7)

A1 and A2 are constants from the initial conditions. (Remember from calcu-lus: When you integrate a second order ODE you get 2 constants)

We can express Eq. 2.7 in terms of a sinusoidal function using Eulersformula.

Substituting this formula, youll get

x(t) = C1 cosk

mt+ C2 sin

k

mt (2.8)

Introduce natural frequency n:

n =

k

m

Thenx(t) = C1 cos nt + C2 sin nt (2.9)

This is the solution to the equation of motion of free undamped vibration.Another form of the solution is in terms of the amplitude C and phase

x(t) = Csin(nt + ) (2.10)

With initial diplacement and velocity x0 and v0

C =

x20 +

v0n

2and

= tan1

x0n

v0 2.3 Examples

See various examples of problem solving in the references

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Chapter 3

Forced Undamped Vibration

Definition:Forced vibration: vibration due to external inputs (forces)Implication: the right hand side of the equation of motion is NOT zero

3.1 Types of forcing function

periodic For our case, harmonic forces, e.g. F = F0 sin t

general/random

shock/impact force that acts on short time, i.e., t < T, T=natural pe-riod of system

Here we only deal with Type I forcing function.

3.2 General response

First step: Draw free body diagram (FBD), then analyze the forces acting

on the mass.Second step: Use Netwons Second Law of Motion:

Fx = max = mx

F0 sin t kx = mx

8

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3.2. GENERAL RESPONSE 9

Bring x and all its derivatives to the left hand side to write the equation of

motion:

mx + kx = F0 sin t (3.1)

As before, We want to solve this equation to get x(t)...

Recall your calculus to understand Eq. 3.1. This is non-homogeneous2nd order ODE, where you have to find the particular solution for x(t).

For the particular solution, x should look like the forcing function, so weset

xp = Xsin t (3.2)

Then the derivatives arexp = X cos t

xp = X2 sin tFind X to complete the solution ...

Substitute Eq. (3.2) and its derivatives into the EOM:

X2 sin t + km

Xsin t =F0m

sin t

ThenX2 + k

mX =

F0m

X2 + km

X =F0m

X =F0m

km 2

Dividing by k/m

X =

F0

k1 (

n)2

We now have xp

xp =F0k

1 ( n

)2sin t (3.3)

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10 CHAPTER 3. FORCED UNDAMPED VIBRATION

3.3 Amplitude ratio (Magnification factor)

The amplitude ratio or equivalently magnification ratio M (Meriam p.616) is defined as:

M =X

st

So,

M =

1

n

21(3.4)

Figure: Refer [2] Figure 8/10.

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Chapter 4

Free Damped Vibration

4.1 Damping Effect

Note: A damper causes energy to flow out of the systemModes of damping:

1. Viscous effect2. friction3. Material hysterisisHere we only deal with Type 1 damping, i.e., where we have F = cx, c =

damping coefficient [N/(m/s)].

4.2 The EOM and Its Solution

See Figure 8/2 (a) in Reference [2]: mass-spring-dashpot systemThe mass displaces to the right from the neutral positionFirst step: Draw free body diagram (FBD), then analyze the forces

acting on the mass.Second step: Use Newtons Second Law of Motion:

Fx = max = mx

kx cx = mxBring x and all its derivatives to the left hand side to write the equation ofmotion:

11

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12 CHAPTER 4. FREE DAMPED VIBRATION

mx + cx + kx = 0 (4.1)Again, We want to solve this equation to have x as a function of time

x(t)...Like before, Eq. (4.1) is a second order ordinary differential equation.

The principle in calculus tells us the solution to Eq. (4.1) has the form of

x(t) = Aet (4.2)

perform some algebraic tricks on Eq. (4.1) so that our solution laterprovides more understanding on the nature of this damped free vibration:

Divide Eq. (4.1) by m, and use n

:

x + c/mx + 2nx = 0 (4.3)

Instead of c/m as coefficients, we introduce damping factor defined as

=c

2mn(4.4)

Now we have the equation of motion for damped free vibration

x + 2nx + 2nx = 0 (4.5)

By substitution of x(t) = Aet, we get

(2 + 2n + 2n)e

t = 0 (4.6)

Now, the solution to x(t) depends on as discussed in the followingsection.

4.3 Damping ratio

The root of the above is

1 = n(+

2 1)2 = n(

2 1)

Case 1: > 1, so that 1, 2 are negativeOVERDAMPEDNO OSCILLATION

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4.3. DAMPING RATIO 13

Case 2: = 1,

The response is

x(t) = C1e(+

21)nt + C2e

(

21)nt (4.7)

CRITICAL DAMPINGNO OSCILLATION

Notes:

Critical damping constant cc is defined as the value that makes the radicalin the root to be zero: (cf. Rao p 140) so that

cc = 2mk

m = 2km = 2mn (4.8)Damping ratio can be given by

= c/cc (4.9)

Response is: (Note that the form is a bit different, due to the equal roots)

x(t) = A1ent + A2te

nt (4.10)

x(t) is decaying over time, no oscillation.Case 2: < 1,

UNDERDAMPEDOSCILLATION

Notice now that 1, 2 are imaginary like in the undamped free vibration.

Using the Eulers formula again, we write

x = ent(B1 cos dt + B2 sin dt) (4.11)

where

d = n1 2

Or in terms of amplitude and phase

x = Cent sin(dt + ) (4.12)

where C =

B21 + B22 and = tan

1(B1/B2).

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14 CHAPTER 4. FREE DAMPED VIBRATION

4.4 Logarithmic decrement

Refer Figure 8/7 in Reference [2]. can be approximated from experiment using the logarithmic decrement

.

= ln

x1x2

=

21 2

Therefore,

=

42 + 2

.

4.5 Summary

E.O.M.

mx + cx + kx = 0

orx + 2nx +

2nx = 0

The solution has 3 different forms for the underdamped case:Form 1:

x(t) = ent(A1eidt + A2e

idt)

Form 2:x(t) = ent(B1 cos dt + B2 sin dt)

Form 3:x(t) = Cent sin(dt + )

where d =

1 2n.

The coefficients are related as the following:Given initial conditions x0 and x0,

B1 = A1 + A2 = x0

B2 = i(A1 A2) = x0 + nx01 2n

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4.5. SUMMARY 15

C =

B

2

1 + B

2

2 =

x

2

0 +

(x0 + nx0)2

(1 2)2n

= tan1(B1/B2) = tan1

1 2nx0

x0 + nx0

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Chapter 5

Forced Damped Vibration

sintF = F0k

c

m

Figure 5.1: A mass-spring-dashpot system with forcing function F =F0 sin t on mass m.

5.1 The EOM and Its Solution

See Figure 5.1: mass-spring-dashpot system with a forcing function F0sin t.You should know how to:i. draw free body diagram (FBD) and analyze the forces acting on the

mass.

ii. use Newtons Second Law of Motion to write down the equation ofmotion.

Fx = max = mx

F0 sin t kx cx = mx

16

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5.2. AMPLITUDE RATIO 17

Bring x and all its derivatives to the left hand side to write the equation of

motion:

mx + cx + kx = F0 sin t

The equation of motion for forced damped vibration is

x + 2nx + 2nx =

F0m

sin t (5.1)

NOTE: With this nonhomogeneous, ordinary differential equation, we arealways interested with the particular solution, i.e, the steady-state solu-tion.

Again, we want to solve this equation to have x as a function of timex(t)...Set xp = Xsin(t ), then we have (with the same process as before)

X =F0/k

{[1 r2]2 + [2r]2}1/2where r =

n.

= tan1

2r

1 r2

.As before, the total solution is the combination of the homogenous solu-

tion and the particular solution.

x(t) = Cent sin(dt + ) + Xsin(t )NOTE: C and are found from the initial conditions, X and are NOT.

5.2 Amplitude Ratio

We use the same definition of M as before.

M ={

[1

r2]2 + [2r]2

}1/2 (5.2)

You must learn Figures 8/11, 8/12, and 8/13 in Reference [2].NOTE: the maximum M for damped system no longer occurs at r=1.Two steps to reduce vibration:1. At resonance, increase damping.2. Run the system at higher frequency than the natural frequency.

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18 CHAPTER 5. FORCED DAMPED VIBRATION

5.3 Vibration due to movement of base

Instead of directly applying force to a system, we apply a motion to itsfoundation

m x(t)

y(t)

k c

Base

Figure 5.2: A mass-spring-dashpot system with base excitation function y =Y sin t.

Instead of force, we have here y(t) as the base harmonic movement withY amplitude.

To get the equation of motion you have to be able to draw the FBD withrelative coordinates (Refer to Fig. 3.14 of Reference [3]).

The EOM:

x +c

m(x y) + k

m(x y) = 0 (5.3)

Giveny(t) = Y sin t

Then

x +c

mx +

k

mx =

k

mY sin t +

c

mY cos t (5.4)

which we can rewrite as

x + cm x + kmx = A sin(t ) (5.5)

If we set the xp = Xsin(t ) as before, then

X = Y

1 + (2r)2

(1 r2)2 + (2r)21/2

(5.6)

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5.3. VIBRATION DUE TO MOVEMENT OF BASE 19

and

= tan1

2r3

1 + (42 1)r2

. Eq. (5.6) gives us displacement transmissibility ratio:

X

Y=

1 + (2r)2

(1 r2)2 + (2r)21/2

(5.7)

Recall the amplitude ratio M from Eq. (5.2). The amplitude ofxp X can bewritten as

X = MY(1 + (2r)2)1/2 (5.8)

From the graph of M vs. n

, we can have cases where M = 1, Hence atthese ratios, the amplitude X becomes

X = Y(1 + (2r)2)1/2 (5.9)

NOTE: the maximum M for damped system no longer occurs at r=1, alsodifferent from the above section.

NOTE: The discussion here is quite different from older lecture materials.

5.3.1 Force Transmissibility due to Base Excitation

From the FBD, there is a force F transmitted to the base/support due toreactions from the spring and the dashpot.

Newtons Second Law of motion gives

F = k(x y) + c(x y) = mx

With the form ofxp above,

F = m2Xsin(t

) = FT sin(t

)

ThenFTkY

= r2

1 + (2r)2

(1 r2)2 + (2r)21/2

(5.10)

This ratio is know as force transmissibility.

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20 CHAPTER 5. FORCED DAMPED VIBRATION

5.4 Vibration due to unbalanced mass

Unbalance in rotating machinery is one of the main causes of vibration.The unbalance mass always gives force excitation of

F(t) = me2 sin t (5.11)

where m is the unbalanced mass, e is the eccentricity.The EOM is then:

Mx + cx + kx = F0 sin t = me2 sin t

where M is the total mass of the system.

Or in terms of the damping ratio

x + 2nx + 2nx =

me2

Msin t (5.12)

where the natural frequency n =

k/M.With the particular solution of the usual form:

xp = Xsin(t )The amplitude X is expressed as

X =

me2(k M2)2 + (c2)2 (5.13)

is expressed as

= tan1

c

k M2

.A more useful and interesting form is by looking at the ratio. With

frequency ratio rM Xm e =

r2

(1 r2)2 + (2r)2 (5.14)

If no damping, M Xm e =

r2

1 r2 (5.15)The amplitude is then

X =m e r2

M(1 r2) (5.16)

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5.4. VIBRATION DUE TO UNBALANCED MASS 21

5.4.1 Force Transmissibility due to Unbalanced Mass

Let FT be the resulting force transmitted to the base/ground due to thevibrating system that is caused by the rotating unbalance mass. Then theratio of FT to the static force from the unbalance mass is given by

FTF0

=

1 + (2r)2

(1 r2)2 + (2r)21/2

(5.17)

where F0 = me2.

This ratio is the force transmissibilty ratio (FTR) due to unbalance mass.Compare this with FTR for the base excitation in the Section 5.3.1.

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Chapter 6

Vibrations of 2-DOF System

analysis of 2 dof system allows for understanding of multi-dof but muchreduced complexity

This chapter introduces the concept of mode shape As an application, we introduce different methods of attenuation such

as tunable vibration absorbor or dynamic vibration absorber (DVA)

6.1 Torsional Vibration

To add.

6.2 Free Vibration of a 2 dof System

6.2.1 EOM

The setup of a 2 DOF system may be given as in Fig. 6.1.You must know how to appropriately draw the FBD to get the EOM.EOM for mass 1:

k1x1 + k3(x2 x1) = m1x1EOM for mass 2:

k2x2 k3(x2 x1) = m2x2Arrange the two simultaneous equations:

22

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6.2. FREE VIBRATION OF A 2 DOF SYSTEM 23

k1

1m

3

m2

k2

x 2

1x

k

Figure 6.1: 2 dof spring-mass system.

m1x1 + k1x1 + k3x1 k3x2 = 0 (6.1)m2x2 k3x1 + k2x2 + k3x2 = 0 (6.2)

We can cast these equations into matrix form:m1 00 m2

x1x2

+

k1 + k3 k3k3 k2 + k3

x1x2

=

00

(6.3)

which has the following general form

M x + K x = 0 (6.4)M is mass matrix, K is stiffness matrix.

The motivation is always to determine x1(t) and x2(t).We assume first that m1 and m2 oscillate with the same frequency and

phase angle:

x1 = X1 cos(t + ) (6.5)

x2 = X2 cos(t + ) (6.6)

We substitute the displacements and accelerations as we always do in 1-DOFproblem before.

Well get k1 + k3 m12 k3

k3 k2 + k3 m22

X1X2

=

00

(6.7)

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24 CHAPTER 6. VIBRATIONS OF 2-DOF SYSTEM

This is the form of the standard eigenvalue problem:

A X = 0

Recall your Linear Algebra subject: This is an eigenvalue problem where youhave to find the eigenvalues for the equation to obtain its non-trivial solutionsfor X1, X2. (What is the trivial solutions?)

To find the eigenvalues of A, set det(A) = 0.Recall: For a 2 by 2 matrix, the formula for the determinant is simple:

det

a bc d

= ad bc

Notice that ad bc = 0 for our case is the characteristic equation!Take the determinant of the A matrix of Eq. (6.7):

det(A) = (k1 + k3 m12)(k2 + k3 m22) (k3)(k3)

Characteristic equation:

(k1 + k3 m12)(k2 + k3 m22) k23 = 0

Rewrite:

m1m24 [m1(k1 + k3) + m2(k1 + k2)]2 + k1k2 + k3(k1 + k2) = 0The above equation is called the characteristic equation or frequency equationof the 2-DOF system.

There are 2 roots of this equation which correspond to the 2 naturalfrequencies of the system.

Let = 2. Then

m1m22 [m1(k1 + k3) + m2(k1 + k2)] + k1k2 + k3(k1 + k2) = 0

This is quadratic equation where the roots are

1,2 = (b

b2 4ac)/2a

For the case of m1 = m2 = m; k1 = k2 = k3 = k;

1 = k/m; 2 = 3k/m

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6.2. FREE VIBRATION OF A 2 DOF SYSTEM 25

Meaning that the natural frequencies are

21 = k/m; 22 = 3k/m

The corresponding eigenvectors give the mode shapes of the vibrationsystem.

To find the eigenvectors, use 21 = k/m in Eq. (6.7):

After simplification youll get:1 11 1

X1X2

=

00

(6.8)

To find the eigenvector check for

X =

11

Then 1 11 1

11

=

00

(6.9)

Then this this is the eigenvector for 1FOr the second eigenvector, use 22 = 3k/m in Eq. (6.7):

After simplification youll get:1 11 1

X1X2

=

00

(6.10)

To find the eigenvector check for

X =

11

Then

1 11 1

11

=

00

(6.11)

Then this this is the eigenvector for 2This two eigenvectors describe the mode shapes of the free vibration of

the 2-dof system with the above defined values of m1, m2, k1, k2, k3.

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26 CHAPTER 6. VIBRATIONS OF 2-DOF SYSTEM

6.3 Forced Vibration of a 2 dof System

response of a 2 dof system is similar to a 1 dof system Recall: resonance occurs when the forcing frequency is similar to the

the natural frequency of the system

for a 2 dof system there are 2 resonant frequencies because we have 2natural frequencies

Example:

k3

k1

k2

1m

1x

x2

m2

1sin tf =F1

Figure 6.2: 2 dof with forcing function on mass 1.

A forcing function f1(t) = F1 sin t is applied to mass m1. Determine theresponse of masses 1 and 2 with the changes in the forcing frequency .

Solution:

The steps for the solution is almost exactly the same as before for freevibration. The only different is we have to take the inverse of the matrix aswill be shown.

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6.3. FORCED VIBRATION OF A 2 DOF SYSTEM 27

i. Draw FBD

ii. Write the equation of motion

For mass 1:k1x1 + k3(x2 x1) + f1(t) = m1x1

For mass 2:k2x2 k3(x2 x1) = m2x2

3. Due to the harmonic forcing function the response is expected to beharmonic too

x1 = X1 sin tx2 = X2 sin t (6.12)

so the x1 and x2 are

x1 = X12 sin tx2 = X22 sin t (6.13)

As before, substitute Eqs. (6.12) and (6.13) into the equtions of motion. Wethen get our matrix form of the equations of motion.

k1 + k2 m12 k3

k3 k2 + k3

m2

2X1X2 =

F10 (6.14)

The determinant of the above matrix will give us the characteristic equation

() = (k1 + k3 m12)(k2 + k3 m22) k23 = 0 (6.15)

and this will give us the natural frequencies of the system: 1 and 2.Recall from Linear Algebra:Given

A x = Fthen to solve for x we taken the inverse of A:

x = A1 F

For a 2 by 2 matrix, the formula is simple:a bc d

1

=

d bc a

1

(ad bc)

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28 CHAPTER 6. VIBRATIONS OF 2-DOF SYSTEM

So the solution isX1X2

=

1

()

k2 + k3 m22 k3

k3 k1 + k2 m12

F10

(6.16)

To write in individual X1 and X2

X1 =(k2 + k3 m22)F1

()(6.17)

X2 =k3F1()

(6.18)

Notice that:

the denominator for both terms above is the same.

the denominator when set to equal zero is our characteristic equation!

if is either equal to any of the two natural frequencies, () will bezerothis is how we found 1 and 2 before for free vibration

It means that our X1 and X2 will be infinitywe will get resonance!

If we let: m1 = m2 = m and k1 = k3 = k, k2 = 0, we will get

21 =k/m

22 =3k/m

If we plot the the amplitude ratios versus frequency ratios (see for exampleRao, Mechanical Vibrations, Sec. 5.6, Figure 5.14 :

X1

F1/k

vs. /1

X2F1/k

vs. /1

there is a point which X1 becomes zero. This fact will be the basis in dy-namic vibration absorber (DVA) or tunable vibration absorber.

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6.4. DESIGN STRATEGY FOR DVA 29

x 1

1m

F=F1 sin t

k2

k1

m2

2x

Secondary

system

Primarysystem

Figure 6.3: Secondary system as dynamic vibration absorber

6.4 Design strategy for DVA

Compare Fig. 6.3 with Fig. 6.2. The setup here is equivalent to Fig. 6.2with k = 2 in Fig. 6.2 being zero. Look at Eq. 6.17. To make X1 be zero,we have to set the numerator to be zero. Hence:

k2 + k3 m22 =0

2

=

k2 + k3

m2

We use this fact as a strategy to design the DVA:

1. To absorb vibration at n, so that X1 = 0, design such that

k2m2

=k1m1

2. To absorb vibration at forcing frequency , so that X1 = 0, design

such that k2m2

= 2

For further reading, see for example Rao, Mechanical Vibrations, Section9.11.

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References

[1] R. C. Hibbeler, Engineering Mechanics: Dynamics, Prentice Hall, (LatestEdition; Chapter 22).

[2] J. L. Meriam and L. G. Kraige, Engineering Mechanics: Dynamics, JohnWiley and Sons. Latest Edition.

[3] S. S. Rao, Mechanical Vibrations, Pearson/Prentice Hall, 2004.

[4] R. F. Steidel Jr, An Introduction to Mechanical Vibrations, John Wiley& Sons. (Latest Edition)

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