System Dynamics & Vibrations
m
k
x(t)
f(t)
c
Author: Allyn W. Phillips, PhD
University of Cincinnati
UC Course Nr.: 20-MECH-3080
Revision: 02-May-2025
Copyright  2001-2025
System Dynamics & Vibrations - Lecture Notes
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Table of Contents
#1 – Introduction to System Dynamics & Vibrations ..................................................................... 5
Introductory Example ................................................................................................................. 7
#2 – Newton’s Method ( -or- d'Alembert’s Method ) ..................................................................... 9
Force Balance.............................................................................................................................. 9
Moment Balance ......................................................................................................................... 9
* Sanity Checks......................................................................................................................... 10
Newton Example A ................................................................................................................... 11
Newton Example B ................................................................................................................... 13
Enhanced 1-DOF Newton Example.......................................................................................... 14
#3 – Homogeneous Solution of the Equation of Motion .............................................................. 16
Homogeneous Example ............................................................................................................ 20
* Note on Consistent Units ....................................................................................................... 21
Enhanced 1-DOF Homogeneous Example ............................................................................... 22
#4 – Estimating Modal Parameters – Time Domain ..................................................................... 23
Log Decrement.......................................................................................................................... 23
Log Decrement Example .......................................................................................................... 25
#5 – Lagrange’s Method ............................................................................................................... 26
Lagrange’s Method - Overview ................................................................................................ 26
Lagrange’s Method - Details .................................................................................................... 27
* Sanity Checks......................................................................................................................... 28
Lagrange Example A ................................................................................................................ 29
Lagrange Example B................................................................................................................. 30
Energy Example A .................................................................................................................... 32
Enhanced 1-DOF Lagrange Example ....................................................................................... 33
Enhanced 1-DOF Kinematic Example...................................................................................... 36
#6 – Steady-State Solution of Equation of Motion ....................................................................... 39
Transfer Function - Overview ................................................................................................... 39
#7 – The Frequency Response Function (FRF) ............................................................................ 42
FRF - Overview ........................................................................................................................ 42
FRF Example A ........................................................................................................................ 49
#8 – Rotating Unbalance ............................................................................................................... 50
Rotating Example A .................................................................................................................. 52
#9 – Steady-State Applications ..................................................................................................... 55
Force Transmissibility .............................................................................................................. 55
Displacement Transmissibility.................................................................................................. 56
Vibration Isolation .................................................................................................................... 56
Conclusion ................................................................................................................................ 57
Transmissibility Example A ..................................................................................................... 58
#10 – The Eigenvalue Problem ..................................................................................................... 61
Eigenvalue/Eigenvector Example ............................................................................................. 63
Final Notes ................................................................................................................................ 65
#11 – Multiple Degree of Freedom Systems ................................................................................ 66
MDOF Example ........................................................................................................................ 66
Enhanced 2-DOF Lagrange Example ....................................................................................... 73
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Enhanced 2-DOF Non-Linear Example.................................................................................... 76
#12 – Estimating Modal Parameters – Frequency Domain .......................................................... 81
Quadrature................................................................................................................................. 81
#13 – Transient Solution of Equation of Motion .......................................................................... 85
Impulse Response ..................................................................................................................... 85
Arbitrary Excitation .................................................................................................................. 86
* Convolution Process .............................................................................................................. 87
* Convolution Example ............................................................................................................ 87
#14 – Numerical Solution of Equation of Motion ........................................................................ 93
Euler’s Method.......................................................................................................................... 93
State-Space Expansion .............................................................................................................. 93
Numerical Solution ................................................................................................................... 95
* #15 – System Analogies (Introduction) ..................................................................................... 96
Analogous Systems: .................................................................................................................. 96
Mechanical-Electrical ("intuitive analogy"): 1st Order Relationships .................................... 96
#16 – System Analogies (Overview) ............................................................................................ 97
Analogous Systems (summary) ................................................................................................ 97
Analogous Systems (relationships) ........................................................................................... 99
#17 – System Analogies (Circuit Concepts) ............................................................................... 101
Circuits Review....................................................................................................................... 101
Complex Impedance ............................................................................................................... 102
#18 – System Analogies (Models) .............................................................................................. 106
Analogous Systems (single loop)............................................................................................ 106
Analogous Systems (multiple loop) ........................................................................................ 107
Low Pass Filter (example) ...................................................................................................... 109
Analogous Systems (detailed solution procedure) .................................................................. 110
#19 – Multiple Degree of Freedom – Applications .................................................................... 114
Hybrid Systems (flux coupling) .............................................................................................. 114
Appendix A – Unit Systems (aka ‘g’ is a unit) ........................................................................... 117
Unit System Conversions:....................................................................................................... 117
Conversion: ............................................................................................................................. 117
SI Gravimetric Example: ........................................................................................................ 118
Standard Gravity: .................................................................................................................... 118
Appendix B – Complex Numbers (review) ................................................................................ 119
Real / Imaginary...................................................................................................................... 119
Complex Math ........................................................................................................................ 119
Mag / Phase ............................................................................................................................. 119
True Angle .............................................................................................................................. 119
* Appendix C – Harmonic Function (review) ............................................................................ 120
* Appendix D – Rolling & Cable Constraints ............................................................................ 121
* Appendix E – Supplement Math Hints .................................................................................... 123
Appendix F – MDOF Detail Overview ...................................................................................... 125
General Solution Approach..................................................................................................... 125
Complementary (Transient) Solution: .................................................................................... 125
Particular Solution Approach .................................................................................................. 128
Total Solution.......................................................................................................................... 129
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Example Solution:................................................................................................................... 129
Example Solution - Particular Part: ........................................................................................ 130
Example Solution - Complementary Part: .............................................................................. 131
Complementary Solution Approach - Adjoint Matrix ............................................................ 133
Complementary Solution Approach - Eigenvalue/Eigenvector Method ................................ 135
Weighted Orthogonality Concept ........................................................................................... 137
Note: sections marked with an asterisk (*) are under revision.
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#1 – Introduction to System Dynamics & Vibrations
All mechanical systems vibrate or undergo oscillatory motion. However, in order to discuss the
behavior of vibrating systems, it is first necessary to establish some nomenclature. There are
several terms used to describe vibrations, among which are: oscillatory motion, periodic motion,
harmonic motion, period, and frequency.
Oscillatory motion is any pattern of motion where the system under observation moves back and
forth across some equilibrium position, but does not necessarily have any particular repeating
pattern.
Periodic motion is a specific form of oscillatory motion where the motion pattern repeats itself
with a uniform time interval. This uniform time interval is referred to as the period and has units
of seconds per cycle. The reciprocal of the period is referred to as the frequency and has units of
cycles per second. This unit combination has been given a special unit symbol and is referred to
as Hertz (Hz).
Harmonic motion is a specific form of periodic motion where the motion pattern can be
described by either a sine or cosine. This motion is also sometimes referred to as simple
harmonic motion. Because the sine or cosine technically uses angles in radians, the frequency
term expressed in the units of radians per second ( rad sec ). This is sometimes referred to as the
circular frequency. The relationship between the frequency in Hz and the frequency in rad sec is
simply the relationship, 2 rad cycle .
Natural frequency is the frequency at which an undamped system will tend to oscillate due to
initial conditions in the absence of any external excitation. Because there is no damping, the
system will oscillate indefinitely.
Damped natural frequency is frequency that a damped system will tend to oscillate due to initial
conditions in the absence of any external excitation. Because there is damping in the system, the
system response will eventually decay to rest.
Resonance is the condition of having an external excitation at the natural frequency of the
system. In general, this is undesirable, potentially producing extremely large system response.
Degrees-of-freedom is the number of independent coordinates necessary to describe the
configuration (state) of a system. It can also be expressed as the number of coordinates used to
describe the configuration of a system minus the number of independent constraint equations
between those coordinates. For example, a point in space has three degrees-of-freedom ( x, y, z )
and a rigid body in space has six ( x, y, z, x , y , z ) .
A decibel is a log scale measure of relative power. Specifically, it is ten times the log base ten of
 P 
the power to a reference power, dB = 10log10 
. If the quantity being referenced is not
 P 
ref


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 V2 
power based, then the quantities are first squared, yielding dB = 10log10  2  or
V 
 ref 
 V 
dB = 20log10 
.
 V 
ref


An octave is a frequency range where the upper frequency is twice the lower frequency.
A decade is a frequency range where the upper frequency is ten times the lower frequency.
Bandwidth and frequency span both terms describe a range of frequencies from low to high.
Bandwidth is particularly associated with the energy of the response of a system and in this
course is used in reference to damping (see section eight.)
Having established a small set of basic nomenclature, it is now possible to provide a brief
overview of the direction of this course. In this course, System Dynamics & Vibrations, much of
the effort will be spent upon bringing together the many concepts and techniques developed in
previous courses and applying them to the solution of vibrations problems.
Note that, it is possible to solve simple, but representative, problems utilizing only information
previously learned. To help clarify which courses and knowledge are used, the following
introductory example has been developed.
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Introductory Example
For the figure given, assume that the object rolls without
slipping. The objective will be to develop the equation of
motion and determine the natural frequency of the system.
(A more complete example would involve a general
multi-degree of freedom system, where the objective
would be to determine the equations of motion, the
system poles, mode shapes, and steady-state response.)
f (t )
Rolls w/o slipping
 (t )
k
x(t )
r
m, J cg
Step 1: Draw the free-body diagram for each independent
rigid body in the system. Be sure to include all external
forces and moments and their points of application. (Recall that the internal force resulting from
the spring is due to the actual relative motion of the ends of the spring. Therefore the coordinate
x is measured from the free length of the spring.) [Statics & Particle Dynamics, Kinematics &
Kinetics]
Step 2: Write down all the force and moment
balance equations.
f
 F ) f − k x − f = mx
 F ) N − mg = 0
M ) f r + f r = J 
kx
f
x
mg
y
cg
f
=
J cg 
mx
cg
ff
N
Step 3: Identify all relevant coordinate constraint
equations.
x = r and therefore x = r & x = r
Step 4: Choose the desired independent coordinates and eliminate all the dependent coordinates
by substituting the relevant constraint equations.
 F ) f − k x − f = mx
 F ) N − mg = 0
f
x
y
M )
f r + f f r = J cg
cg
x
r
Step 5: Evaluate the set of simultaneous differential equations to yield the equations of motion.
 F ) N = mg
y
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M )
cg
F &M )
x
cg
J


f f = −  f − cg2 x 
r 

 
J

f − k x −  −  f − cg2 x   = mx
r 
 
Step 5a: Identify the actual equation of motion.
J cg 

m+ 2 x + kx = 2 f
r 

Step 6: Solve the simultaneous set of differential equations of motion. The total solution ( xT (t ) )
involves the sum of two parts: the particular solution ( x p (t ) ) and the homogeneous solution (
xh (t ) ).
Step 6a: The particular solution involves knowing the exact form of the forcing function ( f (t ) ).
Step 6b: A second order homogeneous constant coefficient differential equation has complex
exponentials as its solution.
xh (t ) = Cet and therefore xh (t ) = Cet & xh (t ) = C  2et

J cg  2
 t
  m + 2   + k Ce = 0
r 


therefore the system poles (eigenvalues) are:
1,2 =  j
k
m+
J cg
r2
and the homogeneous solution is:
xh (t ) = C1e1t + C2e2t
Step 6c: Formulate the complete solution ( xT (t ) = x p (t ) + xh (t ) ) and evaluate the initial
conditions to eliminate the constants of integration.
Note: if this example had involved more than one independent dynamic coordinate, the solution
would have additionally required the solution of an eigenvalue/eigenvector problem.
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#2 – Newton’s Method ( -or- d'Alembert’s Method )
Force Balance
 F = M q ( -or-  F − M q = 0 )
G
G
Moment Balance
 M P = J P + rG / P  M qP ( -or-  M P − J P − rG / P  M qP = 0 )
P
G
Typical problems with Newton’s ( -or- d'Alembert’s ) formulation:
• Be sure to establish the number of degrees of freedom first and formulate all terms in
only those variables. Clearly identify which degrees of freedom are relative coordinates
versus absolute coordinates. Also, clearly identify what will be the positive direction of
motion for each coordinate. Watch out for rotational/translational problems. State any
constraint relationships that relate independent and dependent coordinates.
• Evaluate the static balance for the problem in order to determine whether the orientation
of the system in the gravitational field will affect the equations of motion (Are the
weights of the objects balanced by an initial static deflection in the springs?). When in
doubt, perform a static force balance to determine the appropriate constraint equation.
• For displacement, velocity and acceleration terms, be sure to develop absolute or relative
displacement, velocity and accelerations of appropriate points as required. Watch out for
2-D and 3-D vector motions.
• Be sure to draw the appropriate free body diagrams for each mass (or combination of
masses) in the system.
o Whenever the system is separated in order to draw a free body diagram, replace
the separation with the appropriate internal forces/moments (equal and opposite
forces/moments on each side of the separation).
o Do not move forces/moments arbitrarily from one mass to another. The internal
forces account for the effects of one mass on another.
• Develop one equation of motion for each degree of freedom of the system using
Newton's ( -or- d'Alembert’s ) method. Be sure to watch for moving reference frame
issues. Also, check that the units are the same for each term in an equation (Forces +
Moments: NOT!)
• If necessary, once the exact equations of motion have been determined, linearize the
equations of motion by neglecting nonlinear terms in the equations of motion. Note that
the linear equations of motion may not adequately describe the original equations of
motion if some of the terms that have been neglected are not insignificant.
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* Sanity Checks
The following are a list of “sanity checks” that help create valid FBDs and also help detect
common mistakes.
•
•
•
•
•
Establish the two-sided FBDs using the "natural coordinates". (Do NOT apply constraints
at this point.)
o Acceleration/inertial forces are always in the positive coordinate direction.
o Stiffness and damping forces always point in the negative coordinate direction.
o Cable end tensions will be equal and opposite. Cable tensions around a pulley are
(generally) not equal.
Always assemble the sum of forces/moments in the positive direction of the associated
coordinate/DOF. (Write in terms of the "natural coordinates" of the FDBs.)
Identify (the independent) constraints; then apply constraints to the sum of
forces/moments equations.
o Avoid creating negative constraints.
Combine equations to find final EOM by substituting into the equation associated with
the desired final coordinate/DOF.
For the final EOM, the effective M, C, and K value/expression(s) will be inherently nonnegative.
o Mass, damping, and stiffness are positive, semi-definite quantities.
In addition, don’t forget the key process…
•
•
•
(initially) ignore any potential constraints; instead, write/draw everything from the given
body's perspective (using each body's natural coordinate); start with the direct
connections to the current body, after that deal with the other end of the connection
(next) repeat the above for each body
(lastly) identify any kinematic constraints
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Newton Example A
In order to solve Newton’s Equation,
 F = mx ,
m
i
for the given single degree-of-freedom system, it
is first necessary to draw the complete free-body
diagram. This involves identifying all internal and
external forces acting upon the degree-of-freedom.
(Recall that the internal force resulting from the
spring is due to the actual relative motion of the
ends of the spring. Therefore the coordinate x is
measured from the free length of the spring.)
k
All the internal and external forces are
combined to yield the actual equation of
motion.
mx
=
m
kx
f(t)
c
f
f − kx − mg − cx = mx
Collecting all the coordinate-based terms
to one side of the equation and all the
externally applied forces to the other
yields:
x(t)
mg
m
cx
mx + cx + kx = f − mg
Since the mg term ends up on the righthand side with the external force, the coordinate x can
be redefined to be from the static equilibrium position. As a result, the mg term can be
eliminated as shown.
Evaluating the static equilibrium equation (i.e. x = 0 , x = 0 & f = 0 ) yields,
kx = −mg
Therefore, the static deflection (  ) can be evaluated as
=−
mg
k
The coordinate system reference can be shifted to the static equilibrium position by defining
xs = x − 
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Since the static deflection (  ) is constant, obviously the velocity and acceleration is unchanged
by the coordinate shift.
xs = x and xs = x
Substituting into the original solution,
mxs + cxs + k ( xs +  ) = f − mg
But recall that k  = −mg , therefore
mxs + cxs + kxs = f
Note: Often the solution solved about the static equilibrium position will eliminate the ‘mg’ term
when there are springs to carry the load. When this occurs, the ‘mg’ term will end up on the
right-hand side of the equation with the applied external forces.
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Newton Example B
Again, in order to solve Newton’s Equation,
 M = J  + L  mL ,
0
cg
for the given single degree-of-freedom system, it is
first necessary to draw the complete free-body
diagram. This involves identifying all internal and
external forces acting upon the degree-of-freedom. In
this case, there is no specific external moment applied
at point ‘o’. Also, recall the parallel axis theorem,
J 0 = J cg + mL2 . (See above for the general
L
m, J cg

expression.)
All the internal and external moments are combined to
form the actual equation of motion. Notice that neither
the support force ‘T’ nor the centripetal acceleration
force ‘ mL 2 ’ contribute to the moment about point ‘o’.
(However, ‘T’ would contribute to the moments about
the ‘cg’.)
T
=

mg
−mgL sin  = ( J cg + mL2 )
mL
mL
2
J cg
collecting all the coordinate-based terms to one side of the equation and all the externally applied
moments to the other yields
( J + mL ) + mgL sin  = 0
2
cg
This time, since the mg term ends up on the left-hand side (multiplied by the coordinate), it is
not possible to eliminate the mg term from the solution. So, the exact equation of motion
remains,
( J + mL ) + mgL sin = 0
2
cg
However, because the mgL sin  term is non-linear in the desired coordinate, it is possible to
reduce the solution to a linear approximation by using the first order Maclaurin Series
expansion, sin    for  1 , as follows.
( J + mL ) + mgL = 0
2
cg
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Enhanced 1-DOF Newton Example
System Figure:
f
x
k
θ
m
c
m1, Jcg
R
k1
Free Body Diagrams:
kx
f
T
=
mg N
Jcgθ
T
0
Rx
=
0
m 1g
Ry
k1Rθ
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Equations for Body #1:
 F ) − kx − cx + f − T = mx
 F ) N − mg = 0
x
y
Equations for Body #2:
 F ) R +T = 0
 F ) R − m g + k R = 0
 M ) − ( k R ) R + TR = J 
x
x
y
y
o
1
1
1
cg
Constraint Equation:
x = R
Solution:
(1) solve  M o of body #2 for T
T=
J cg
R
 + k1 R
(2) using constraint equation, convert to independent coordinate x
J
T = cg2 x + k1 x
R
(3) combine with  Fx for body #1
 J cg

−kx − cx + f −  2 x + k1 x  = mx
R

(4) identify Equation of Motion (EOM) by rearranging to collect dynamics coordinate on the left
and external forces on the right
J cg 

 m + 2  x + cx + ( k + k1 ) x = f
R 

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#3 – Homogeneous Solution of the Equation of Motion
For the single degree of freedom
(SDOF) system shown, the equation of
motion is:
f(t)
x(t)
mx(t ) + cx(t ) + k x(t ) = f (t )
k
m
c
This differential equation can be
solved by any of several methods,
including: Laplace Transforms,
Fourier Transforms, or by an assumed solution. In this case, assuming a solution appropriate for
a second-order, constant coefficient differential equation.
The general multi-degree of freedom problem can be formulated in matrix form as:
 M  x(t ) + C  x(t ) +  K  x(t ) =  f (t )
Using a solution appropriate for a linear, constant matrix coefficient, differential equation,
(namely complex exponentials), we have:
 x(t ) =  X  e st &  f (t ) = F  e st
NOTE: that X, F and s are complex scalars.
By focusing on the homogeneous solution (i.e. f = 0 ), we first evaluate the derivatives of the
assumed solution,  x(t ) =  X  e st :
 x(t ) =  x(t ) =  X  sest
d
 x(t ) =  x(t ) =  X  s 2e st
dt
d
dt
2
2
By substituting the assumed velocity and acceleration into the original matrix expression and
collecting terms, we have: (NOTE: e st is non-zero for all time and so may be eliminated.)
 M  s 2 + C  s +  K   X  = 0
Later in Section 10, you will recognize this as a form of eigenvalue problem (or a null space
solution involving s.) However, recognize that the solution  X  = 0 is the trivial solution and
does not contribute any useful information. Therefore, since the system matrix becomes rank
deficient for certain values of s, the problem reduces to a polynomial root solver. (The
determinant of the system matrix is zero.)
 M  s 2 + C  s +  K  = 0
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This determinant results in a polynomial whose roots are the poles of the system. This is known
as the characteristic polynomial. Often the determinant notation is dropped (but still written
equal to the scalar zero implying a determinant) and the matrix characteristic polynomial is
expressed as,
 M  s 2 + C  s +  K  = 0
Returning to the scalar (1 DOF) problem as initially presented, we have:
ms 2 + cs + k = 0
Since the determinant of a scalar is simply the scalar itself, the characteristic polynomial is
simply:
ms 2 + cs + k = 0
Since this is a quadratic equation, there are two roots, namely:
1,2 =
−c  c 2 − 4mk
2m
This results in the homogeneous (transient) solutions form of:
x(t ) = A1e1t + A2e2t
The characteristic behavior of this system solution is determined by the relationship of m, c, & k.
c 2  4mk
c 2 = 4mk
c 2  4mk
→
→
→
Two distinct real roots.
Two equal real roots.
One pair of complex conjugate roots.
For the condition of c 2  4mk , no oscillation occurs. This is called over-damped. Since no
oscillation occurs, this is not particularly interesting from a vibration point of view and we will
not investigate it any further.
For the condition of c 2 = 4mk , this is the boundary between oscillatory and non-oscillatory
behavior. This is called critically-damped. We will return to this later.
The interesting oscillatory behavior occurs when c 2  4mk . This is called under-damped. When
a system is under-damped, the system will respond to initial conditions with harmonic oscillatory
behavior.
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First, we recognize the roots of the characteristic equation as the system poles ( r ). For the given
SDOF system, the roots have the following form:
−c
k  c 
r = 1,2 =
j
−

2m
m  2m 
2
Examining the two parts (real and imaginary) of the poles, we can write:
r =  r  j r
Where,
−c
is the damping for mode r (for single degree-of-freedom systems, it is often
2m
written as  d ) and
r =
2
k  c 
r = 
−
 is the damped natural frequency for mode r (for single degree-ofm  2m 
freedom system, it is often written as  d )
Since 1 and 2 are complex conjugates, the homogeneous solution for the SDOF problem can
be written as:
x(t ) = Xe t + X *e  t
*
k
. This is called the undamped natural frequency, written as  r
m
(for single degree-of-freedom system, it is often written as  n .) This is the frequency at which
the system will oscillate if there is no damping in the system.
If we let c → 0 , then  r →
Returning to the critically damped condition, where c 2 = 4mk , we can now define a few more
vibration parameters.
Defining the critical damping as:
cc = 2 mk
Allows us to define the damping ratio as:
c
cc
We can now express the damping and damped natural frequency as functions of the undamped
natural frequency and damping ratio, as follows:
 =
r = −
System Dynamics & Vibrations - Lecture Notes
c
= − r  r
2m
-18-
02-May-2025 03:33 PM
2
k  c 
2
r = 
−
 =  1 −  r r
m  2m 
The single degree of freedom equation can now be written as:
x(t ) + 2 r  r x(t ) +  r2 x(t ) =
f (t )
m
While each of these parameters has been developed from a single degree of freedom (1 DOF)
point of view, the concepts are equally applicable to the multi-degree of freedom case, as well.
For this, we need to use a nomenclature that reflects the fact that there is more than one system
pole (with its conjugate). We do this by using the subscript r (indicating resonance) with each of
the modal parameters, as follows.
r =  r + j r is the system pole for mode r, where  r is the damping and  r is the damped
natural frequency for mode r. From this, we define r =  r2 +  r2 as the undamped natural

frequency for mode r. Then, the damping ratio is simply defined as,  r = − r . The general
r
homogeneous solution then takes on the following form.
 x(t ) =  X r e +  X *r e
r t
*r t
r
The outgrowth of this approach is that it is no longer necessary to know the actual mass, stiffness
and damping of the system in order to identify its modal characteristics. This is a significant
advantage when applied to experimental modal analysis because, as will be shown later, the
system poles ( r ’s) can be obtained directly from measured Frequency Response Functions.
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Homogeneous Example
Given a single degree-of-freedom mass/spring/damper system (MCK) with 5kg mass, 20 N s m
damping, and 1000 N m stiffness: identify the pole of the system ( r ), the undamped natural
frequency (  r ), and the damping ratio (  r ).
To calculate the system pole, evaluate the characteristic equation,
ms 2 + cs + k = 0
5s 2 + 20s + 1000 = 0
which has the following roots. Note that the roots of the characteristic equation are the system
poles.
−20 N s m
1000 N m  20 N s m 
r =
j
−

2  5kg
5kg
 2  5kg 
r = −2 rad s  j 200 rad s − ( 2 rad s )
2
2
2
2
r = −2  j14 rad s
The pole gives the damping (  r ) and the damped natural frequency (  r ) directly. Recall, that to
express the answer in units of Hertz, it is necessary to divide by 2 .
1
 r = −2 rad s -or- − Hz = −0.318Hz

7
 r = 14 rad s -or-  Hz = 2.228Hz

For this problem, the undamped natural frequency (  r ) may be calculated from either the
magnitude of the system pole, or by evaluating the characteristic equation for c = 0 . Both
approaches yield the same solution.
r =  r =  −2  j14 rad s =  200 rad s2 = 14.14 rad s -or- 2.251Hz
2
r = 
k
1000 N m
2
=
= 200 rad s2 = 14.14 rad s -or- 2.251Hz
m
5kg
Again, for this problem, the damping ratio may be calculated by either of two approaches, either
directly from the system pole or from the system’s mechanical properties (MCK).
r = −
r
r
=−
−2 rad s
= 0.1414 -or- 14.14%
14.14 rad s
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
r =
c
c
20 N s m
=
=
= 0.1414 -or- 14.14%
cc 2 mk 2 5kg 1000 N m
Recognize one important point in the solution method. The direct (MCK) solutions are only valid
for a single degree-of-freedom (SDOF) system while the solution methods based upon the
system pole are valid for a system with any number of degrees-of-freedom (MDOF).
* Note on Consistent Units
Solving the equations (particularly the frequency) must be done in a consistent units system. This
means that the set of units being used must satisfy the force-mass-acceleration unity relationship.
i.e. 1 force unit = 1 mass unit * 1 acceleration unit
For example:
• newton-kilogram-meter-second
• poundforce-slug-foot-second
1 N = 1 kg * 1 m/s2
1 lbf = 1 slug * 1 ft/s2
Note: While there are other consistent unit systems, most people (students and non-students)
have difficulty identifying the base unity relationship.
System Dynamics & Vibrations - Lecture Notes
-21-
02-May-2025 03:33 PM
Enhanced 1-DOF Homogeneous Example
Starting with the Equation of Motion (EOM) from the Enhanced 1-DOF Newton Example.
J cg 

 m + 2  x + cx + ( k + k1 ) x = f
R 

The characteristic equation becomes
J cg  2

 m + 2  s + cs + ( k + k1 ) = 0
R 

Assume some values for the physical properties:
m = 0.5 kg
J cg = 0.2 kg  m 2
R = 0.5 m
k = 100, 000 N m
k1 = 400, 000 N m
c = 80 N  s m
Substitute values into the characteristic equation
1.3s 2 + 80s + 500, 000 = 0
Solving for the roots of the characteristic equation
1,2 = −30.8  619.4 j rad s
Thus
r = 619.4 rad s  98.6 Hz
 r = −30.8 rad s
 r = r = 620.2 rad s  98.7 Hz
−
 r = r = 0.0496  5 %
r
Alternatively, using the SDOF relationships
k
500, 000
n =  =
=
= 620.17 rad s
m
1.3
cc = 2 k  m = 2 500, 000 1.3 = 1612.5 N  s m
c
80
 = =
= 0.0496
cc 1612.5
d = r =  1 −  2 = 620.17 1 − 0.04962 = 619.4 rad s
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
#4 – Estimating Modal Parameters – Time Domain
There are many techniques, both simple and advanced, for estimating modal parameters
(frequency, damping, mode shape, and scaling) from measured experimental data. This section
will focus upon the single degree-of-freedom time domain technique known as log decrement.
Log Decrement
The log decrement method is based entirely upon the characteristics of a single degree-offreedom free decay response.
4
3
2
Amplitude
1
0
-1
-2
-3
-4
0
1
2
3
4
5
6
7
8
Time (sec)
9
10
11
12
13
14
15
Recalling the form of solution, x(t ) = Xert , it is possible to estimate the system pole (
r =  r + j r ) directly from the observed response.
Estimating the damped natural frequency (  r ) is easy. First estimate the period of one oscillation
(  ), the reciprocal of the period, 1 , is the damped natural frequency (  ) in Hz.

r
 r = 2  rad s -or- 1 Hz
Estimating the damping (  r ) requires a little more effort. The development proceeds as follows.
Begin by evaluating the position, x (t ) , for two time instants one cycle apart.
x(t1 ) = Xert1 = Xe rt1 e jrt1
x(t2 ) = Xert2 = Xe rt2 e jrt2
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Next evaluate the log decrement (  ) as:
 x(t ) 
 Xe t e j t 
 = ln  1  = ln   t j t 
 Xe e

 x(t2 ) 
r1
r1
r 2
r 2
However, since the points are defined one cycle apart, t2 = t1 +  . Substituting into the above
expression yields,
 Xe rt1 e jrt1 
 = ln   r (t1+ ) jr (t1+ ) 
e
 Xe

Next, recall that e j = cos + j sin  . Hence,
e r( 1 ) = e
j t +
(
j r t1 + 2
r
) = e j t +2 = e j t e j 2 = e j (t )
r1
r 1
r1
So the expression for the log decrement can be reduced to
 e rt1 
 = ln   r (t1+ )  = ln e rt1 e− r (t1+ ) = ln ( e− r ) = − r
e

(
)
Recalling that  r = − r r ,  r = 1 −  r2 r and  r = 2
 . Substituting into the above
expression yields,
 = − r =  r  r =  r  r 2  =
r
Therefore the log decrement equals,
=
1 −  r
2
r
=
2 r
1 −  r2
Multiple Cycle Solution
2 r
1 −  r2
If the damping ratio is “small” (eg.   0.1 -or10%), then the above expression can be
simplified to:
If time points multiple (n) periods apart
are used, the expression for the log
decrement is modified as follows:
1  x(t ) 
2 r
 = ln  0  =
n  x(tn ) 
1 −  r2
where: n is the number of cycles.
Which simplifies to:
1  x(t ) 
 = ln  0   2 r
n  x(tn ) 
  2 r
(Error less than 1% when   0.1 )
System Dynamics & Vibrations - Lecture Notes
2 r  r
-24-
02-May-2025 03:33 PM
Log Decrement Example
Using the figure given above, estimate both the damped natural frequency and the damping ratio.
First, select two convenient time points to estimate the period (  ). Between t = 0 seconds and
t = 9 seconds, there are 11 cycles.
=
9sec
= 0.8182 sec cycle
11cycles
Therefore the damped natural frequency is
r =
11cycles
= 1.222 Hz -or- 7.679 rad s
9sec
Second, select two convenient time points to estimate the log decrement (  ). Between t  1.8
seconds and t  6.7 seconds, there are 6 cycles with x(t  1.8)  3 and x(t  6.7)  2 .
1  x(t )  1  x(t  1.8)  1  3 
 = ln  0  = ln  0
 = ln   = 0.0676
n  x(tn )  6  x(tn  6.7)  6  2 
Since the log decrement (  ) is so “small”, first try the simplified approximation for damping
ratio (  ).
   2 = 0.0676 2 = 0.0108 -or- 1.08%
(Error check:   0.1 so the approximation is adequate.)
Just for completeness, since   0.1 , 1 −  2  1 , therefore
r =
r
1−
2
r
  r = 7.679 rad s -or- 1.222Hz
 r = − r r = −0.0108  7.679 rad s = −0.0829 rad s -or- −0.0132Hz
System Dynamics & Vibrations - Lecture Notes
-25-
02-May-2025 03:33 PM
#5 – Lagrange’s Method
Lagrange’s Method - Overview
Lagrange’s Equation
d  T  T U D
+
+
= Fi

−
dt  qi  qi qi qi
Where:
• T = Kinetic Energy
• U = Potential Energy
• D = Dissipative Energy
• Fi = Externally applied force/moment
•
qi = Generalized coordinate
Generalized Coordinates
q = x, y, z, x , y , z
Kinetic Energy
N
Nm
j
1
1
T =  M i x 2 +  J i 2
i =1 2
i =1 2
Potential Energy1
N
Nk
Nm
kT
1
1
U =  Ki x 2 +  KTi  2 +   ( M i g ) x
i =1 2
i =1 2
i =1
Dissipative Energy
NC
N
CT
1
1
D =  Ci x 2 +  CTi  2
i =1 2
i =1 2
1
The sign (±) of the gravitational potential term is determined by whether or not a positive coordinate motion
produces a positive potential energy change.
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Lagrange’s Method - Details
Typical problems with Lagrange formulations:
• Be sure to establish number of degrees of freedom first and formulate all energy terms in
only those variables. Clearly identify which degrees of freedom are relative coordinates
versus absolute coordinates. Watch out for rotational/translational problems.
• For kinetic energy terms, be sure to formulate absolute velocities before taking
derivatives. Watch out for 2-D and 3-D vector motions.
• For potential energy terms, be sure that the actual deflection, described by relative and/or
absolute coordinates, in spring elements is described. Watch out for 2-D and 3-D vector
motions.
• There should be only one total kinetic energy equation, one total potential energy and one
total dissipative energy equation for the system. The kinetic, potential and dissipative
energy equations should involve only the N generalized coordinates and the constants
(mass, damping, stiffness) of the system.
• Apply the Lagrange Equation once for each generalized coordinate. For N degrees of
freedom, N generalized coordinates will yield N equations of motion.
• If necessary, linearize the equations of motion by neglecting nonlinear terms in the
equations of motion. Note that the linear equations of motion may not adequately
describe the original equations of motion.
NOTE: For the single degree-of-freedom (only one coordinate), conservative (no damping), unforced problem (no external forces), there is a simplified method called the Energy Method.
d
(T + U ) = 0
dt
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
* Sanity Checks
The following are a list of “sanity checks” that help define valid energy equations and also help
detect common mistakes.
•
•
•
Assemble the energy equation using the "natural coordinates". (Do NOT apply
constraints at this point.)
o All energy equations are inherently positive, except for the possible +/- associated
with gravity.
Identify (the independent) constraints; then apply constraints to the energy equations.
o Avoid creating negative constraints.
For the final EOM, the effective M, C, and K value/expression(s) will be inherently nonnegative.
o Mass, damping, and stiffness are positive, semi-definite quantities.
System Dynamics & Vibrations - Lecture Notes
-28-
02-May-2025 03:33 PM
Lagrange Example A
In order to solve the Lagrange Equation,
d  T  T U D
+
+
= Fi ,

−
dt  qi  qi qi qi
m
for the given single degree-of-freedom system, it is
first necessary to formulate the energy equations:
kinetic, potential and dissipative.
k
x(t)
f(t)
c
First, formulating the kinetic energy yields, T = 12 mx 2
Second, formulating the potential energy yields,
U = 12 kx 2 + mgx
Finally, formulating the dissipative energy yields, D = 12 cx 2
Next, it is necessary to evaluate each of the Lagrange terms,
d  T  d   1 2  d

 =  ( 2 mx )  = ( mx ) = mx
dt  x  dt  x
 dt
T  1 2
=
=0
2 mx
x x
U  1 2
=
2 kx + mgx = kx + mg
x x
D  1 2
=
= cx
2 cx
x x
(
(
)
)
(
)
Which are combined to form the actual equation of motion.
mx − 0 + kx + mg + cx = f
mx + cx + kx = f − mg
Since the mg term ends up on the right-hand side with the external force, the coordinate x can
be redefined to be from the static equilibrium position. As a result, the mg term is eliminated as
shown.
mx + cx + kx = f
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Lagrange Example B
Again, in order to solve the Lagrange Equation,
d  T  T U D
+
+
= Fi ,

−
dt  qi  qi qi qi
L
for the given single degree-of-freedom system, it is
necessary to formulate the energy equations: kinetic,
potential and dissipative.
m, J cg

First, formulating the kinetic energy yields,
( )
T = 12 J cg 2 + 12 m L
2
Second, formulating the potential energy yields, U = mgL (1 − cos )
Finally, formulating the dissipative energy yields, D = 0
Next, it is necessary to evaluate each of the Lagrange terms,
d  T  d  

= 
dt    dt  
( J  + m ( L ) )  = dtd ( J  + mL  ) = J  + mL 
T

=
J  + m ( L ) ) = 0
  (
1
2
2
cg
2
1
2
2
cg
2
1
2
cg
1
2
2
cg
2
U

=
( mgL (1 − cos ) ) = mgL sin
 
D 
=
( 0) = 0
 
Which are combined to form the actual equation of motion.
J cg + mL2 − 0 + mgL sin  + 0 = 0
This time, since the mg term ends up on the left-hand side (multiplied by the coordinate), it is
not possible to eliminate the mg term from the solution. So the exact equation of motion
remains,
( J + mL ) + mgL sin = 0
2
cg
However, because the mgL sin  term is non-linear in the desired coordinate, it is possible to
reduce the solution to a linear approximation by using sin    as follows.
System Dynamics & Vibrations - Lecture Notes
-30-
02-May-2025 03:33 PM
( J + mL ) + mgL = 0
2
cg
System Dynamics & Vibrations - Lecture Notes
-31-
02-May-2025 03:33 PM
Energy Example A
In order to solve the Energy Equation,
d
(T + U ) = 0 ,
dt
m
for the given single degree-of-freedom system, it is first
necessary to formulate the energy equations: kinetic and
potential.
x(t)
k
Formulating the kinetic energy yields, T = 12 mx 2 , and
formulating the potential energy yields, U = 12 kx 2 + mgx .
Evaluating the Energy equation yields,
d
d
(T + U ) = ( 12 mx 2 + 12 kx 2 + mgx ) = mxx + kxx + mgx = 0
dt
dt
Observe the odd form of the equation. There is an extra velocity term, x , that must be canceled
out to yield the actual equation of motion.
mx + kx + mg = 0
mx + kx = −mg
Since the mg term ends up on the right-hand side with the external force, the coordinate x can
be redefined to be from the static equilibrium position. As a result, the mg term is eliminated as
shown.
mx + kx = 0
NOTE: Comparing the solution generated by Lagrange Method to that of the Energy Method2
shows that the Energy Method is not simply a single degree-of-freedom Lagrange Solution, but
rather a related, by not equivalent technique. Recognize also that the application of the Energy
Method is extremely limited, i.e. single degree-of-freedom, conservative and un-forced. The
Lagrange Method has no such limitations and is in fact a completely general solution method,
comparable in usefulness to Newton’s Method.
2
Note that the Energy Method is presented for completeness. You may encounter it in your studies of other
reference material, but because it is not a general technique, it will not be used in this course.
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Enhanced 1-DOF Lagrange Example
System Figure:
f
x
k
θ
m
c
R
m1, Jcg
k1
Constraint Equation:
x = R
Lagrange’s Equation:
d  T  T U D
+
+
=F

−
dt  q  q q q
Solution:
(1) develop energy equations in terms of the given (natural) coordinates
1
1
Kinetic Energy : T = mx 2 + J cg 2
2
2
1
1
2
Potential Energy : U = kx 2 + k1 ( R )
2
2
1 2
Dissipative Energy : D = cx
2
(2) using the constraint equation, express the energy equations in terms of the desired
independent coordinate x
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
J cg 
1
1 J cg 2 1 
Kinetic Energy : T = mx 2 +
x =  m + 2  x2
2
2
2R
2
R 
1
1
1
Potential Energy : U = kx 2 + k1 x 2 = ( k + k1 ) x 2
2
2
2
1
Dissipative Energy : D = cx 2
2
(3) evaluate the partial derivatives for each of the energy equations
J cg 
J cg 
T 
d  T  
= m+ 2  x →

 = m+ 2  x
x 
R 
dt  x  
R 
T
=0
x
U
= ( k + k1 ) x
x
D = cx
(4) substituting the partials into the Lagrange equation
J cg 

 m + 2  x − 0 + ( k + k1 ) x + cx = f
R 

(5) identify Equation of Motion (EOM) by simplifying the Lagrange equation
J cg 

 m + 2  x + cx + ( k + k1 ) x = f
R 

Alternative Solution:
Alternatively, if the desired independent coordinate is  then solution step #1 is unchanged
(2) using the constraint equation, express the energy equations in terms of the desired
independent coordinate 
2
1
1
1
Kinetic Energy : T = m R + J cg 2 = mR 2 + J cg  2
2
2
2
1
1
1
2
2
Potential Energy : U = k ( R ) + k1 ( R ) = ( k + k1 ) R 2 2
2
2
2
1
Dissipative Energy : D = cR 2 2
2
( )
(
)
(3) evaluate the partial derivatives for each of the energy equations
T
d  T 
2
= mR 2 + J cg  →

 = mR + J cg 

dt   
(
)
System Dynamics & Vibrations - Lecture Notes
(
)
-34-
02-May-2025 03:33 PM
T
=0

U
= ( k + k1 ) R 2
x
D = cR 2
(4) substituting the partials into the Lagrange equation
( mR2 + J cg ) − 0 + ( k + k1 ) R2 + cR2 = moment
(5) identify Equation of Motion (EOM) by simplifying the Lagrange equation
( mR2 + J cg ) + cR2 + ( k + k1 ) R2 = moment
System Dynamics & Vibrations - Lecture Notes
-35-
02-May-2025 03:33 PM
Enhanced 1-DOF Kinematic Example
System Figure:
k1
θ1
R
m1, Jcg
x1
m
x
massless
Evaluating the actual rotation of the disc involves identifying the amount of cable motion.
x1
θ1
θ1
R
R
x1
System Dynamics & Vibrations - Lecture Notes
x1
-36-
x
02-May-2025 03:33 PM
Constraint Equation(s):
2x1
R
x = x1 + R1
1 =
Lagrange’s Equation:
d  T  T U D
+
+
=F

−
dt  q  q q q
Solution:
(1) develop energy equations in terms of the given (natural) coordinates
1
1
1
Kinetic Energy : T = M 1 x12 + J cg12 + mx 2
2
2
2
1
Potential Energy : U = k1 x12 − M1 gx1 − mgx
2
Dissipative Energy : D = 0
(2) using the constraint equations, express the energy equations in terms of the desired
independent coordinate x
2
2

1
1  M 4 J cg
 x 1
 2x  1
Kinetic Energy : T = M 1   + J cg   + mx 2 =  1 +
+ m  x2
2
2
2  9 9R
3 2
 3R  2

2
1  x
1k 
M

 x
Potential Energy : U = k1   − M 1 g   − mgx =  1  x 2 −  1 + m  gx
2 3
2 9 
3
 3

Dissipative Energy : D = 0
(3) evaluate the partial derivatives for each of the energy equations


T  M 1 4 J cg
d  T   M 1 4 J cg
=
+
+ m x →
+
+ m x

=
2
2
x  9 9 R
dt  x   9 9 R


T
=0
x
U k1
M

= x − 1 + m g
x 9
 3

D
=0
x
(4) substituting the partials into the Lagrange equation
 M 1 4 J cg

k
M

+
+ m x − 0 + 1 x − 1 + m g = f

2
9
 3

 9 9R

System Dynamics & Vibrations - Lecture Notes
-37-
02-May-2025 03:33 PM
(5) identify Equation of Motion (EOM) by simplifying the Lagrange equation and eliminating
the ‘mg’ terms on the right-hand side (due to static equilibrium offset)
 M 1 4 J cg

k
M

+
+ m x + 1 x = f +  1 + m g

2
9
 3

 9 9R

SE
or
 M 1 4 J cg

k
+
+ m x + 1 x = f

2
9
 9 9R

System Dynamics & Vibrations - Lecture Notes
-38-
02-May-2025 03:33 PM
#6 – Steady-State Solution of Equation of Motion
The general solution to the equation of motion involves both the homogeneous and the particular
solution. While the form of the homogeneous solution is dictated by the nature of the governing
second order constant matrix coefficient differential equation, the form of the particular solution
is controlled by the form of the forcing function (excitation).
Transfer Function - Overview
Mathematically, there are infinite possibilities, but for practical applications, the form can be
restricted to harmonic functions. By limiting the form of the excitation to harmonic functions, the
concept of steady-state solution can be defined. For vibration purposes, the steady-state solution
is the resulting response after all the initial condition transients have decayed. When solving the
response for the steady-state, the solution is most easily formulated in either the Laplace Domain
(s) or the Frequency Domain (  ). (NOTE: it will be shown later that the Frequency Domain
solution can be developed from the Laplace Domain by letting s = j .)
Working with a representative single degree-of –
freedom system (SDOF), the equation of motion is:
mx(t ) + cx(t ) + k x(t ) = f (t )
m
x(t)
f(t)
For a harmonic excitation, the forcing function can
be represented as:
f (t ) = Fe st
k
c
For the assumed solution, the response will have
the compatible form,
x(t ) = Xe st , x(t ) = sXe st and x(t ) = s 2 Xe st
where both F & X are complex scalars.
Substituting, the above solution forms into the differential equation of motion produces a
solution that must be valid for every value of s. This results in the following Laplace Domain
solution.
ms 2 Xe st + csXe st + kXe st = Fe st
By collecting common terms, the expression reduces to:
( ms + cs + k ) Xe = Fe
2
System Dynamics & Vibrations - Lecture Notes
st
-39-
st
02-May-2025 03:33 PM
At this point the response/excitation ratio can be
calculated. This ratio is called the Transfer
Function. It is normally written as H ( s ) .
H (s)
X
1
( s) = 2
F
ms + cs + k
Multi-Degree of Freedom
Transfer Function
The same basic development is
applicable to the multi-degree of freedom
system, as well.
Starting from the equation
While the Transfer Function is a convenient
 M   x(t ) + C  x(t ) +  K  x(t ) =  f (t )
mathematical model for defining the input-output
and combining with
relationship, well suited to controls applications,
 x(t ) =  X  e st &  f (t ) = F  e st
it is less appealing for vibrations purposes
yields the matrix equation
primarily because it a continuous mathematical
 M  s 2 + C  s +  K   X  e st = F  e st
expression. However, because the complex
surface defined by the H ( s ) expression is
By recognizing the basic definition of the
analytic, any slice through the surface contains all
Transfer Function, the expression
the information necessary to reconstruct the entire becomes,
−1
surface. This recognition leads to the use of the
H ( s ) =  M  s 2 + C  s +  K 
Fourier Transform, that while its mathematical
background is different, it is effectively the
Laplace Surface evaluated at s = j . (Another
argument for the equivalence of the Laplace Transform and the Fourier Transform lies in the fact
that both transforms start with the same Time Domain function, i.e. g (t ) = L−1 (G ( s)) = F −1 (G( ))
.)
Evaluating the Transfer Function at s = j yields another input-output form known as the
Frequency Response Function (FRF).
H ( )
Multi-Degree of Freedom
Frequency Response Function
X
1
( ) =
2
F
− m + j c + k
By contrast with the Transfer Function, the
Frequency Response Function is well suited for
vibration applications, particularly experimental
applications. While there is no discrete equivalent
to the Laplace Transform, there is a discrete
Fourier Transform that yields essentially the same
information as the continuous integral transform.
This is particularly important when working with
discrete, sampled time data.
Just as the Transfer Function has a multidegree of freedom form, so does the
Frequency Response Function.
By evaluating the multi-degree of
freedom (matrix) Transfer Function at
s = j , the multi-degree of freedom
Frequency Response Function results.
H ( ) =  − 2  M  + j C  +  K 
−1
By plotting both the Transfer Function and the Fourier Transform (over a limited range), the
equivalence of both transforms can be shown.
System Dynamics & Vibrations - Lecture Notes
-40-
02-May-2025 03:33 PM
One interesting aspect of the difference, however, is that while the Transfer Function requires the
plotting of two surfaces, the Frequency Response Function requires only two curves, clearly a
much more convenient representation.
1
0.9
0.8
1
0.7
Magnitude
Magnitude
0.8
0.6
0.4
0.2
0.6
0.5
0.4
0.3
0
10
0.2
5
0.1
1
0.5
0
-8
-6
-4
-8
-6
-4
-0.5
-10
Omega [rad/sec]
0
-10
0
-5
-1
-2
0
2
Omega [rad/sec]
4
6
8
10
Sigma [rad/sec]
4
3
4
2
1
Phase [rad]
Phase [rad]
2
0
0
-2
-1
-4
10
-2
5
-3
1
0.5
0
0
-5
Omega [rad/sec]
-4
-10
-0.5
-10
-1
Sigma [rad/sec]
-2
0
2
Omega [rad/sec]
4
6
8
10
Note that in addition to displacement per unit force input-output relationships, other steady-state
input-output relationships can be developed. Some of these alternative relationships will be
explored later as steady-state applications.
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
#7 – The Frequency Response Function (FRF)
In general, for most vibrations problems, the input/output relationship known as the Frequency
Response Function (FRF) will be used. This function describes the input/output relationship on a
frequency-by-frequency basis.
FRF - Overview
Although not generally referred to as a Frequency Response Function, the effect of a graphic
equalizer on the sound from a stereo system can be described by the frequency response function.
In this case, the input is the raw audio and the output is the modified audio. By adjusting the
amplification or attenuation (and relative phasing) of the various bands, the equalizer can give
the resulting audio a variety of sound characteristics, from flat response to enhancing male or
female voices to concert hall sound effects.
Recall that the Frequency Response Function can be expressed by evaluating the Transfer
Function at s = j .
H ( )
Multi-Degree of Freedom
Frequency Response Function
X
1
( ) =
2
F
− m + j c + k
By contrast with the Transfer Function, the
Frequency Response Function is well suited for
vibration applications, particularly experimental
applications. While there is no discrete equivalent
to the Laplace Transform, there is a discrete
Fourier Transform that yields essentially the same
information as the continuous integral transform.
This is particularly important when working with
discrete, sampled time data.
Just as the Transfer Function has a multidegree of freedom form, so does the
Frequency Response Function.
By evaluating the multi-degree of
freedom (matrix) Transfer Function at
s = j , the multi-degree of freedom
Frequency Response Function results.
H ( ) =  − 2  M  + j C  +  K 
−1
Input-Output Model
For the frequency response function, the following pictorial model is helpful. In fact, when used
for determining the steady state response of a system, its application is extremely simple. It is
merely a multiplication in the frequency domain.
X ( ) = H ( )  F ( )
System Dynamics & Vibrations - Lecture Notes
F ( )
-42-
H ( )
X ( )
02-May-2025 03:33 PM
Complex Exponential vs. Sine/Cosine Relationship
The relationship between the complex exponential form and the sine/cosine form of the forcing
function can be developed as follows:
Starting with the complex exponential form for the forcing function,
f (t ) = Fe jt + F *e− jt
and recalling the Euler identity,
ei = cos + i sin 
the relationship is easily developed. (Remember that F is complex.) Expanding the expression
for f (t ) ,
f (t ) = ( FR + jFI )(cos t + j sin t ) + ( FR − jFI )(cos t − j sin t )
Collecting real and imaginary terms yields,
f (t ) = ( FR cos t − FI sin t ) + j ( FI cos t + FR sin t ) + ( FR cos t − FI sin t ) − j ( FI cos t + FR sin t )
Observing that the imaginary terms cancel and that the real terms are equivalent yields,
f (t ) = 2 FR cos t − 2 FI sin t
Comparing with a traditional differential equation solution of f (t ) = A cos  t + B sin  t shows that
A = 2 FR and B = −2 FI . Despite the use of complex exponentials in the development, clearly the
resulting time domain waveform is real.
FRF Characteristics
m = 5 kg ; c = 20 Ns/m ; k = 1000 N/m
H ( ) = H * (− )
Because of this complex conjugate
relationship, usually only the positive
frequency part is plotted.
System Dynamics & Vibrations - Lecture Notes
90
0
-90
-2
10
Magnitude [m/N]
This figure presents a typical single
degree-of-freedom frequency response
function. The FRF is present two sided
(both positive and negative frequency)
to clearly show that the FRF for positive
and negative frequencies are complex
conjugates.
Phase [deg]
180
-3
10
-4
10
-5
10
-20
-43-
-15
-10
-5
0
5
Frequency [Hz]
10
15
20
02-May-2025 03:33 PM
Predicted Response3
It is important to remember that although the FRF is defined for all frequencies, the steady-state
response (output) for a linear system will always be at the same frequency as the excitation
(input). This prediction comes by manipulating the definition of the frequency response function,
as shown.
X ( ) = H ( )  F ( )
For a single degree-of-freedom system, the predicted steady-state response has the following
form.
X ( ) =
F ( )
−m + j c + k
2
Peak Response
Another important aspect of the frequency response function involves the frequency of
maximum response. It will be designated by  p .
Starting with the single degree-of-freedom frequency response function,
H ( ) =
1
−m + j c + k
2
it is possible to calculate the frequency of maximum (peak) response. Formulating the magnitude
of the frequency response yields,
H ( ) =
1
( k − m ) + (c )
2 2
2
In order to identify the frequency of maximum response, it is necessary to evaluate the derivative
of the frequency response magnitude with respect to frequency as follows.

d
d 
( H ( ) ) = d 
d





2
2
( k − m 2 ) + ( c ) 
1
TODO: Show example F(w), show resulting X(w). (FRF continuous, F&X discrete) Express F(w) = CONST ….
Show units of FRF = 1/k, compliance. Work example. Show MDOF FRF for comparison.
3
System Dynamics & Vibrations - Lecture Notes
-44-
02-May-2025 03:33 PM
1
− 
d
d 
2
2 2
( H ( ) ) = d  ( k − m ) + ( c ) 2 
d


)
(
)
(
(
3
−
2
d
1
d
2
2
2 2
H
(

)
=
−
k
−
m

+

c
k − m 2 ) + ( c )
( ) 2
(
)
(
)
(
d
2
d
)
(
)
3
−
2
d
1
d
d
2 2
2
H ( ) ) = − ( k − m 2 ) + ( c )
k − m 2 ) + 2 ( c )
( c ) 
(
(
 2 ( k − m )
d
2
d
d


) (
(
3
−
2
d
1
2
2
H ( ) ) = − ( k − m 2 ) + (c )
2 ( k − m 2 ) ( −2m ) + 2 (c ) c
(
d
2
)
In order to satisfy the maximum (or minimum), the derivative of the magnitude of the frequency
d
H ( p ) = 0 , must equal zero. Since only the second term in the above
response function,
d
relation can be equal to zero (except at  =  ), the solution can be reduced to solving the
following equation.
(
)
( 2 ( k − m ) ( −2m ) + 2 ( c ) c ) = 0
2
p
p
p
( −2 ( km − m  ) + c ) = 0
2
2
p
2
p
By dividing by the mass, m 2 , the expression can be simplified by recognizing that  2 =
2 =
k
,
m
c
and  p = 0 cannot be the peak response for all possible MCK combinations.
m
2
k
 c
−2  −  2p  +   = 0
m
 m
−2 (  2 −  2p ) + ( 2 ) = 0
2
 2 −  2p − 2 2 2 = 0
 2p = (1 − 2 2 ) 2
Therefore, the frequency of maximum (peak) response is as follows.
System Dynamics & Vibrations - Lecture Notes
-45-
02-May-2025 03:33 PM
 p = 1 − 2 2 
For comparison, recognize that the damped natural frequency is  r = 1 −  2 r .
Resonant Response
Starting with the definition of resonance,  =  , (i.e. the response of the system to an excitation
at the undamped natural frequency of the system.), the amplitude of the system response can be
calculated.
Starting with the single degree-of-freedom frequency response function,
H ( ) =
1
−m + j c + k
2
from the definition of resonance,  =  , (and k = m 2 ) the resonant response, H () , equals
H ( ) =
1
1
1
1
1
=
=
=
=
jc j cc j 2 km j 2 k j 2 k

If the damping is “small” (i.e.   0.1 ), then the peak response can be approximated as
H ( p )  H () =
1
j 2 k
Half Power Method
After defining the resonant frequency, it is
possible to estimate the damping,  , from the
shape of the frequency response function in the
vicinity of the resonance. Again, assuming that
the damping is “small” (i.e.   0.1 ), the halfpower bandwidth points can be calculated and
then damping can be estimated from the
bandwidth and the resonance location.
( )
H p
2
L  pH
Starting with the definition of the half-power
bandwidth points,
System Dynamics & Vibrations - Lecture Notes
( )
H p
-46-
02-May-2025 03:33 PM
H ( p ) = H (L ) = H (H )
1
2
and recognizing that for a lightly damped system,  p   , then expanding the single degree-offreedom frequency response yields,
1
1
1
=
k − m 2 + j c
2 j 2 k
Manipulating the expression yields a form from which the half-power bandwidth points may be
calculated.
1 1
=
2 2
1
1−

2

2
+ j
2 2 = 1 −
c
k
1
=
1−

2

2
+ j 2


2

+ j 2
2


Squaring both sides yields
2
 2 
2
2 4
2
8 = 1 − 2  + 4 2 2 = 1 − 2 2 + 4 + 4 2 2

 

  
2
 
 
2
2
  + ( 4 − 2 )   + (1 − 8 ) = 0


4
2
 
Solving for   yields,

2
2
− ( 4 2 − 2 ) 
 
  =

( 4 − 2 ) − 4 (1 − 8 )
2
2
2
2
 
2
2
  = 1 − 2  2 1 + 

 
2
Assuming  is small (i.e. 
approximated by,
1 ), then  2  0 , and the half-power bandwidth points are
 
   1  2

2
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Taking the difference between the points yields,
 H2 −  L2

2
  +  L   H −  L 
 H

  4
    
Using the approximation,  H +  L  2 , allows reducing the expression to,
 H − L 

  2
  
Therefore the damping factor can be approximated as,

H − L
2
This approximation is also often written as  
BW
f
or  
.
2 fc
2 fc
While this method appears particularly simple, in practice, there are several difficulties utilizing
it effectively. Specifically, when using real sampled data, the ability to determine the actual
magnitude and location of H ( p ) is limited by the measured data. The actual sampled data will
not in general include the exact frequency (  p ) and so both the location and magnitude will be
in error. Since the magnitude of H ( p ) is used to estimate the half-power bandwidth points, the
location of  L and  H will be in error. In general, the errors tend to be the following: under
estimating the magnitude of H ( p ) , under estimating the magnitude of the half-power points,
estimating  L too low, and estimating  H too high. The compound effect of these errors is that
generally the damping factor (  ) is estimated too high.
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
FRF Example A
Given a single degree-of-freedom (SDOF) system with m = 5kg , c = 20 N s m & k = 1000 N m ,
assuming a harmonic excitation of magnitude 20N @ 10Hz , determine the response of the
system to the given force.
Solution
First, identify the actual steady-state operating frequency.
 = 10Hz -or- 20 rad s = 62.832 rad s
Then, evaluate the SDOF frequency response function.
1
−m + j c + k
1
H (20 rad s ) =
2
rad
−5kg  ( 20 s ) + j 20 rad s  20 N s m + 1000 N m
1
H (20 rad s ) =
( −18,739 + j1, 257 ) N m
H ( ) =
2
H (20 rad s ) = ( −53.125 − j3.563) 10 −6 m N
Finally, determine the response in the frequency domain.
X (20
rad
X ( ) = H ( )  F ( )
−6 m
s ) = ( −53.125 − j 3.563 )  10
N  20 N
X (20 rad s ) = ( −1.063 − j 0.071) 10−3 m -or- 1.06 − 176.16 mm -or- 1.06 − 3.075rad mm
Therefore, the time domain response of the system to the given forcing function, f (t ) , is
expressed by4:
f (t ) = 20cos(20 t ) N
x(t ) = 1.06cos(20 t − 3.075)mm
4
TODO: Need to show equivalence of two sided FRF solution.
System Dynamics & Vibrations - Lecture Notes
-49-
02-May-2025 03:33 PM
#8 – Rotating Unbalance
For many situations, the source of excitation is an unbalance in a rotating machine part. This
internal (self) excitation is a function of the operation of the mechanism. However, the
techniques developed are applicable to this condition, as well.
By defining the position of the unbalance mass ( me ) as x(t ) + e(t ) and the mass of the system
moving in translation only as m − me , we can write the
equation of motion for this representative single degreeof-freedom system as:

e me
x(t)
me ( x(t ) + e (t )) + (m − me ) x(t ) + cx(t ) + k x(t ) = 0
m
Rearranging terms yields:
mx(t ) + cx(t ) + k x(t ) = −mee (t )
k
c
Comparing this solution with the previously developed
equation of motion shows that the expression mee (t ) is
equivalent in form to the original external forcing function f (t ) . As such we can use the same
harmonic function assumed solution form to convert this to a steady-state solution.
By using both,
x(t ) = Xe st , x(t ) = sXe st & x(t ) = s 2 Xe st
and
e(t ) = Ee st , e(t ) = sEe st & e (t ) = s 2 Ee st
where both X & E are complex scalars.
Substituting, the above solution forms into the differential equation of motion produces a
solution that must be valid for every value of s. This results in the following Laplace Domain
solution.
ms 2 Xest + csXe st + kXe st = −me s 2 Ee st
By collecting common terms, the expression reduces to:
( ms + cs + k ) Xe = −m Es e
2
st
2 st
e
Again, comparison with the Laplace Domain steady-state solution shows that F  − me Es 2 .
Calculation of the effective Transfer Function yields:
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
H ( s) =
X
−s 2
( s) = 2
me E
ms + cs + k
Or expressed in the Frequency Domain as:
H ( ) =
X
2
( ) =
me E
− 2m + j c + k
In the Frequency Domain, the effective forcing function is F  me E 2 . As can be seen from the
form, the magnitude of the force is linearly proportional to both the unbalance mass and
eccentricity, however, it is proportional to the square of the rotational frequency  . Because of
this, the unbalance is normally expressed in mass times length units ( mee ) and hence the above
frequency response form (displacement / unbalance).
9
8
10
7
8
Magnitude
Magnitude
6
6
4
2
5
4
3
0
10
5
2
1
0.5
0
1
0
-5
-0.5
-10
Omega [rad/sec]
-1
0
-10
Sigma [rad/sec]
-8
-6
-4
-2
0
2
Omega [rad/sec]
4
6
8
10
-8
-6
-4
-2
0
2
Omega [rad/sec]
4
6
8
10
4
3
4
2
1
Phase [rad]
Phase [rad]
2
0
-2
0
-1
-4
10
-2
5
1
0.5
0
Omega [rad/sec]
-3
0
-5
-0.5
-10
-1
-4
-10
Sigma [rad/sec]
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Rotating Example A
A machine of 100 kg mass has a 20 kg rotor with 5 mm eccentricity. The mounting springs have
k = 85 kN m and the damping is negligible. The operating speed is 600 rpm and the unit is
constrained to move vertically. (a) Determine the dynamic amplitude of the machine. (b)
Redesign the mounting so that the dynamic amplitude is reduced to one half of the original
value, but maintaining the same natural frequency.
Solution
Given the following parameters:
m = 100kg , me = 20kg , e = 5mm , k = 85 kN m , c
 = 600rpm

0 N s m &
x(t)
m
(a) Starting with the frequency response function previously
developed,
H ( ) =
e me
k
c
X
2
( ) =
mee
− 2m + j c + k
Solve for the dynamic amplitude X by rearranging the expression and evaluating its magnitude,
mee  2
X ( ) =
− 2 m + j c + k
The additional parameters needed are the unbalance and the rotational frequency in rad s .
mee = 20kg  5 10−3 m = 0.1kg  m
600rpm
 = 2
= 20 rad s = 62.83 rad s
60 s min
Evaluating the dynamic amplitude with c = 0 yields,
X =
mee  2
(0.1kg  m)(20 rad s ) 2
=
k − m 2 85,000 N m − 100kg (20 rad s ) 2
X =
394.78 kgm s2
= −0.001274m
−309,784 kg s2
System Dynamics & Vibrations - Lecture Notes
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or 1.274mm
02-May-2025 03:33 PM
(b) There are two options for reducing the dynamic amplitude by half and yet keeping the same
natural frequency, increasing mass and increasing damping.
X reduced = 12 X original = 0.6372mm
=
k
85,000 N m
=
= 29.15 rad s
m
100kg
Option 1: increasing mass.
Manipulating the expression for dynamic amplitude
X=
mee  2
mee  2
mee  2
=
=
k − m 2 m( mk −  2 ) m(2 −  2 )
yields an expression for the increased mass.
mee  2
(0.1kg  m)(20 rad s ) 2
m=
=
= 200kg
X (2 −  2 ) −0.6372mm((29.15 rad s ) 2 − (20 rad s ) 2 )
However, keeping the same  implies a corresponding increase in k . Therefore,
k = m 2 = 200kg (29.15 rad s )2 = 170,000 N m or 170 kN m
Notice how the solution makes sense intuitively. The operating speed is above the natural
X
frequency of the system. In that region, the frequency response function magnitude (
)
mee
1
, so it would be expected that to reduce the dynamic amplitude by half would
m
require approximately doubling the mass.
approaches
Option 2: increasing damping.
Again, manipulating the expression for dynamic amplitude
X =
mee  2
−m 2 + j c + k
X −m 2 + j c + k = mee  2
−m + j c + k =
2
System Dynamics & Vibrations - Lecture Notes
-53-
mee  2
X
02-May-2025 03:33 PM
( k − m ) + ( c) =
2 2
( c) 2 =
2
mee  2
2
2
X
mee  2
X
2
2
− ( k − m 2 )
2
yields an expression for the necessary damping
mee  2
X
c=
( 0.1kg  m(20
− ( k − m 2 )
2
2
rad
(0.006372m)
c=
2
2
s
2
)2 )
2
− ( 85,000 N m − 100kg (20 rad s ) 2 )
2
(20 rad s ) 2
155.85 103 kg m s4
2
2
2
− ( −309,784 kg s2 )
−9 2
383.87 109 kg s4 − 95.966 109 kg s4
406.0

10
m
c=
=
= 8,540 kg s
2
2
rad
rad
3,947.8 s2
3,947.8 s2
2
2
or more familiarly, c = 8,540 N s m . This can be expressed as a fraction of critical damping (
c
 = ) by recalling the expression for the critical damping value, cc = 2 km .
cc
 =
c
8,540 N s m
=
= 1.465 or 146.5%
2 km 2 85, 000 N m 100kg
Notice that to reduce the dynamic response of this system utilizing only damping requires an
over-damped condition of nearly 150%.
For this problem, the mass addition is most likely to be the easiest and cheapest practical
solution.
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
#9 – Steady-State Applications
In this section, two different but related steady-state applications will be developed. The first is
the force transmissibility relationship. For transmissibility, the interest is in determining the ratio
of the force transmitted to the base ( FT ( ) ) to the applied force ( F ( ) ) as a function of
frequency. The second application is displacement transmissibility. For this case, the interest is
in determining the ratio of the response of the system ( X ( ) ) to the motion of the base ( Y ( ) )
as a function of frequency. (This is sometimes referred to as response ratio.) Just as in the earlier
development of the Frequency Response Function (FRF), for both applications, the focus will be
upon harmonic excitation. A related concept, vibration isolation, is also addressed.
Force Transmissibility
FT
( ) relationship. We begin with the
F
equation of motion for our representative single degree-of -freedom system. (NOTE: we assume
negligible motion of the base.)
For force transmissibility, we want to determine the
mx(t ) + cx(t ) + k x(t ) = f (t )
m
In addition, we need the restraining force at the
other end of the spring/damper.
x(t)
f(t)
cx(t ) + k x(t ) = fT (t )
k
Applying the same transformations developed
earlier, namely that the forcing functions and
responses can be express as complex exponentials,
and collecting terms, we get:
( ms + cs + k ) Xe = Fe
2
and
st
c
fT(t)
st
( cs + k ) Xest = FT est
Finally, by taking the ratio of the second equation to the first, we get the desired transmissibility
relationship.
FT
cs + k
( s) = 2
F
ms + cs + k
Or expressed in the Frequency Domain as:
System Dynamics & Vibrations - Lecture Notes
-55-
02-May-2025 03:33 PM
FT
j c + k
( ) =
2
F
− m + j c + k
Displacement Transmissibility
X
( ) relationship. We begin with
Y
the equation of motion for our representative single degree-of –freedom system. (NOTE: this
time the base motion is NOT negligible, however there is no explicitly applied external force.)
Because there is motion for both the structure and the
base, we need to use the difference in motion across the
spring/damper.
m
x(t)
mx(t ) + c( x(t ) − y (t ) ) + k ( x(t ) − y (t ) ) = 0
For displacement transmissibility, we want to determine the
Transforming the equation as before, we get:
ms 2 Xe st + c( sXe st − sYe st ) + k ( Xe st − Ye st ) = 0
k
c
y(t)
Collecting terms, and moving the base motion to the
other side of the equation yields,
(ms 2 + cs + k ) Xe st = (cs + k )Ye st
Finally, by cross dividing the relevant terms, we get the desired isolation relationship.
X
cs + k
( s) =
2
Y
ms + cs + k
Or expressed in the Frequency Domain as:
X
j c + k
( ) =
2
Y
− m + j c + k
Vibration Isolation
Isolation is essentially the flip side of transmissibility; however, the emphasis is upon the
reduction in transmission. It is often expressed as a percent reduction.
ISO = 1 − TR
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Conclusion
Comparing the expression for transmissibility to the expression for isolation, we find that they
are identical. This interesting result implies that the process of determining the force transmitted
to the base is equivalent to determining the base motion transmitted to the system.
10
8
8
7
6
Magnitude
Magnitude
9
10
4
6
5
4
2
3
0
10
2
5
1
1
0.5
0
0
-5
0
-10
-0.5
-10
Omega [rad/sec]
-1
-8
-6
-4
-2
0
2
Omega [rad/sec]
4
6
8
10
-8
-6
-4
-2
0
2
Omega [rad/sec]
4
6
8
10
Sigma [rad/sec]
4
3
4
2
1
Phase [rad]
Phase [rad]
2
0
-2
0
-1
-2
-4
10
5
-3
1
0.5
0
0
-5
Omega [rad/sec]
-4
-10
-0.5
-10
-1
Sigma [rad/sec]
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Transmissibility Example A
A vertical single-cylinder diesel engine of 500 kg mass is mounted on springs with k = 200 kN m
and dampers with  = 0.2 . The rotating parts are well balanced. The mass of the equivalent
reciprocating parts is 10 kg and the stroke is 200 mm. Find the dynamic amplitude of the vertical
motion, the transmissibility, and the force transmitted to the foundation, if the engine is operated
at (a) 200 rpm; (b) 600 rpm.
Solution
Given the following parameters:
stroke
m = 500kg , mR = 10kg , stroke = 200mm , k = 200
 = 0.2
kN
m
x(t)
&
m
(a) 200 rpm
k
c
Starting with the single degree-of-freedom frequency
response function previously developed,
H ( ) =
X
1
( ) =
2
F
− m + j c + k
Solve for the dynamic amplitude X by rearranging the expression and evaluating its magnitude,
X ( ) =
F ( )
− m + j c + k
2
The additional parameters needed are the equivalent force, the damping and the operating
frequency expressed in rad s . From the mechanical unbalance development, the equivalent force
can be expressed as Feq = mee 2 . Recognizing that the reciprocating mass is the effective mass,
me = mR , and that the effective eccentricity is half the stroke, e = stroke , yields,
2
200rpm 20 rad
=  s = 20.94 rad s
60 s min
3
2
Feq = mee = 10kg  0.1m  (20.94 rad s ) 2 = 438.7 N
 = 2
c = 2 km = 2  0.2  200,000 N m  500kg = 4000 N s m
Evaluating the dynamic amplitude yields,
System Dynamics & Vibrations - Lecture Notes
-58-
02-May-2025 03:33 PM
X =
Feq
k − m + j c
2
X =
=
438.7 N
200,000 m − 500kg (20.94 rad s ) 2 + j (20.94 rad s )(4000 N s m)
N
438.7 N
= 0.0051m or 5.1mm
( −19,324 + j83, 776 ) kg s2
Evaluating the transmissibility yields,
TR =
k + j c
200,000 N m + j (20.94 rad s )(4000 N s m)
=
k − m 2 + j c
200,000 N m − 500kg (20.94 rad s ) 2 + j (20.94 rad s )(4000 N s m)
TR =
( 200,000 + j83,776 ) N m = 2.52
( −19,324 + j83,776 ) N m
From the definition of transmissibility, namely TR =
FT
, the force transmitted to the foundation
F
can be determined.
FT = TR  F = TR  Feq = 2.52  438.7 N = 1106 N
Notice that due to the characteristics of the system, the force transmitted to the foundation is
approximately 2.5 times the force applied to the system.
(b) 600 rpm
Again solving for the dynamic amplitude X using the rearranged expression and evaluating its
magnitude,
X ( ) =
F ( )
− m + j c + k
2
The additional parameters needed are the equivalent force, the damping and the operating
frequency expressed in rad s . From the mechanical unbalance development, the equivalent force
can be expressed as Feq = mee 2 . Recognizing that the reciprocating mass is the effective mass,
me = mR , and that the effective eccentricity is half the stroke, e = stroke , yields,
2
600rpm
= 20 rad s = 62.83 rad s
60 s min
Feq = mee 2 = 10kg  0.1m  (62.83 rad s ) 2 = 3,948 N
 = 2
c = 2 km = 2  0.2  200,000 N m  500kg = 4000 N s m
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Evaluating the dynamic amplitude yields,
X =
Feq
k − m + j c
2
X =
=
3,948 N
200,000 N m − 500kg (62.83 rad s ) 2 + j (62.83 rad s )(4000 N s m)
3,948 N
= 0.0022m or 2.2mm
( −1.774 106 + j 251.3 103 ) kg s2
Evaluating the transmissibility yields,
TR =
k + j c
200,000 N m + j (20.94 rad s )(4000 N s m)
=
k − m 2 + j c 200,000 N m − 500kg (62.83 rad s ) 2 + j (62.83 rad s )(4000 N s m)
( 200,000 + j 251.3 10 )
3
TR =
N
m
( −1.774 10 + j 251.3 10 )
6
3
From the definition of transmissibility, namely TR =
N
= 0.179
m
FT
, the force transmitted to the foundation
F
can be determined.
FT = TR  F = TR  Feq = 0.179  3,948N = 708N
Notice that although the equivalent applied force is 900% larger than for the first case, the force
actually transmitted to the foundation is about 30% smaller.
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
#10 – The Eigenvalue Problem
In many areas of engineering, a mathematical concept referred to as the eigenvalue problem
arises. While the general application of eigenvalue methods has been used to solve a wide variety
of mathematical, scientific and engineering problems, in this course, we will focus upon its
application to the area of system dynamics and vibrations. Because of this, we will not examine
the eigenvalue/eigenvector problem in detail, but only its most basic characteristics.
Mathematically, the eigenvalue problem can be expressed as:
 A x =   x
In other words, given a matrix  A , the object is to find a vector  x that when multiplied times
the matrix  A does not change direction, but only magnitude (and possibly sense.) If this vector
can be found, it is referred to as an eigenvector of  A , and the scalar  , which describes the
change in magnitude and sense, is referred to as an eigenvalue of  A . (Note that in general,
there will be as many eigenvalue/eigenvector pairs as the size of the matrix. Also, it is only
defined for square matrices.)
The solution is essentially a two-step process. First identify the eigenvalues of the matrix, and
then solve for their associated eigenvectors.
Solving for the eigenvalues involves manipulating the equation form as follows.
 A x −   x = 0
 A x −   I  x = 0
 A −   I   x = 0
At this point, we recognize the form as a null space solution problem. There are three ways in
which the equation can be satisfied, two of them,  A −   I  =  0 and  x = 0 , are trivial
solutions and are rejected. The third is that the matrix  A −   I  is singular (or rank deficient)
for certain values of  . These values of  are the eigenvalues of the matrix  A . To solve for
the  ’s, recognize that a singular (or rank deficient) matrix has a determinant of zero.
 A −   I  = 0
Expanding this determinant results in a polynomial, known as the characteristic polynomial,
whose roots are the eigenvalues  .
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
After identifying the eigenvalues, go back and solve the linear equations,  A −   I   x = 0 ,
for each eigenvalue  . However there is a problem, since the coefficient matrix is singular (or
rank deficient) when evaluated at the eigenvalues, we cannot simply solve the equation set. In
general, the matrix will be rank deficient by one, indicating that one variable is free, so the most
straight-forward solution approach is to simply set one of the vector  x elements equal to one
and solve as normal. (We will not get into the issues involved in the repeated root problem. In
other words, where two or more of the eigenvalues  are equal.)
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Eigenvalue/Eigenvector Example
Given the following matrix, find its associated eigenvalues and eigenvectors.
 4 −1

 −2 5 
 A = 
First identify the eigenvalues,
 4 −1
1 0   4 −  −1 
−


0 1  =  −2 5 −   = 0
 −2 5 

 

 4 −  −1 
 −2 5 −   = (4 −  )(5 −  ) − (−1)(−2) = 0


 2 − 9 + 18 = 0
 A −   I  = 
Which has the roots, 1 = 3 and 2 = 6 .
Next identify the eigenvector associated with each eigenvalue.
 A −   I   x = 0
  4 −1
1 0    x1   4 −  −1   x1  0 
− 


   = 
 =  
0 1    x 2   −2 5 −    x 2  0 
  −2 5 
For the first eigenvalue, 1 = 3 ,
−1   x1  0 
 4 − 1
 −2 5 −    x  = 0 
 

1  2 
 4 − 3 −1   x1   1 −1  x1  0 
 −2 5 − 3  x  =  −2 2   x  = 0 

 2 
 2  
Notice the rank deficient nature of the resulting coefficient matrix.
Set one element of the vector  x equal to one (unity) to solve. For this example, we will use the
first element. (Note that we could actually set it equal to any arbitrary value except zero and still
solve. Occasionally an eigenvector will contain a zero, but we cannot arbitrarily pick that value.
Finally, if after picking a value, the resulting equations cannot be solved, simply try again
picking a different element.)
Letting x1 = 1 ,
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
 1 −1  1  0 
 −2 2   x  = 0 

 2  
We can solve either equation for x2 . Let’s pick the first for simplicity.
1
1 −1  x  = 1 − x2 = 0
 2
Which yields, x2 = 1 . Therefore, the first eigenvalue/eigenvector pair is:
1
1
1 = 3 ,  x1 =  
Repeating the process for the second eigenvalue, 2 = 6 , yields,
−1   x1  0 
 4 − 2
 = 
 −2
5 − 2   x 2  0 

 4 − 6 −1   x1   −2 −1  x1  0 
 −2 5 − 6   x  =  −2 −1  x  = 0 

 2 
 2  
Again letting x1 = 1 , and solving for x2 , yields,
 −2 −1  1  0 
 −2 −1  x  = 0 

 2  
1
 −2 −1  x  = −2 − x2 = 0
 2
Which yields, x2 = −2 . Therefore, the second eigenvalue/eigenvector pair is:
1

−2 
2 = 6 ,  x2 = 
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Final Notes
Looking at the form of the eigenvalue problem one more time, it can be shown that it can
actually be used to solve certain first order differential equations directly, those of the form
y ' = Ay . Assuming that the vector  x is of the form  x(t ) =  X  et , then the eigenvalue
problem can be written as:
 A x =   x
 A x(t ) =   x(t )
 A  X  e  t =  X   e  t
 A X  =   X 
which is really no different than the example problem solved above.
There are a few issues, which will be covered later, applying this method to the second order
differential equations that arise in structural dynamics, specifically the need to use state-space
expansion to convert the set second order differential equations to a larger system of first order
differential equations.
Finally, when using MATLAB to solve eigenvalue problems, while the eigenvalues should be
the same, the eigenvector will in general be scaled differently. MATLAB uses unity vector
length for its scaling.
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
#11 – Multiple Degree of Freedom Systems
MDOF Example
k1
Starting with the simple two degree-of-freedom system
shown, the objective will be to determine the equations
of motion and the natural frequency and mode shapes.
m1
The equations of motion can be found by either of two
techniques, Newton’s Method or Lagrange’s Method.
k2
x1 (t )
f1 (t )
x2 (t )
f 2 (t )
Equations of Motion – Newton’s Method
As has been shown, Newton’s Method involves
identifying all forces, both internal and external, acting
on each body and then expressing those forces in terms
of the given coordinates. Therefore, the first action must
be the development of the free body diagrams for each
body.
The given two degree-of-freedom system has
the two free body diagrams, show at right.
f1
Evaluating Newton’s equation,  fi = ma ,
m2
=
m1
i
for each body yields a set of equations
describing the motion of the system.
m1 g
f1 − k1 x1 − k2 ( x1 − x2 ) + m1 g = m1 x1
f 2 − k2 ( x2 − x1 ) + m2 g = m2 x2
body #1
k1 x1
k2 ( x1 − x2 )
m1
m1 x1
k2 ( x2 − x1 )
Collecting and rearranging the terms into
standard form yields,
=
m2
m1 x1 + ( k1 + k2 ) x1 − k2 x2 = f1 + m1 g
m2 g
m2 x2 + k2 x2 − k2 x1 = f 2 + m2 g
f2
body #2
m2
m2 x2
Or expressed in matrix form as,
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
 m1 0   x1   k1 + k2
 0 m   x  +  −k

2 2

2
−k2   x1   f1   m1g 
  =  +

k2   x2   f 2  m2 g 
Recall that when the gravitational term ( mi g ) ends up on the force side of the equation, the
associated coordinate ( xi ) can be redefined from the static equilibrium position thereby
eliminating the mi g term from the equation. In this case, both terms are eliminated.
 m1 0   x1   k1 + k2
 0 m   x  +  −k

2 2

2
−k2   x1   f1 
 = 
k2   x2   f 2 
Equations of Motion – Lagrange’s Method
As has been shown, Lagrange’s Method involves identifying the kinetic, potential and dissipative
energies for the entire system, then evaluating the Lagrange Equation for each degree of
freedom.
d  T  T U D
+
+
= Fi

−
dt  qi  qi qi qi
First, identifying the kinetic energy term,
T = 12 m1x12 + 12 m2 x22
And the potential energy term,
U = 12 k1x12 + 12 k2 ( x2 − x1 ) − m1gx1 − m2 gx2
2
Provides all the information necessary to determine the equations of motion. Note that there are
no dissipative terms for this problem, so D = 0 .
Second, evaluating the various partial derivatives for each degree of freedom, yields,
(First degree of freedom)
d  T  d   1
d
2
2 
1
( m1x1 ) = m1x1

= 
2 m1 x1 + 2 m2 x2  =
dt  x1  dt  x1
 dt
U
 1 2 1
2
=
2 k1 x1 + 2 k 2 ( x2 − x1 ) − m1 gx1 − m2 gx2 = k1 x1 + k 2 ( x2 − x1 )( −1) − m1 g
x1 x1
(
)
(
System Dynamics & Vibrations - Lecture Notes
)
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02-May-2025 03:33 PM
(Second degree of freedom)
d  T  d  

= 
dt  x2  dt  x2
U

=
x2 x2

( m x + m x )  = dtd ( m x ) = m x
1
2
2
1 1
2
2 2
1
2
2 2

2 2
( k x + k ( x − x ) − m gx − m gx ) = k ( x − x ) − m g
1
2
2
1 1
1
2
2
2
2
1
1
1
2
2
2
2
1
2
Finally, collecting the various terms for each degree of freedom yields the equations of motion.
m1 x1 + ( k1 + k2 ) x1 − k2 x2 = f1 + m1 g
m2 x2 + k2 x2 − k2 x1 = f 2 + m2 g
Or expressed in matrix form as,
 m1 0   x1   k1 + k2
 0 m   x  +  −k

2 2

2
−k2   x1   f1   m1g 
  =  +

k2   x2   f 2  m2 g 
Again recall, when the gravitational term ( mi g ) ends up on the force side of the equation, then
the associated coordinate ( xi ) can be redefined from the static equilibrium position thereby
eliminating the mi g term from the equation. In this case, both terms are eliminated.
 m1 0   x1   k1 + k2
 0 m   x  +  −k

2 2

2
−k2   x1   f1 
 = 
k2   x2   f 2 
Notice that the actual equations of motion identified are identical, regardless of whether
Newton’s Method or Lagrange’s Method was used.
Natural Frequency and Mode Shape
In order to identify the natural frequency and mode shape for a set of simultaneous differential
equations of motion, it is necessary to solve an eigenvalue problem. By transforming the set of
equations to the Laplace Domain ( s ), the differential equations of motion are transformed into a
set of simultaneous algebraic equations.
The transformation to the Laplace Domain can be easily accomplished by using an assumed
solution approach.
x ( t ) = Xe st & f ( t ) = Fe st
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Taking derivatives of the assumed solutions yields,
x ( t ) = sXe st & x ( t ) = s 2 Xe st
Recognize that the vector form of the assumed solution is simply,
 x =  X  e st &  f  = F  e st
 x = s  X  e st &  x = s 2  X  est
Substituting into the equation of motion yields,
 m1 0  2  X 1  st  k1 + k2
 0 m  s  X  e +  −k

2
 2

2
−k2   X 1  st  F1  st
 e =  e
k2   X 2 
 F2 
Because e st  0 for all time, the solution reduces to,
 m1 0  2  X 1   k1 + k2
 0 m  s  X  +  −k

2
 2 
2
−k2   X 1   F1 
 = 
k2   X 2   F2 
Since the objective is to calculate natural frequency and mode shape, it is necessary to solve the
homogeneous portion of the solution. Collecting terms and setting the force side of the equation
to zero yields,
  m1 0  2  k1 + k2

 s +  −k

2
  0 m2 
− k 2    X 1  0 
  =  
k2    X 2  0 
The solution is now in one of the eigenvalue/eigenvector forms. The simplest solution approach
is to recognize that in order for the equation to be valid, either of two conditions must hold. First,
the vector  X  is identically zero (the trivial solution) or second, the determinant of the
coefficient matrix is zero (i.e. the coefficient matrix is singular for certain values of s .) The
second is the only useful solution and will be pursued here.
Setting the determinant of the coefficient matrix to zero yields,
 m1 0  2  k1 + k2
 0 m  s +  −k

2

2
−k2 
=0
k2 
This determinant is called the characteristic polynomial and the solution effectively becomes the
problem of finding the roots of the characteristic polynomial.
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
 m1s 2

 0
0   k1 + k2
+
m2 s 2   −k2
 m1s 2 + k1 + k2

− k2

−k2 
=0
k2 
− k2 
 =0
m2 s 2 + k2 
( m s + k + k )( m s + k ) − ( −k )( −k ) = 0
2
2
1
1
2
2
2
2
2
( m1m2 ) s + ( m1k2 + m2k1 + m2k2 ) s + ( k1k2 ) = 0
4
2
In general, the characteristic polynomial is complete (has all terms), but because this example
problem does not include damping, all the odd powers of s are eliminated. This makes finding
the roots ( r ) of the equation somewhat easier.
 =
2
r
− ( m1k2 + m2k1 + m2k2 ) 
( m1k2 + m2k1 + m2k2 ) − 4 ( m1m2 )( k1k2 )
2 ( m1m2 )
2
At this point, continued manipulation of the arbitrary variables does not contribute to the
understanding of the solution, but only proves pure obstinacy.
Returning to the eigenvalue/eigenvector problem above and substituting some numerical values
should prove more interesting. Let m1 = 2 kg , m2 = 1 kg , k1 = 3 N m and k2 = 2 N m .
  2 0  2 3 + 2 −2    X 1  0 

 s +  −2
   X  = 0 
0
1
2



  2   

Evaluating the characteristic polynomial yields,
( 2s + 5)( s + 2) − ( −2)( −2) = 0
( 2s + 9s + 6 ) = 0
2
2
4
2
Which has the solution,
 =
2
r
r2 
−9  92 − 4 ( 2 )( 6 )
2 ( 2)
=
−9  33
4
−9  5.745
2
 −0.814, −3.686 rad s 2
4
Finally, taking the square root of r2 yields the four roots of the original quartic equation.
r   j 0.814,  j 3.686  0.902 j, 1.920 j rad s
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Notice that the roots occur in complex conjugate pairs. This will always be the case for
physically real systems (i.e. mass, stiffness & damping are positive and real.)
Which means the natural frequencies are,
r = 0.902 &  1.920 rad s
Recall the original development of the general solution to the homogeneous problem. It was
stated then that the solution could be written as pairs of complex conjugate terms in the
following form,
 x(t ) =  ( Qr  r e t + Qr*  r e t )
# dof
*
r
*
r
r =1
(Ignore for the moment the Qr term. It is simply a scaling term to account for the arbitrary
scaling of  r that results from the eigenvalue solution. In essence, it is the constant of
integration for the differential equation.)
Now that the eigenvalues (natural frequencies) are known, it is possible to solve for the
eigenvectors (mode shapes). Because the solution is a sum of complex conjugate terms, it is only
necessary to solve for the positive frequency vectors.
Again, starting with the above homogeneous equation,
  2 0  2 3 + 2 −2    X 1  0 

 s +  −2
   X  = 0 
0
1
2



  2   

Substitute the roots ( r ) for s and expand the equations,
( 2 + 5) X − 2 X = 0
−2 X + (  + 2 ) X = 0
2
r
1
1
2
r
2
2
Since the values of r are chosen to make the equation set rank deficient, there is an arbitrary
choice of one value (either X 1 or X 2 .) For this example, choosing X 1 = 1 yields the following
equation set,
X2 =
− ( 2r2 + 5 )
−2
−2
X2 =
− ( r2 + 2 )
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Substituting the first root ( 1 = 0.902 j ) into either equation yields X 2 = 1.686 . Substituting the
second root ( 2 = 1.920 j ) into either equation yields X 2 = −1.186 .
Therefore, the natural frequencies and mode shapes for this two degree-of-freedom system are,
 1 

1.686 
1 = j 0.902 rad s
 1 = 
2 = j1.920 rad s
 2 = 
and

1 

−1.186 
plus, their complex conjugates.
(Final note: if the complete homogeneous (free response) solution was desired, the final step
would be to solve for the Qr ’s by evaluating the initial conditions  x(0) and  x(0) .)
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
Enhanced 2-DOF Lagrange Example
System Figure:
x1
k1
θ1
f
x2
a
R1
k4
k3
m1, J1
m2
r1
b
c4
k2
Constraint Equation:
x1 = r11
Lagrange’s Equation:
d  T  T U D
+
+
=F

−
dt  q  q q q
Solution:
(1) develop energy equations in terms of the given (natural) coordinates
1
1
1
Kinetic Energy : T = m1 x12 + J112 + m2 x22
2
2
2
1 2
Dissipative Energy : D = c4 x2
2
1
1
1
1
2
2
2
Potential Energy : U = k1 ( x1 + a1 ) + k2 ( x1 − b1 ) + k3 ( x2 − x1 ) + k4 x22
2
2
2
2
(2) using the constraint equation, express the energy equations in terms of the desired
independent coordinates x1 and x2
1
1 J1 2 1
Kinetic Energy : T = m1 x12 +
x + m2 x22
2 1
2
2 r1
2
1
Dissipative Energy : D = c4 x22
2
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
2
2
1  a
1  b
1
1
2
Potential Energy : U = k1 1 +  x12 + k2 1 −  x12 + k3 ( x2 − x1 ) + k4 x22
2  r1 
2  r1 
2
2
(3) evaluate the partial derivatives for each of the energy equations (for each of the independent
coordinates x1 and x2 )
J 
T 
=  m1 + 21  x1 →
x1 
r1 
J1 
d  T  

 =  m1 + 2  x1
dt  x1  
r1 
T
=0
x1
D
=0
x1
2
2
 b 
U   a 
=  k1 1 +  + k2 1 −   x1 + k3 ( x2 − x1 )( −1)
x1   r1 
 r1  

T
= m2 x2
x2
→
d  T 

 = m2 x2
dt  x2 
T
=0
x2
D
= c4 x2
x2
U
= k3 ( x2 − x1 )(1) + k4 x2
x2
(4) identify Equations of Motion (EOMs) by substituting the partials into the Lagrange equation
and simplifying
2
  a 2


 b
J1 


m
+
x
+
k
1
+
+
k
1
−
+
k
 1
 1  1


2
3 x1 − k3 x2 = f1

r12 
r1 
r1 





m2 x2 + c4 x2 + ( k3 + k4 ) x2 − k3 x1 = f 2
(5) expressed in matrix form
2
  a  2



 b
J1 



k
1
+
+
k
1
−
+
k
m
+
0


 1
  x1  0 0   x1   1 
2
3
2 

r
r
+
+
r






1


1

1 
0 c  x

 x


 2 
4 2



0
m2 
− k3


System Dynamics & Vibrations - Lecture Notes
-74-

  x1   f1 
  =  
  x2   f 2 
k
+
k
( 3 4 )
− k3
02-May-2025 03:33 PM
Numerical Solution:
Let:
m1 = m2 = 5 kg
R1 = 1 m
r1 = 0.25 m
a = 0.75 m
b = 0.5 m
J1 = 3 kg  m2
c4 = 2 N  s
m
k1 = k2 = k3 = k4 = 1000 N
m
The equations have the form:
 M  x + C  x +  K  x =  f 
53 0   x1  0 0   x1  18000 −1000  x1   f1 
 0 5   x  + 0 2   x  +  −1000 2000   x  =  f 

 2 
 2 
 2  2
The undamped problem can be solved by assuming [C] = 0
1 = 17.36 j rad s
0.4423
 1 = 

0.8969
2 = 20.93 j rad s
−0.1879 
 2 = 

 0.9822 
System Dynamics & Vibrations - Lecture Notes
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02-May-2025 03:33 PM
System Dynamics & Vibrations
Enhanced 2-DOF Non-Linear Example
Newton’s Method
System Figure:
c
k
m
x
θ
m1
Free Body Diagrams:
kx
=
mg
T
m 1L 2
T
m 1L
=
m1
m 1g
System Dynamics & Vibrations - Lecture Notes
-76-
02-May-2025 03:33 PM
System Dynamics & Vibrations
Note: the mass ‘m’ is constrained to move vertical only; and the reaction force ‘T’ is along the
pendulum rod only because the mass ‘m1’ is a point mass and thus its moment of inertia ‘Jcg’ is
zero.
Equations for Body #1:
 F ) mg − kx − cx + T cos ( ) = mx
x
Equations for Body #2:
 F ) − T cos ( ) + m g = m x − m L sin ( ) − m L cos ( )
 F ) − T sin ( ) = m L cos ( ) − m L sin ( )
2
x
1
y
1
1
1
1
2
1
Solution:
Determining the exact equations of motion.
(1) substitute  Fx of body #2 into  Fx for body #1; (thus eliminating T)
( m + m1 ) x − m1L sin ( ) − m1L 2 cos ( ) + cx + kx = ( m + m1 ) g
(2) combine body #2 equations by using − (  Fy ) cos ( ) + (  Fx ) sin ( ) ; (thus eliminating T)
−m1 x sin ( ) + m1 L + m1 g sin ( ) = 0
Determining the linearized equations of motion.
(3) linearize by using the Maclaurin series expansion of the trigonometric functions and then
eliminating any non-linear terms; ( sin ( )   ; cos ( )  1)
( m + m1 ) x + cx + kx = ( m + m1 ) g
m1L + m1 g = 0
(4) express the linear Equations of Motion (EOMs) in matrix form
0  ( m + m1 ) g 
( m + m1 ) 0   x   c 0   x   k
 + 


 + 
=

m1 L    0 0    0 m1 g  
0

 0
Note: the linear EOM is not representative of the real system. The EOMs are completely
decoupled!
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System Dynamics & Vibrations
Lagrange’s Method
System Figure:
c
k
m
x
θ
m1
Velocity Diagrams:
L
Note: each diagram represents the absolute velocity for that body.
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System Dynamics & Vibrations
Solution:
(1) develop energy equations in terms of the given (natural) coordinates
2
2
1
1
Kinetic Energy : T = mx 2 + m1  x − L sin ( ) + L cos ( ) 

2
2 
1
Dissipative Energy : D = cx 2
2
1 2
Potential Energy : U = kx − mgx − m1 g x − L (1 − cos ( ) )
2
(
) (
(
)
)
Determining the exact equations of motion.
(2) evaluate the partial derivatives for each of the energy equations (for each of the independent
coordinates x and  )
T
d  T 
2
= mx + m1 x − L sin ( ) →

 = ( m + m1 ) x − m1L sin ( ) − m1L cos ( )
x
dt  x 
T
=0
x
D
= cx
x
U
= kx − ( m + m1 ) g
x
(
)
T
= m1 x − L sin ( ) ( − L sin ( ) ) + m1 L cos ( ) ( L cos ( ) )

T
d  T 
2
→
= −m1 xL sin ( ) + m1 L2 →

 = −m1 xL sin ( ) + m1 L  − m1 x L cos ( )

dt   
T
= m1 x − L sin ( ) − L cos ( ) − m1L2 2 sin ( ) cos ( )

D
=0

U
= m1 gL sin ( )

(
)
(
)(
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)
)
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(3) identify Equations of Motion (EOMs) by substituting the partials into the Lagrange equation
and simplifying
( m + m1 ) x − m1L sin ( ) − m1L 2 cos ( ) + cx + kx = ( m + m1 ) g
; actual RHS = ( m + m1 ) g + f
−m1 Lx sin ( ) + m1 L  + m1 gL sin ( ) = 0
2
; actual RHS = Moment
Determining the linearized equations of motion.
(4) linearize by using the Maclaurin series expansion of the trigonometric functions and then
eliminating any non-linear terms; ( sin ( )   ; cos ( )  1)
( m + m1 ) x + cx + kx = ( m + m1 ) g
m1L2 + m1 gL = 0
(5) express the linear Equations of Motion (EOMs) in matrix form
0  ( m + m1 ) g 
0   x   c 0  x   k
( m + m1 )
+
 + 


=

2 
m1 L    0 0    0 m1 gL  
0

 0
Note: the linear EOM is not representative of the real system. The EOMs are completely
decoupled!
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System Dynamics & Vibrations
#12 – Estimating Modal Parameters – Frequency Domain
There are many techniques, both simple and advanced, for estimating modal parameters
(frequency, damping, mode shape, and scaling) from measured experimental data. This section
will focus upon the single degree-of-freedom frequency domain technique known as quadrature.
Quadrature
The quadrature method is based entirely upon the characteristics of a single degree-of-freedom
free frequency response function. The quadrature method has many advantages over the log
decrement technique presented in section five. In the frequency domain, many lightly damped,
multi degree-of-freedom systems behave much like a single degree-of-freedom system in the
region around a resonance. This allows the single degree-of-freedom techniques previously
developed, to be used to estimate the modal parameters for multi degree-of-freedom systems.
180
90
0
-90
2
10
1.5
Imaginary part [m/N]
Phase [deg]
Examining the principle characteristic of a lightly damped, single degree-of-freedom system at
resonance: the magnitude peaks, the phase passes through −90 , the imaginary part peaks, and
the real part passes through zero. These characteristics, in addition to the shape of the frequency
response in the vicinity of the resonance, can be used to estimate the natural frequency and
damping.
1
10
0
1
0.5
0
-0.5
-1
-1.5
1.5
-1
10
1
Real part [m/N]
Magnitude [m/N]
10
-2
10
-3
10
0.5
0
-0.5
-1
-1.5
-4
10
0
1
2
3
4
5
6
Frequency [Hz]
7
8
9
0
10
1
2
3
4
5
6
Frequency [Hz]
7
8
9
10
Specifically, recall the definition of resonance (excitation at the undamped natural frequency of
the system). At resonance, the frequency response for a single degree-of-freedom system reduces
to H ( ) = 1
. This is purely imaginary, implying that the response is 90 behind the
j c
forcing function. Therefore, the imaginary part of the frequency response is maximum and the
real part is zero. Also, from section eight, the half-power method may be used to estimate the
damping as  = f
.
2 fc
Examining a multi degree-of-freedom frequency response, many of the same characteristics can
be observed, in particular, the change of phase through resonance, the peak in the magnitude and
imaginary parts, and the zero crossing in the real part. From these characteristics, in addition to
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the shape of the frequency response in the vicinity of the resonance, the natural frequency and
damping can be estimated. But additionally, for multi degree-of-freedom systems, the mode
shape can also be estimated.
H11 .. H21 .. H31
H11 .. H21 .. H31
0.3
180
0.2
Imag [m/N]
Phase [deg]
90
0
0.1
0
-0.1
-90
-0.2
-180
0
2
4
6
8
10
0
12
0
4
6
8
10
12
0.3
10
H11
H21
H31
-1
10
H11
H21
H31
0.2
-2
Real [m/N]
Magnitude [m/N]
2
10
-3
10
0.1
0
-0.1
-4
10
-0.2
0
2
4
6
Frequency [Hz]
8
10
0
12
2
4
6
Frequency [Hz]
8
10
12
While the mathematical form of a multi degree-of-freedom frequency response function is more
complicated than for a single degree-of-freedom, it is still possible in many situations to
reasonably apply single degree-of-freedom theory. Notice that there are several differences
between a single and multiple degree-of-freedom frequency response function. One important
difference is that there is now more than one frequency response function for the system.
Additionally, notice that there are now multiple peaks and the phase is no longer restricted to the
range 0, −180 . As a result, the phase at resonance can be 90 . (It must still lose phase
through resonance, though.)
The reason that there are multiple frequency response
functions is easy to show. Simply recall the definition of
the frequency response function, H ( )  X ( )
k1
c1
F ( ) .
For the given figure, there are force (input) locations and
two response (output) locations. It is then clear that there
m1
are four possible frequency response functions that can
be measured (or calculated.) Therefore, the expression
for the frequency response must be extended to express
x1 (t )
f1 (t )
which force/response pair is being presented. The
k
c
2
2
frequency response function for multi degree-ofX ( )
freedom system is written as H pq ( )  p
,
Fq ( )
m2
where p is the output (response) degree of freedom and
q is the input (excitation) degree of freedom. Using this
definition, the multi degree-of-freedom frequency
x2 (t )
f 2 (t )
response functions plotted above make more sense. The
plots represent a three degree-of-freedom system. If all the FRFs were plotted, there would be
nine FRFs displayed, however only the three FRFs associated with the first input degree-offreedom were actually shown, hence the notation used in the legend ( H11 , H 21 , H 31 ).
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Expanding the region around the third peak allows a more clear observation that the multi
degree-of-freedom frequency response shown looks much like a single degree-of-freedom in the
vicinity of the resonance.
H11 .. H21 .. H31
H11 .. H21 .. H31
180
0.03
0.02
Imag [m/N]
Phase [deg]
90
0
0.01
0
-0.01
-90
-0.02
-180
8
8.2
8.4
8.6
8.8
9
9.2
9.4
9.6
9.8
-0.03
10
-1
8.2
8.4
8.6
8.8
9
9.2
9.4
9.6
9.8
10
0.03
H11
H21
H31
-2
10
H11
H21
H31
0.02
Real [m/N]
Magnitude [m/N]
10
8
-3
10
-4
0.01
0
-0.01
10
-0.02
-5
10
8
8.2
8.4
8.6
8.8
9
9.2
Frequency [Hz]
9.4
9.6
9.8
10
-0.03
8
8.2
8.4
8.6
8.8
9
9.2
Frequency [Hz]
9.4
9.6
9.8
10
To estimate the modal parameters:
• Natural Frequency – To identify the natural frequency (  r ), it is first necessary to
identify one of the following:
o The 90 phase change location (frequency)
o The peak imaginary response location (frequency)
o The zero real response location (frequency)
• Damping – To estimate the damping, take the peak FRF magnitude, multiply by the
square root of two ( 12 H ( ) ), find the locations (frequencies), one above and one
•
below, that have that FRF magnitude, the locations are called the half-power points.
Divide the difference of the two frequencies by twice the peak frequency (natural
frequency).  r = f
2 fc
Mode Shape – To estimate the mode shape, either
o Take the value of the peak imaginary part (at the natural frequency) for each
output (response) location keeping the input (force) location fixed.
– OR –
o Take the value of the peak imaginary part (at the natural frequency) for each input
(force) location keeping the output (response) location fixed.
For the third mode shown above:
The natural frequency is estimated, using the peak imaginary response, as 9.068 Hz.
3 = 9.068Hz
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The peak FRF magnitude equals 0.0204 m/N, therefore the half-power magnitude would be
0.0144 m/N. So the half-power points are estimated as 9.0383 Hz and 9.0982 Hz, which result in
an estimate of damping ratio of 0.0033.
3 =
f 9.0982 Hz − 9.0383Hz
=
= 0.0033 -or- 0.33%
2 fc
2 ( 9.068Hz )
The peak imaginary values, for each output (response) location for the first input (force) location,
are estimated as [-0.0164 0.0204 –0.0091] m/N.
 H11 = -0.0164 m N ,  H 21 = 0.0204 m N &  H 31 = -0.0091m N
Therefore the mode shape is estimated as,
−0.80 
 1.00 
−0.0164 




 3 =  0.0204  -or-  1.00  -or- −1.24
 −0.0091
 −0.45
 0.55 






So the modal parameter estimate for the third mode is:
 1.00 


3 = 9.068Hz ,  3 = 0.33% &  3 = −1.24
 0.55 


NOTE: By convention, modes are numbered from lowest frequency to highest frequency.
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System Dynamics & Vibrations
#13 – Transient Solution of Equation of Motion
The last piece of the vibration problem is to solve for the transient solution of the differential
equation of motion. In general, since the excitation (forcing) function can be literally anything, a
closed form solution is not practical and numerical techniques are used. (The numerical methods
will be discussed in the next section.) However, a study of the general solution for certain forcing
functions is important to the basic understanding of the complete solution for the equation of
motion.
Impulse Response
The impulse response is the response of the
system to an impulsive excitation. An impulse
is frequently a large magnitude, short duration
force.
f (t )
F̂
The impulse ( F̂ ) of a force is defined as
 f (t )dt .
Fˆ



t
 f (t )dt
If the impulse is unity ( Fˆ = 1 ) and the duration of the pulse (  ) is allowed to approach zero (
 → 0 ) then the force ( f (t ) ) is referred to as a unit impulse. The unit impulse is equivalent to
the mathematical expression called a unit delta function (  (t ) ). The delta function is defined by:
when t  
 (t −  ) = 0

  (t −  )dt = 1
when 0    
0
Further, the delta function has a number of interesting and useful properties. Of particular
interest is the result of multiplication by another function.

 f (t ) (t −  )dt = f ( )
0
Recalling the impulse/momentum relationship ( f (t )dt = m dv ) and integrating both sides shows
that an impulse acting upon a mass results in a change in velocity ( v ) without a significant
change in position. Therefore,
v = F̂
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The free response of a single degree-of-freedom system to initial conditions is given by:
 x(0) −  r x(0)

x(t ) = e rt 
sin( r t ) + x(0)cos( r t ) 
r


Assuming the system is initially at rest, then combining the free response with the change in
momentum yields,
x(t ) =
Fˆ  rt
ˆ (t )
e sin( r t )  Fh
m r
where h(t ) =
1  rt
e sin( r t )
m r
The unit impulse response function ( h(t ) ) is important for the analysis of transients!
Arbitrary Excitation
Having developed the unit impulse response (
h(t ) ), it is now possible to establish the
response ( x (t ) ) of a system to an arbitrary
excitation
( f (t ) ).
f (t )
First identify the strength of the impulse at
time  ( Fˆ = f ( ) ) and its contribution to
the response ( x (t ) )) at time
t ( x(t ) = f ( ) h(t −  ) ).
Recognized that t is constant for the
above expressions, it is  that is
f (t )
variable.


t
f ( )
Because the system is linear, the
f ( )  h(t −  )
principal of superposition applies and

t −
it is possible to add up all the impulse
contributions. Adding up all the
 =t
contributions of the individual
impulses yields the expression for the
response of a system to an arbitrary excitation, (assuming zero initial conditions.)
t
x(t ) =  f ( )h(t −  )d
0
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This expression is called the convolution integral. As can be seen from the form, the forcing
function ( f ( ) ) could be any arbitrary function and a general closed form solution is
impractical.
* Convolution Process
The following summarizes the discrete convolution process.
•
•
•
•
•
define time instant, t, to be evaluated; e.g. t = 3
align h(0) to f(t); e.g. h(0) with f(3)
draw h(t) in reverse back to t = 0; e.g. align f(0) with h(3)
compute product of f & h over time interval 0 to t
x(t) is area of product
* Convolution Example
The following are the images for the convolution example presented in class.
Initial Figures
Solution
As shown in the figures below, the response, x(t), for t=0,1,2,… is x(t) = [0 0 2 5 2 -4 -4 1 -1] m
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System Dynamics & Vibrations
Since the integral is from 0 to 0, x(0) = 0 m.
Note: for the following figures, the expression g(2..3) means to take the value of the function g
over the time interval from 2 to 3.
x(1) = f(0..1) * h(1..0) = 0*1 = 0 m
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x(2) = f(0..1) * h(2..1) + f(1..2) * h(1..0) = 0*2 + 2*1 = 2 m
x(3) = f(0..1) * h(3..2) + f(1..2) * h(2..1) + f(2..3) * h(1..0) = 0*0 + 2*2 + 1*1 = 5 m
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x(4) = f(0..1) * h(4..3) + f(1..2) * h(3..2) + f(2..3) * h(2..1) + f(3..4) * h(1..0) = 0*(-2) + 2*0 +
1*2 + 0*1 = 2 m
The solution of the rest of the values follows the same pattern/process.
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#14 – Numerical Solution of Equation of Motion
The general solution of the differential equation of motion with initial conditions and an arbitrary
forcing function is most easily done numerically. While, there are numerous techniques for
solving linear, simultaneous, ordinary differential equations, this section will focus upon only
one method, Euler’s Method, both for its simplicity and its representative character.
Euler’s Method
Euler’s Method is the simplest first-order method. It can be used to solve differential equations
of the form, y = f ( x, y ) . It is an explicit method, which means that the derivative ( y ) is an
explicit function of the variable x and all previous values of both x and y .
Euler’s Method is very simple. Given the
current value of the function ( yn ), the current
derivative of the function ( yn ) and the step
y ( x)
size ( x ), simply evaluate:
yn+1
yn
yn+1  yn + yn x
xn
Clearly the accuracy of this method is
determined by the step size x .
ynxn
xn
x
xn
xn
x
State-Space Expansion
One critical issue for the application of numerical integration to the differential equation of
motion is that the equation of motion is second-order, while the integration techniques require
first-order differential equations. This is not however a significant limitation. By a technique
known as state-space expansion, a high-order differential equation can be converted into a set of
simultaneous lower-order differential equations. For the equation of motion, this means
converting the single second-order differential equation into a pair of simultaneous first-order
differential equations.
Starting with the original second-order differential equation.
f (t ) = mx + cx + kx
Augment the equation with the identity relationship 0 = mx − mx .
f (t ) = mx + cx + kx
0 = mx − mx
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Rearranging the equations yields,
mx = f (t ) − cx − kx
mx = mx
Converting to a vector/matrix equation set yields,
mx   f (t ) − cx − kx 
 =

mx
mx  

x
Substituting the state vector, y =   , into the above expression yields,
x
 x
 f (t ) − cx − kx 
d  x
d
m   = m   = m  y = m  y = 

mx
dt  x 
dt
 x


f
(
t
)
−
cx
−
kx


m  y = 

mx


Dividing by m and manipulating the right-hand side similarly yields,
k   f (t )   c
k   f (t )   c
k
 f (t ) c
− x − x 
−  x

 − x − x  
 −
 y =  m m m  =  m  +  m m  =  m  +  m m  
x

  0  
  0   1
x
x
0   
k
k
 f (t )   c
 f (t )   c
−  x 
− 

 −
 −
 y =  m  +  m m   =  m  +  m m  y
x
 0   1
0     0   1
0 
Therefore, the state-space expansion of the original equation becomes,
k
 f (t )   c
− 

 −
 y =  m  +  m m  y -or 0   1
0 
k
 f (t )   c
−   x
d x 
 −
m  
  =  m + m
x
dt  x  


0   
 0   1
At this point, the original second-order differential equation has been manipulated into two,
simultaneous, first-order differential equations. In this form, it is suitable for use in the numerical
integration scheme presented. (Actually, it can be used with any first-order numerical integration
scheme.)
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Numerical Solution
Combining the results of the two preceding sections, allows the development of an iteration
scheme for numerically evaluating the response of the system to an arbitrary input (force).
Constructing the state vector from the current conditions,
x 
 xn 
 yn =  n 
Evaluating the state equation of motion yields the current derivative,
 f (tn )   c
−
 yn =  m  +  m
 0   1
k

m  yn
0 
−
Finally, evaluating the numerical integration rule (in this case, Euler’s Method) yields,
 yn+1 =  yn + t  yn -or xn+1   xn 
 xn 
  =   + t  
 xn+1   xn 
 xn 
This provides a simple marching scheme to numerically integrate the equations of motion
applying both initial conditions and an arbitrary force. In general, a more precise numerical
integration rule may be substituted for the last step*.
*
TODO: Add 2DOF non-linear example.
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* #15 – System Analogies (Introduction)
Analogous Systems:
An analogous system is one that exhibits mathematically equivalent behavior to another
physically different system.
There are many different useful analogous system models, both intuitive and topological, which
can be formulated for mechanical (translational and/or rotational), electrical, fluidic, thermal, etc.
systems.
(Note: the related issue of coupled hybrid systems will be covered later.)
Mechanical-Electrical ("intuitive analogy"): 1st Order Relationships
where:
• m, c, k, f, x, v, and F ̂are the mechanical terms
• L, R, C, V, q, i, and φ are the electrical terms
• B is the electro-mechanical coupling term
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#16 – System Analogies (Overview)
The input/output relationship, known as the Frequency Response Function (FRF), is not only
applicable to mechanical systems, but has broad application to other physical disciplines. System
analogies, based upon the similarity between the governing differential equations, can be drawn
between mechanical, electrical, fluidic, and other physical systems. One method for developing
these analogs is called the impedance method and is explored in this section.
Analogous Systems (summary)
Analogous System:
An analogous system is one that exhibits mathematically equivalent behavior to another
physically different system.
For the mechanical-electrical analogy, there are two possible analogs. One that is “intuitive” and
one that is “topological”. Although both will be presented, it the “intuitive” analogy that will be
emphasized.
Mechanical-Electrical Analogy:
“Intuitive” Mechanical - Electrical Analogy
force
voltage
velocity
current
[displacement charge]
damper
resistor
spring
capacitor
mass
inductor
“Topological”
current
voltage
---------resistor
inductor
capacitor
“Intuitive” Analogy Characteristics:
force causes velocity
damper dissipates mechanical
energy (to heat)
mass stores mechanical
energy (kinetic)
1 2
mv
2
springs stores mechanical
energy (potential)
1 2
kx
2
force*velocity = mechanical
power (instantaneous)
System Dynamics & Vibrations - Lecture Notes
-
voltage causes current
resistor dissipates electrical
energy (to heat)
inductor stores electrical
energy (magnetic field)
1 2
Li
2
capacitor stores electrical
energy (electric field)
1 1  2
 q
2C 
voltage*current = electrical
power (instantaneous)
-
-
-
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Related Analogies:
Besides the mechanical-electrical analogy presented above, there are other related analogies. It
should be noted that the thermal analogy doesn’t exactly fit since thermal systems are inherently
first order and a thermal resistor conducts heat.
Mechanical
force (f)
velocity (v)
[displacement (x)
damper(c)
spring (k)
mass (m)
Electrical
voltage (V)
current (i)
charge (q)
resistor (R)
capacitor (C)
inductor (L)
Hydraulic
pressure (P)
volumetric flow (Q)
volume (V)]
resistor (R)
tank (C)
inertance (I)
|
|
|
|
|
|
|
Thermal
temperature (T)
heat flow (q)
resistance (Rx)
Thermal Analogy:
The thermal analogy is presented simply because it is (mostly) a DC equivalent and the
corresponding relationship, V = IR , is easily presented.
Just as the voltage drop across an electrical resistor (R) can be evaluated as:
V1
V = V1 − V2 = IR
Where the resistance (R) is a function of both material and geometric properties.
L
R=
A
The heat flow through a thermal resistor (RT) can be evaluated as:
T = T1 − T2 = qRT
R
V2
I
RT
T1
T2
q
Where the thermal resistance (RT) is a function of material and geometric properties, as
well as, the type of heat transfer.
L
1
or
kA
hk A
1
Convection (RC) is
hc A
1
Radiation (RR) is
hr A
Note that ρ and k are properties of the material; and h is a function of the properties.
Conduction (RT) is
The following example illustrates that the heat flow through a home window is a combination of
the convective and radiant heat flow on the outside and the convective heat flow on the inside.
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System Dynamics & Vibrations
conduction
radiation
convection
[exterior]
[interior]
convection
RR
RT
RC
RC
Analogous Systems (relationships)
The following is a brief presentation of the analogous relations for mechanical and electrical
systems.
Mechanical Relationships:
force
spring
momentum
damper
displacement
force:
= kx = cv
momentum:
=  f dt = mv
System Dynamics & Vibrations - Lecture Notes
mass
velocity
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Electrical Relationships:
voltage
field
capacitor
resistor
charge
inductor
current
q
= iR
C
=  vdt = iL
=
voltage:
field:
Mechanical-Electrical Analogies:
force
cv
d
m (v)
dt
-
k  vdt
-
-
voltage
Ri
d
L (i )
dt
1
idt
C
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System Dynamics & Vibrations
#17 – System Analogies (Circuit Concepts)
Circuits Review
The following is a brief review of the basic circuit equations and concepts.
Element Impedance:
R
Resistor
v = iR
i=
c
v
R
ZR = R
Capacitor
v=
1
idt
C
L
i=C
d
(i )
dt
i=
d
(v)
dt
ZC =
1
jC
Inductor
v=L
1
vdt
L
Z L = j L
Ohm’s Law:
V = IR
V = IZ
; DC circuits
; AC circuits
Kirchoff’s Laws:
KVL
v = 0
; around loop
i = 0
; at node
i
i
KCL
i
i
To solve: identify the elements, choose loop direction(s), assume polarity consistent with loop
Combining Impedances:
Series
R1
Z1
R2
Z EQ = Z1 + Z 2
REQ = R1 + R2
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Z2
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System Dynamics & Vibrations
R1
Parallel
Z1
R2
Z2
1
1
1
= +
REQ R1 R2
1
1
1
= +
Z EQ Z1 Z 2
KVL Example:
L
R
Given an RLC loop circuit, find the voltage across the
capacitor.
v
find: out
vin
+
Vin
i
~
c
Vout
ZL
ZR
Impedance Method
ZR = R
1
ZC =
jC
+
Z L = j L
Sum the voltage drops around the loop.
iZ R + iZ L + iZC − vin = 0
vin = iZ R + iZ L + iZC
Vin
~
i
ZC
vout = iZ C
Ratio output to input
vout
iZ C
ZC
=
=
vin iZ R + iZ L + iZ C Z R + Z L + Z C
Substitute definitions
1
vout
jC
=
vin R + j L + 1
=
jC
1
1
jCR + ( j ) CL + 1 1 −  CL + jCR
2
=
2
Complex Impedance
The following is a brief presentation and development of the concept of complex impedance.
Complex Impedance:
Electrically, complex impedance is effectively a complex valued resistance with units of ohms (
 ) just like a resistor.
Z = Z e j
-orwhere: R is resistance and X is reactance
Z = R + jX
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Vout
System Dynamics & Vibrations
Complex impedance extends DC concepts to AC circuits. Thus, Ohm’s law, voltage division,
current division, Thevenin eq’v, Norton eq’v, etc. are still generally valid by replacing R with
Z.
Ohm’s Law:
V = IZ = I Z e j
; Z acts like resistance, that is, voltage drops across the element
;  is the phase, that is, the current lags voltage by the phase
Complex Voltage & Current:
Both the voltage and current can be expressed in terms of either real and imaginary or magnitude
and phase.
j t +
V = VR + jVI = V e ( V )
I = I R + jI I = I e j (t +I )
Since the element impedance can be expressed as
V
Z=
I
Then the voltage-current relationship across an impedance element can be expressed as
j t +
V e ( V ) = I e j (t +I ) Z e j
V e(
j t +V )
= I Z e j (t +I + )
Therefore
V = I Z
V = I + 
Device Impedances:
The following provides a simplified development/proof to the elemental impedances for a
resistor (R), capacitor (C), and inductor (L).
Resistor
vR ( t ) = iR ( t ) R
Let
vR ( t ) = V p sin (t )
vR ( t ) V p sin (t )
=
=R
iR ( t ) I p sin (t )
Therefore
ZR = R
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For AC, the voltage and current are in phase.
Capacitor
iC ( t ) = C
Let
d
( vC ( t ) )
dt
vC ( t ) = V p sin (t )
d
( vC ( t ) ) = Vp cos (t )
dt
Thus
iC ( t ) = CV p cos (t )
V p sin (t )
vC ( t )
=
=
iC ( t ) CV p cos (t )
sin (t )


C sin  t + 
2

Therefore
ZC =
1 − j 2
1
e
=
C
jC
For AC, the voltage lags the current by 90 .
Inductor
vL ( t ) = L
Let
d
( iL ( t ) )
dt
iL ( t ) = I p sin (t )
d
( iL ( t ) ) = I p cos (t )
dt
Thus


L sin  t + 
I  cos (t )
vL ( t )
2

=L p
=
iL ( t )
V p sin (t )
sin (t )
Therefore
j

Z L = Le 2 = j L
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System Dynamics & Vibrations
For AC, the voltage leads the current by 90 .
Combining Impedances:
Series
Z EQ = Z1 + Z 2 + Z3 +
+ ZN
Parallel
1
1
1
1
= +
+ +
Z EQ Z1 Z 2 Z 3
+
1
ZN
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System Dynamics & Vibrations
#18 – System Analogies (Models)
Analogous Systems (single loop)
The following examples present a comparison between the intuitive and topological analogies.
force-voltage:
(velocity-current)
For this analogy ( f  V , v  i ), the equations are intuitive, but it results in a topological
change.
L
R
Evaluating the single loop KVL equation  V = 0 results in:
loop
+
di 1
V = iR + L +  idt
dt C
Vin
Substituting the analogous quantities ( L → m , R → c ,
i
~
c
1
→k
C
) yields:
m
dv
f = vc + m + k  vdt
dt
Or expressed in a more familiar form.
f = mx + cx + kx
x
c
f
k
force-current:
(velocity-voltage)
For this analogy ( f  i , v  V ), the topology is preserved, but the equations are non-intuitive.
Evaluating the three KCL nodal equations  i = 0 results in:
L
R
V1
node
V2 − V1
R
V −V 1
0 = 1 2 +  (V3 − V2 )dt
R
L
1
d
0 =  (V2 − V3 )dt + C ( V 4=0− V3 )
L
dt
0=i+
Vin
~
i
1
1
→ c , → k ) yields:
R
L
d
v 4=0− v3 )
(
dt
System Dynamics & Vibrations - Lecture Notes
c
V4=0
0 = c ( v1 − v2 ) + k  ( v3 − v2 )dt
0 = k  ( v2 − v3 )dt + m
V3
+
Substituting the analogous quantities ( C → m ,
0 = f + c ( v2 − v1 )
V2
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System Dynamics & Vibrations
Or expressed in a more familiar form.
0 = f + c ( x2 − x1 )
m
0 = c ( x1 − x2 ) + k ( x3 − x2 )
0 = k ( x2 − x3 ) − mx3
x3
k
Or alternatively.
c ( x1 − x2 ) = f
x2
c
c ( x2 − x1 ) + k ( x2 − x3 ) = 0
mx3 + k ( x3 − x2 ) = 0
x1
f
force-current:
(velocity-voltage)
As a comparison, the standard SDOF mechanical system results from a parallel RLC circuit.
Again, for this analogy ( f  i , v  V ) preserving the topology.
Evaluating the three KCL nodal equations  i = 0 results in:
V1
node
V =0− V 1
d
0 = i + C ( V 2=0− V1 ) + 2 1 +  ( V 2=0− V1 )dt
dt
R
L
1
1
Substituting the analogous quantities ( C → m , → c , → k )
R
L
yields:
d
0 = f − m ( v1 ) − cv1 − k  v1dt
dt
Or expressed in a more familiar form.
0 = f − mx − cx − kx
f = mx + cx + kx
+
Vin
~
i c
R
L
V2=0
m
x
c
k
Analogous Systems (multiple loop)
The following examples present a comparison between the intuitive and topological analogies for
the same two loop electrical circuit, which results in two different MDOF mechanical system
analogies.
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f
System Dynamics & Vibrations
Intuitive (non-topological):
[force-voltage (velocity-current)]
For this analogy ( f  V , v  i ), the equations are intuitive, but it results in a topological
change.
Evaluating the two KVL loop equations,  V = 0 , results in:
1
( i1 − i2 ) dt − V1 = 0
C
di
1
L 2 + i2 R2 +  ( i2 − i1 ) dt = 0
dt
C
i1 R1 +
Establish the analogous quantities:
i1 → x1
V→ f
R1 → c1
R2 → c2
L
R1
loop
+
Vin
~
i1
R2
i2
C
i2 → x2
1
→k
C
L→m
Substituting the analogous quantities yields:
c1 x1 + k ( x1 − x2 ) − f1 = 0
mx2 + c2 x2 + k ( x2 − x1 ) = 0
x1
x2
c2
k
Or in more familiar form:
c1 x1 + k ( x1 − x2 ) = f1
m
f1
c1
mx2 + c2 x2 + k ( x2 − x1 ) = 0
Which results in the adjacent analogous mechanical system.
Topological (non-intuitive):
[force-current (velocity-voltage)]
For this analogy ( f  i , v  V ), the equations are topological, but it results in non-intuitive
equations.
is = i1
Evaluating the three KCL nodal equations,  I = 0 , results in:
V1
node
 V −V 
i1 +  2 1  = 0
 R1 
 V1 − V2  1
=0
d
V 4 − V2 = 0

 +  (V3 − V2 ) dt + C
dt
 R1  L
)
(
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R1
L
V2
+
Vs
~
V3
R2
C
V4
02-May-2025 03:33 PM
System Dynamics & Vibrations
 V =0 − V 
1
(V2 − V3 ) dt +  4 3  = 0
L
 R2

Establish the analogous quantities:
i1 → f1
V1 → x1
1
1
→ c1
→ c2
R2
R1
V2 → x2
1
→k
L
V3 → x3
C →m
Substituting the analogous quantities yields:
f1 + c1 ( x2 − x1 ) = 0
c1 ( x1 − x2 ) + k ( x3 − x2 ) + m ( − x2 ) = 0
k ( x2 − x3 ) + c2 ( − x3 ) = 0
x1
Or in more familiar form:
c1 ( x1 − x2 ) = f1
f1
x2
c1
m ( x2 ) + c1 ( x2 − x1 ) + k ( x2 − x3 ) = 0
x3
c2
k
m
c2 ( x3 ) + k ( x3 − x2 ) = 0
Which results in the adjacent analogous mechanical system.
Low Pass Filter (example)
The following example shows the equivalence between an electrical and mechanical low pass
filter.
Electrical:
Solve the KVL equation in terms of the voltage drop across each element.
with VC = Vout
Vin − VR − VC = 0
dV
with i = C out
Vin − iR − Vout = 0
dt
 dV 
Vin −  C out  R − Vout = 0
dt 

R
Vin
i
c
Rearranging results in:
dV
RC out + Vout = Vin
dt
Using the differential operator D =
d
( ) in the time domain and D = j in the frequency
dt
domain yields:
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Vout
System Dynamics & Vibrations
Vout
1
1
=
=
Vin RCD + 1 j RC + 1
And by observation, the time constant of the electrical system is  = RC .
Mechanical:
Solve the FBD in terms of the force balance across each element.
fin − f c − f k = 0
with f out = kx
fin − cv − kx = 0
Evaluating the impedance of f o = kx results in:
f out = kx → x =
v=
f out
k
→ v=
fin
v [= ]
f out
or
k
c
fout [=kx]
k
1 d
( fout )
k dt
Substituting the impedance relation yields:
 f 
fin − c  out  − f out = 0
 k 
Rearranging results in:
c
f out + f out = fin
k
Using the differential operator D =
d
( ) in the time domain and D = j in the frequency
dt
domain yields:
f out
1
1
=
=
c
c
fin
D + 1 j + 1
k
k
And by observation, the time constant of the mechanical system is  =
c
.
k
Analogous Systems (detailed solution procedure)
The following presents the basic procedure for converting from one system to the “intuitive”
analogous system.
intuitive analogy:
push/effort: force (f)
flow/motion: velocity (v)
-
System Dynamics & Vibrations - Lecture Notes
voltage (V)
current (i)
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System Dynamics & Vibrations
inertia:
dissipative:
stiffness:
mass (m)
damper (c)
stiffness (k)
-
inductance (L)
resistance (R)
capacitance (1/C)
procedure:
1) label the flow/motion with the sign for each element and its analogy
- pay attention to direction
2) formulate the system equations at each node or loop
-  f = ma (mechanical)
-
V = 0 (electrical)
loop
3) select an element and construct the analogous model
- follow flow/motion
- enforce FBD/KVL element and flow/motion relationships
4) verify the results
- compare the analogous system equations
example:
The following is a three degree-of-freedom mechanical to electrical analogy example. (►Note:
the hashed bar at DOF dofs 1 and 2 are massless nodes.)
x1
x3
c1
k1
x2
F2
m
F1
c2
k2
step #1
Develop the FBD for each nodal DOF.
FBD #1
c1 ( 1 - 3 )
F1
k2 (x1 - x2)
=
0
FBD #2
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c1 ( 3 - 1 )
k1 (x3 - x2)
m
3
=
FBD #3
k1 (x2 - x3)
F2
c2 2
=
0
k2 (x2 - x1)
step #2
Develop the equations for each nodal DOF.
EQ’N #1
F1 − c1 ( x1 − x3 ) − k2 ( x1 − x2 ) = 0
EQ’N #2
−c1 ( x3 − x1 ) − k1 ( x3 − x2 ) = mx3
EQ’N #3
−k1 ( x2 − x3 ) − k2 ( x2 − x1 ) − F2 − c2 x2 = 0
step #3
Construct the analogy.
Present the equivalence table for the mechanical to electrical analogs.
i1
V1
F1 v1
c1 v1 − v3
i1 − i3 R1
i3
L
m v3
i3 − i2 C1
k1 v3 − v2
i1 − i2 C2
k2 v1 − v2
c2 v2
F2 − v2
-
i2
−i2
R2
V2
Substitute into mechanical equations.
EQ’N #1
V1 − R1 ( i1 − i3 ) −
1
( i1 − i2 ) dt = 0
C2 
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EQ’N #2
R1 ( i1 − i3 ) −
di
1
( i3 − i2 ) dt − L 3 = 0

C1
dt
EQ’N #3
1
1
( i3 − i2 ) dt +  ( i1 − i2 ) − V2 − R2i2 = 0

C1
C2
Build circuit: start with EQ’N #1, add EQ’N #2, then add EQ’N #3.
L
#
3
+
+
c1
i3
R1
+
V1
R2
i2
+
+
i1
~
#
+
+
1
~
V2
c2
#
2
step #4
Verify loop equations.
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System Dynamics & Vibrations
#19 – Multiple Degree of Freedom – Applications*
In general, multiple degree of freedom (MDOF) systems are not simply rectilinear and
translational, but rather involve coupling between translation and rotation, absolute and relative
coordinates, mass and mass moment of inertia. This section explores the analysis of these more
challenging, but common, MDOF systems. In addition, some practical MDOF applications (e.g.
tuned vibration absorbers, etc.) and mixed discipline (i.e. electro-mechanical) systems are
presented.
Hybrid Systems (flux coupling)
The following is a brief presentation of how mechanical and electrical systems are coupled
through the magnetic flux coefficient.
Mechanical-Electrical Relationships:
=
f V
φ
d
/dt
=
=
=
c R
B
k 1/ C
m L
*
*
=
x q
*
v i
d
/dt
where:
• m, c, k, f, x, v, and F̂ are the mechanical terms
• L, R, C, V, q, i, and φ are the electrical terms
• B is the electro-mechanical coupling term
System State
*
q : charge
x : displacement
 : magnetic flux
V : voltage
i : current
F̂ : impulse
f : force
v : velocity
This section under development; see reference: Tse, Morse, & Hinkle (ch. 4 & 6)
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System State Equations
dx = vdt
dFˆ = fdt
dq = idt
d = Vdt
dV = Rdi
df = cdv
1
dx = df
k
dFˆ = mdv
dq = CdV
d = Ldi
Flux Intensity:
In an electro-mechanical system, the force acting on a conductor of length l , carrying a current
i in a magnetic field with flux intensity  is:
f (t ) =  l i (t )
where:
• f (t ) = Force (newtons)
•
•
•
 = Flux intensity (weber-meter-2)
l = Conductor length (meters)
i (t ) = Current (amperes)
f
β
i
Flux Coupling:
The magnetic flux coupling constant B is defined as:
B=l
This coupling factor relates the mechanical characteristics to the electrical characteristics.
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Flux EMF:
The inverse relationship is that an open circuit voltage or electromotive force (EMF) is induced
in a conductor moving in a magnetic field as follows:
vEMF (t ) =  l x(t )
where:
• vEMF (t ) = Induce voltage (volts)
•
x (t ) = Velocity (meters-sec-1)
+
v
β
-
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Appendix A – Unit Systems (aka ‘g’ is a unit)
Unit System Conversions:
As engineers, we are frequently required to work in both the SI and British units systems. Not
only that, both systems can be expressed in customary and gravimetric forms.
The following are the units for each unit system (both base and derived.)
SI (Système
Internationale)
British (Imperial)
customary
(force,mass,length,time)
gravimetric
(force,mass,length,time)
N, kg, m, s
kgf, kgm, gs2, s
lbf, slug, ft, s
lbf, lbm, gs2, s
All unit systems are in practice derived from (or must satisfy) Newton’s formula.
F = ma
SI (Système
Internationale)
British (Imperial)
customary
gravimetric
1 N = 1 kg * 1 m/s2
1 kgf = 1 kgm * 1 g0
1 lbf = 1 slug * 1 ft/s2
1 lbf = 1 lbm * 1 g0
Notice that for each system, one unit of force equals one unit of mass multiplied by one unit of
acceleration. Also note that for each term, there is a value (numeric) part and unit (symbol) part.
Thus, although ‘g’ is often presented as a unit conversion factor, it is in reality a unit in its own
right. Consider, for example, the expression that “a roller coaster pulled 4 g’s pulling out of the
drop.” For this, ‘4’ is the value; and ‘g’ is the unit.
The common statement: g = 9.81 m/s2 is technically false. The correct expression is: 1 g = 9.81
m/s2. Again, value plus unit.
The common relationship is:
1 g = 9.81 m/s2 = 32.2 ft/s2 = 386 in/s2
Conversion:
Conversion is always handled by multiplying by unity relationships. Always!
1=
9.81 m s2
32.2 ft s 2
386 in s 2
9.81 m s 2
1g
1g
1g
=
=
=
=
=
=
=
1g
9.81 m s2
1g
32.2 ft s2
1g
386 in s 2 32.2 ft s 2
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For example, to convert from ‘slugs’ to ‘lbm’, begin with the force balance relationships above:
1 lb f = 1 slug (1 ft s2 ) = 1 lbm (1 g )
The manipulate equation form and multiply by the appropriate unity relationship(s):
 1g 
 1 g   32.2 ft s2 
1 slug = 1 lbm  ft  = 1 lbm  ft  
 = 32.2 lbm
 1 s2 
 1 s2   1 g 
Thus, technically there are *NO* unit conversion factors; only unity relationships. And what is
really being done is to change the expression from one unit system to another.
SI Gravimetric Example:
Below is an image taken from a dishwasher packaging where both the British and SI gravimetric
system have been used.
Standard Gravity:
For completeness, technically, the standard gravity definition is:
1 g0 = 9.80665 m/s2 = 32.1740 ft/s2 = 386.088 in/s2
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Appendix B – Complex Numbers (review)
Real / Imaginary
C = 3+ 4 j
C* = 3 − 4 j
; complex conjugate
Complex Math
C C * = ( 3 + 4 j )( 3 − 4 j )
C C * = 3 ( 3) + 3 ( −4 j ) + 4 j ( 3) + 4 j ( −4 j )
C C* = 9
− 12 j
+ 12 j
+ 16 = 25
1 1 C
3− 4 j
=
=
= 0.12 − 0.16 j
*
C C C
25
*
Mag / Phase
C = C C
; note
or
C = C e jC
C  0 and −  C  
3+ 4 j = 5
(3 + 4 j ) = 0.93 rad
C = 3 + 4 j = 5e0.93 j
C * = 3 − 4 j = 5e−0.93 j
True Angle
* watch the quadrant!
−3 + 4 j = 5e2.21 j
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* Appendix C – Harmonic Function (review)
Harmonic Functions are sinusoidal (or co-sinusoidal) functions of time. (Technically, they can
also be composed of the sum of sinusoidal and/or co-sinusoidal functions.)
For example, 𝑥(𝑡) = 𝑋 sin(𝜔𝑡 + 𝜙), is an harmonic function where ‘𝑋’ is the amplitude, ‘𝜔’ is
the frequency in rad/sec, and ‘𝜙’ is the phase angle in rad.
Its derivatives would be: 𝑥̇ (𝑡) = 𝜔𝑋 cos(𝜔𝑡 + 𝜙) and 𝑥̈ (𝑡) = −𝜔2 𝑋 sin(𝜔𝑡 + 𝜙)
𝑑
Note that 𝑑𝜃 (sin(𝜃)) = cos(𝜃) only if ‘𝜃’ is expressed in radians. It is not true if ‘𝜃’ is
expressed in degrees.
Also note that: max(𝑥(𝑡)) = |𝑋|
For the summation of harmonic functions: 𝑥(𝑡) = 𝑋1 sin(𝜔1 𝑡) + 𝑋2 sin(𝜔2 𝑡)
•
beating happens when 𝜔1 ≈ 𝜔2
•
if 𝑋 = 𝑋1 = 𝑋2 & 𝜀 = 𝜔2 − 𝜔1 then 𝑥(𝑡) = 2𝑋 cos (2 𝑡) sin ((𝜔1 + 2) 𝑡)
𝜀
𝜀
where:
o 2𝑋 is the magnitude
𝜀
o cos ( 𝑡) is the beating
2
𝜀
o and sin ((𝜔1 + 2) 𝑡) is the carrier
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* Appendix D – Rolling & Cable Constraints
(Basic) Rolling Constraint
• 𝑥 =𝑟𝜃
• (actual) contact path equals arc cord
Rolling Constraints
• 𝑥𝐴 = 𝑥 + 𝑅 𝜃
• 𝑥𝐵 = 𝑥 + 𝑟 𝜃
• 𝑥𝐶 = 𝑥
• 𝑥𝐷 = 𝑥 − 𝑟 𝜃
• 𝑥𝐸 = 𝑥 − 𝑅 𝜃
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(Basic) Cable Constraint
• 𝑥 =𝑟𝜃
• 𝑦 =𝑥+𝑟𝜃
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* Appendix E – Supplement Math Hints
Working with 'absolute values' and 'inequalities' requires care. For example, what does the
following expression imply:
|X| = 5
The most common response is that X = ±5, but that answer is based upon the assumption
that X is inherently real. If 'X' is complex, that answer is incomplete. Actually, all that is known
is that the magnitude of X is 5. The angle can be anything. And thus the answer lies on a circle
of radius, 5, in the complex plane.
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=Consider another equation:
|3x-2| = 7
If we assume a real solution, then:
+(3x-2) = 7
==>
3x-2 = 7
-->
x=3
-(3x-2) = 7
==>
3x-2 = -7
-->
x = -5/3
and
What if it was an inequality:
|3x-2| ≤ 7
Then the result is not a value, but a range:
+(3x-2) ≤ 7
==>
3x-2 ≤ 7
-->
x≤3
-(3x-2) ≤ 7
==>
3x-2 ≥ -7
-->
x ≥ -5/3
and
thus
-5/3 ≤ x ≤ 3
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=The previous examples assume that the value for 'x' was real, but what if the solution includes
complex numbers (or is just very difficult to manipulate.)
|3x-2| ≤ 7
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If that is the case, then it can be solved by squaring the magnitude of both sides and checking the
viability of the solutions.
|3x-2|² ≤ 7²
Which for complex numbers means multiplying be the complex conjugate.
(3x-2) (3x-2)* ≤ 49
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=One way to check the viability of the solution and the direction of the inequality is to test a
nearby value. For example, if evaluation of the equality of the function… (expr) ≤ (#)… results
in k = 5000, should that be k ≤ 5000 or k ≥ 5000?
To test, try either k = 4900 and/or k = 5100 to see which satisfies the original expression. (Note:
this test will often identify sign manipulation errors as well.)
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=Finally, be careful when dividing by expressions like ' k - mω² ' because …
when ω = Ω the expression is zero, and …
when ω > Ω the expression is inherently negative (and thus equivalent to dividing by a
negative number.)
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Appendix F – MDOF Detail Overview
In general, this section will not be covered during the course System Dynamics & Vibrations.
Instead, it is provided as a reference for those who wish to know a little more about MDOF
systems, but who for various reasons may not end up taking the following course, Experimental
Vibrations.
General Solution Approach
The matrix equation of motion for a general multi-degree-of-freedom system can be written as
(Time Domain):
 M  x(t ) + C  x(t ) + [ K ]x(t ) =  f (t )
The solution of this linear, constant matrix coefficient, second order differential equation follows
the solution approach for the simpler single degree of freedom problem. The solution takes on
the following form involving complementary and particular parts:
 x(t ) = xc (t ) + x p (t )
 xc (t ) is the complementary portion of the solution and depends on the system characteristics
and initial conditions. The complementary portion of the solution is sometimes referred to as the
transient portion of the solution.
x (t ) is the particular portion of the solution and depends upon the system characteristics and
p
harmonic forcing functions. The particular portion of the solution is sometimes referred to as the
steady state portion of the solution.
Frequently, one portion of the solution will be of interest due to the application under study. If
both portions are of interest, the initial conditions must not be applied until the total solution is
formed (both complementary and particular portions).
Complementary (Transient) Solution:
The complementary solution is found by transforming the original differential equation, in
homogeneous form (temporarily removing the forcing function(s)), from the time domain to the
frequency domain (transforming from differential to algebraic equations) by using Fourier
transforms, Laplace transforms or by assuming a solution that is appropriate for the second order,
linear, constant coefficient matrix differential equation.
The system of equations has a complementary solution of the assumed form:
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 xc (t ) =  X  e st
 xc (t ) = s  X  e st = s  xc (t )
 xc (t ) = s 2  X  e st = s 2  xc (t )
where:
• s =  + j = complex valued frequency
Note that the derivative of the response vector is simply the same vector multiplied by the
complex frequency.
Substituting the above assumed relationships into the matrix equation of motion:
 M  s 2  X  est + C  s  X  e st + [ K ] X  e st = 0
The non-trivial solution of the above equation leaves:
 s 2  M  + s C  +  K   X  = 0
The above equation can be solved in terms of 2N characteristic values and 2N characteristic
vectors. Due to the underdamped nature of vibration problems, there will always be N pairs of
characteristic values and N pairs of characteristic vectors. The characteristic values are the
complex-valued natural frequencies (modal frequencies)
r and the characteristic vectors are the complex-valued modal vectors  r  . Since the
characteristic values and vectors are often found by placing the previous equation in an
eigenvalue form, the characteristic values are often referred to as eigenvalues and the
characteristic vectors as eigenvectors.
More commonly, the characteristic values and vectors are found by way of rudimentary
mathematic manipulations as follows:
The characteristic values can be found for the above algebraic equation, for non-trivial solutions
of  X  , from the matrix characteristic equation as follows:
s 2  M  + s C  +  K  = 0
The above matrix characteristic equation is of model order two (2) with coefficient matrices of
size N ( N  N ). Therefore, this characteristic equation will yield 2N modal frequencies.
Expanding the matrix characteristic equation completely yields a high order ( 2N ) polynomial
characteristic equation with scalar coefficients:
 2 N s 2 N +  2 N −1s 2 N −1 +  2 N −2 s 2 N −2 +
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The characteristic values (complex valued modal frequencies ( r =  r + j r )) are found as the
roots of this characteristic, high order ( 2N ), scalar polynomial.
Once the modal frequencies ( r ) have been determined, the characteristic vectors (modal vectors
 r  ) can be found from the following relationship:
 s 2  M  + s C  +  K   X  = 0
Evaluating at s = r :
r2  M  + r C  +  K   r  = 0
Note that this system of linear equations is always rank deficient by at least one since the
equation system is being evaluated at one of the characteristic frequencies ( r ).
This process must be repeated for each modal frequency to determine each modal vector.
Once the modal frequencies (complex valued, in general) and modal vectors (complex valued, in
general) are determined, the final form of the complementary solution can be formulated as
follows:
 xc (t ) =  ( r  r  e t +  r*  r* e t )
N
r
*
r
r =1
Note that, in the above equation, the unknown coefficients  r appear in complex conjugate
pairs. This will always be true in the underdamped case. It is not necessary to assume that the
conjugate relationship exists; this will result when the solution method is followed.
If there is no forcing function (the particular solution is zero), the unknown (complex valued)
coefficients in the above equation (  r ) can be determined by applying the initial conditions to
the above equation and/or the derivative of the above equation. The derivative of the above
equation is shown below.
 xc (t ) =  ( r r  r  e t +  r*r*  r* e t )
N
r
*
r
r =1
If there is a forcing function, the solution for the unknown (complex-valued) coefficients must
wait until the particular solution has been found. The initial conditions apply to the complete
solution and the unknown coefficients must be found after the particular solution has been added
to the complementary solution.
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Particular Solution Approach
The particular solution is found by assuming a solution form for the response consistent with the
forcing function characteristic. Since the forcing function (steady-state) is some form of
harmonic (sine plus cosine terms), the forcing function and associated response can always be
put into the following form (use the Euler identity for sine and cosine). This approach to solving
for the particular solution is known as the method of undetermined coefficients.
f (t ) = A cos( at ) + B sin( at )
 e jat + e − jat 
 e jat + e − jat 
f (t ) = A 
 + B

2
2j




j at
− j at
j at
− j at
e +e

e +e

f (t ) = A 
 − jB 

2
2




Collecting like terms:
B
B
A
A
f (t ) =  − j  e jat +  + j  e − jat
2
2
2
2
Therefore, representing the characterisitics of the force at  a with a complex-valued magnitude
yields the general form for the forcing function:
 f (t ) = F  e j t + F * e− j t
a
a
The response(s) to the previous forcing function(s) will be of the same form:
x (t ) =  X  e +  X e
x (t ) = j  X  e − j  X e
x (t ) = −  X  e −   X e
j at
*
− j at
p
j at
p
a
p
2
a
*
− j at
*
− j at
a
j at
2
a
Substituting the above relationships into the matrix equation of motion and collecting like terms
gives the following form for the positive frequency terms ( e j at ). Note that the portion of the
solution involving the negative frequency terms ( e − jat ) is the complex conjugate of the positive
frequency and provides no new information.
− a2  M  X  + j a C  + [ K ] X  = F 
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Note that the above linear, algebraic matrix equation involves N independent equations, as long
as the system is not undamped with the forcing frequency  a equal to one of the natural
frequencies. Since the complex-valued forcing vector  F  provides N known pieces of
information, this system of equations can be solved for the complex-valued response vector  X 
.
Note also that, if more than one forcing frequency is present, the above particular solution
process must be repeated for each forcing frequency.
An alternative to the traditional method of undetermined coefficients is the frequency response
function approach. In this approach, the forcing function(s) are described, as above, in the
frequency domain. The frequency response function(s) between the forcing degrees-of-freedom
(DOFs) and the response degrees-of-freedom (DOFs) are computed from:
 H ( a ) = − a2  M  + j a C  +  K 
−1
The responses caused by the forcing functions can now be found (in the frequency domain) by:
 X ( a ) =  H ( a )F ( a )
The particular response(s) can now be formulated in the time domain by converting the
frequency domain information back to the time domain using the Euler identities (sine and
cosine).
Note also that, if more than one forcing frequency is present, the above particular solution
process must be repeated for each forcing frequency.
Total Solution
The total solution, therefore, can be formulated as:
 x(t ) = xc (t ) + x p (t )
 x(t ) =  ( r  r  e t +  r*  r* e t ) + ( X  e j t +  X * e− j t )
N
*
r
r
a
a
r =1
Example Solution:
For the following three degree of freedom system, formulate the complete solution for the
following initial conditions and forcing function.
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10 0 0 
 M  =  0 14 0 
 0 0 12 
 50 −30 0 
C  =  −30 55 −25
 0 −25 25 
Initial Conditions:
1.0 
 x(0) =  0 
0
 
0 
 5000 −3000

 K  =  −3000 5500 −2500
 0
−2500 2500 
Forcing Function:
 0 
 f (t ) = −75 sin(30t )
 0 


20.0 
 x(0) =  0 
 0 


Complementary Solution Results:
1 = −0.1848 + 6.0760 j
2 = −1.6417 + 18.0458 j
3 = −3.6795 + 26.8767 j
1.0000 
 1 = 1.5435
1.8763


 1.0000 
 2 =  0.5722 
−0.9933


 1.0000 
 3 = −0.7863
 0.3105 


Example Solution - Particular Part:
 s 2  M  + s C  +  K   X p (s) = F (s)
Formulating the positive frequency portion of the solution at 30 rad s ( s = j 30 ):
0
 −4000 + 1500 j −3000 − 900 j

 0 
 −3000 − 900 j −7100 + 1650 j −2500 − 750 j  X (30 j ) = 37.5 j 
  

 p



0
−2500 − 750 j −8300 + 750 j 
 0 
 −0.0041 + 0.0011 j 
 X p (30 j ) =  0.0036 − 0.0045 j 
−0.0016 + 0.0009 j 


Therefore:
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 −0.0041 + 0.0011 j 
 −0.0041 − 0.0011 j 

 j 30t 
x p (t ) =  0.0036 − 0.0045 j  e +  0.0036 + 0.0045 j  e− j 30t
−0.0016 + 0.0009 j 
−0.0016 − 0.0009 j 




Or:
−0.0082 
−0.0022 


x p (t ) =  0.0072  cos(30t ) +  0.0090  sin(30t )
 −0.0031
 −0.0018




Example Solution - Complementary Part:
While the complementary solution of the matrix differential equation for linear vibration
problems always involves complex conjugate pair solutions, it is easier, when finding the
numerical solution using MATLAB, to ignore this and determine the solution without this
assumption. Once you find the solution, a check for complex conjugate pairs serves as a
verification that no other mistakes have occurred.
Therefore:
 x(t ) =  xc (t ) + x p (t )
6
 x(t ) =  r  r  e
r =1
r t
 −0.0041 + 0.0011 j 
 −0.0041 − 0.0011 j 

 j 30t 

+  0.0036 − 0.0045 j  e +  0.0036 + 0.0045 j  e − j 30t
−0.0016 + 0.0009 j 
−0.0016 − 0.0009 j 




Rewriting in matrix notation:
 x(t ) = e t  1
1
e2t  2
e3t  3
e4t  4
e5t  5
1 
 
 2
 
e6t  6   3 
 4 
 5 
 
 6 
 −0.0041 + 0.0011 j 
 −0.0041 − 0.0011 j 

 j 30t 

+  0.0036 − 0.0045 j  e +  0.0036 + 0.0045 j  e − j 30t
−0.0016 + 0.0009 j 
−0.0016 − 0.0009 j 




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Taking the time derivative to get velocity:
 x(t ) = 1e t  1 2e t  2 3e t  3 4e t  4 5e t  5
1
2
3
4
5
1 
 
 2
 
6e 6t  6   3 
 4 
 5 
 
 6 
 −0.0041 + 0.0011 j 
 −0.0041 − 0.0011 j 

 j 30t


+ j 30  0.0036 − 0.0045 j  e − j30  0.0036 + 0.0045 j  e − j 30t
−0.0016 + 0.0009 j 
−0.0016 − 0.0009 j 




Evaluating displacement and velocity at t=0:
1 
 
 2  −0.0082 
1.0 
 
 3  

 0  =  1  2  3  4  5  6    +  0.0071 
0
 4   −0.0031
 

 5  
 
 6 
1 
 
 2  −0.0639 
20.0 


 3  

 0  = 1  1 2  2 3  3 4  4 5  5 6  6    +  0.2722 
 0 
 4  −0.0542 



 5  
 
 6 
1 
 1.0082 
 
 −0.0071
 2


 0.0031 
 3 

 =  A  
 20.0639 
 4 
 −0.2722 
 5 


 
 0.0542 
 6 
Where:
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 1.0000
 −0.7863

 0.3105
 A =
 −3.6795 − 26.8767 j

 2.8932 + 21.1335 j
 −1.1425 − 8.3456 j
1.0000
1.0000
1.0000
1.0000
0.5772
1.5435
1.5435
0.5772
−0.9933
1.8763
1.8763
−0.9933
−1.6417 − 18.0458 j
−0.1848 − 6.0760 j
−0.1848 + 6.0760 j
−1.6417 + 18.0458 j
−0.9394 − 10.3255 j
−0.2852 − 9.3783 j
−0.2852 + 9.3783 j
−0.9394 + 10.3255 j
1.6307 + 17.9246 j
−0.3467 − 11.4003 j
−0.3467 + 11.4003 j
1.6307 − 17.9246 j


−0.7863

0.3105


−3.6795 + 26.8767 j

2.8932 − 21.1335 j 

−1.1425 + 8.3456 j 
1.0000
Therefore:
1  0.2567 + 0.2265 j 
  0.1890 + 0.2246 j 
 2 

 3  0.0584 + 0.1902 j 
 =

 4  0.0584 − 0.1902 j 
 5  0.1890 − 0.2246 j 
  

 6  0.2567 − 0.2265 j 
Complementary Solution Approach - Adjoint Matrix
An alternative approach to determining the modal vectors in the complementary solution
involves evaluating the adjoint matrix of the system impedance matrix at the modal frequencies (
r ). This method has two advantages: 1) Formulating the adjoint matrix effectively solves the
set of linear equations one time rather than one set of linear equations for each modal frequency;
2) Evaluating the adjoint matrix handles the rank deficient problem in a uniform way (no
assumption of 1.0 for a modal coefficient). The development is as follows:
 s 2  M  + s C  +  K   X  = 0
Define:
 B(s) =  s 2  M  + s C  +  K 
where:
•  B( s ) = System Impedance Matrix
 B(s) B(s) =  I 
−1
 B( s)
 B( s) =
 B( s)
−1
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where:
•
 B(s) is the adjoint of matrix  B(s) .
A
 B(s) B(s) =  B(s) I 
A
Note:  B(r )  = 0 .
Evaluating at s = r gives:
 B(r ) B(r ) = 0
A
Using any column of  B(r ) , the i th column for example B(r )i . Therefore:
A
A
 B(r )B(r )iA = 0
r2  M  + r C  +  K   r  = 0
Note that the columns of the adjoint matrix  B(r ) are all proportional to the r th modal vector.
A
When the mass, damping and stiffness matrices are symmetric (when absolute coordinates are
used), the system impedance matrix  B( s ) is symmetric. Therefore, in this case, the adjoint
matrix of  B (r )  is also symmetric. Thus, in this case, the rows of the adjoint matrix are also
proportional to the modal vector.
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Complementary Solution Approach - Eigenvalue/Eigenvector Method
The homogeneous form of the Laplace domain model can be used as a general representation of
the matrix relationship that yields the system modal characteristics:
s 2  M  X  + s C  X  +  K  X  = 0
Generally, this problem is solved using eigenvalue-eigenvector solution methods once the
problem is put in the standard eigenvalue form:
 A −   I   X  = 0
 A X  =   X 
 A  X  =   B   X 
Undamped Case:
In order to manipulate the system equations into a standard eigenvalue/eigenvector equation
form, one approach is to assume that the undamped case is a reasonable approximation of the
damped case. For many lightly damped situations, this is a reasonable assumption.
 K  X  = −s 2  M  X 
where:
•  = −s2
General Damped Case
 M  x + C  x +  K  x =  f 
This system of equations can be augmented by the identity shown as follows:
 M  x −  M  x = 0
The above two equations can be combined to yield a new system of 2N equations.
Note that all the matrices in the resulting equation are symmetric and the equation is now in a
classical eigenvalue solution form. The notation used in the following equation is consistent with
the notation used in many mathematics and/or controls textbooks.
 A y +  B  y =  f 
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where:
 −  M   0 
  0   M 
 A =  M C   B  =  0

    K 
    
 x
 x
 y =    y =  
 x
 x
 0 
 f  =  
 f 
Forming the homogeneous equation of this system equation:
 A y +  B  y = 0
The solution of the above equation yields the complex-valued natural frequencies (eigenvalues)
and complex-valued modal vectors (eigenvectors) for the augmented
2N equation system. Note that in this mathematical form, the eigenvalues will be found directly
(not the square of the eigenvalue) and the 2N eigenvectors will be 2N in length. The exact
form of the eigenvectors can be seen from the associated modal matrix for the 2N equation
system. Note that the notation   is used for an eigenvector in the 2N equation system and
that the notation   is used for an eigenvector of the N equation system.
MATLAB Solution
For the general, homogeneous system equation, using either an assumed solution or transform
methods, the following equation must be solved:
  A y +  B  y = 0
MATLAB uses the same matrix terminology but refers to a different eigenvalue equation.
 A  y =   B   y
Therefore, using the MATLAB EIG function to solve for the eigenvalues and eigenvectors
requires the following form:
 y,   = EIG ( A, B ) = EIG ( B, − A)
The eigenvalues of this system of equations are the same as for the original mass, stiffness and
damping matrix equation. The eigenvectors of this system of equations yield the modal vectors
of the original mass, stiffness and damping matrix equation through the modal matrix. The modal
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matrix for this system is a matrix made up of the 2N eigenvectors. The modal matrix   for
this damped system can now be assembled.
  = 1 2
r
     
  =  1 1 2 2
 2
  1
 2 N 
r  r
 r
2 N  2 N 
 2 N 
Weighted Orthogonality Concept
Proportionally Damped Case
A set of weighted orthogonality relationships are valid for the system matrices  M  and  K  .
 r  M  s = 0
T
 r  K  s = 0
T
Modal Scaling - Proportionally Damped Case
 r  M  r = M r
T
 r  K  r = K r
T
 r C  r = Cr
T
General Case
A set of weighted orthogonality relationships are valid for the system matrices  A and  B  .
r  As = 0
T
r  Bs = 0
T
Modal Scaling - General Case
r  Ar = M A
T
r
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r  Br = M B
T
r
The terms modal A and modal B are modal scaling factors for the general case of system with
damping. Note that modal A and modal B are related by the complex modal frequency for each
mode.
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