Ch19

VECTOR MECHANICS FOR ENGINEERS:
DYNAMICSDYNAMICS
NinthNinth
EditionEdition
Ferdinand P. BeerFerdinand P. Beer
E. Russell Johnston, Jr.E. Russell Johnston, Jr.
Lecture Notes:Lecture Notes:
J. Walt OlerJ. Walt Oler
Texas Tech UniversityTexas Tech University
CHAPTER
© 2010 The McGraw-Hill Companies, Inc. All rights reserved.
19
Mechanical Vibrations

Vector Mechanics for Engineers: DynamicsVector Mechanics for Engineers: DynamicsNinth
Edition
Contents
19 - 2
Introduction
Free Vibrations of Particles. Simple Harmonic Motion
Simple Pendulum (Approximate
Solution)
Simple Pendulum (Exact Solution)
Sample Problem 19.1
Free Vibrations of Rigid Bodies
Sample Problem 19.2
Sample Problem 19.3
Principle of Conservation of Energy
Sample Problem 19.4
Forced Vibrations
Sample Problem 19.5
Damped Free Vibrations
Damped Forced Vibrations
Electrical Analogues

Edition
Introduction
19 - 3
• Mechanical vibration is the motion of a particle or body which
oscillates about a position of equilibrium. Most vibrations in
machines and structures are undesirable due to increased stresses
and energy losses.
• Time interval required for a system to complete a full cycle of the
motion is the period of the vibration.
• Number of cycles per unit time defines the frequency of the vibrations.
• Maximum displacement of the system from the equilibrium position is
the amplitude of the vibration.
• When the motion is maintained by the restoring forces only, the
vibration is described as free vibration. When a periodic force is
applied to the system, the motion is described as forced vibration.
• When the frictional dissipation of energy is neglected, the motion
is said to be undamped. Actually, all vibrations are damped to
some degree.

Edition
19 - 4
• If a particle is displaced through a distance xm from its
equilibrium position and released with no velocity, the
particle will undergo simple harmonic motion,
( )
0=+
−=+−==
kxxm
kxxkWFma st

δ
• General solution is the sum of two particular solutions,
( ) ( )tCtC
t
m
k
Ct
m
k
Cx
nn ωω cossin
cossin
21
21
+=






+





=
• x is a periodic function and ωn is the natural circular
frequency of the motion.
• C1 and C2 are determined by the initial conditions:
( ) ( )tCtCx nn ωω cossin 21 += 02 xC =
nvC ω01 =( ) ( )tCtCxv nnnn ωωωω sincos 21 −== 

Edition
19 - 5
( )φω += txx nm sin
==
n
n
ω
π
τ
2
period
===
π
ω
τ 2
1 n
n
nf natural frequency
( ) =+= 2
0
2
0 xvx nm ω amplitude
( ) == −
nxv ωφ 00
1
tan phase angle
• Displacement is equivalent to the x component of the sum of two vectors
which rotate with constant angular velocity
21 CC

+
.nω
02
0
1
xC
v
C
n
=
=
ω

Edition
19 - 6
( )φω += txx nm sin
• Velocity-time and acceleration-time curves can be
represented by sine curves of the same period as the
displacement-time curve but different phase angles.
( )
( )2sin
cos
πφωω
φωω
++=
+=
=
tx
tx
xv
nnm
nnm

( )
( )πφωω
φωω
++=
+−=
=
tx
tx
xa
nnm
nnm
sin
sin
2
2


Edition
Simple Pendulum (Approximate Solution)
19 - 7
• Results obtained for the spring-mass system can be
applied whenever the resultant force on a particle is
proportional to the displacement and directed towards
the equilibrium position.
for small angles,
( )
g
l
t
l
g
n
n
nm
π
ω
π
τ
φωθθ
θθ
2
2
sin
0
==
+=
=+
:tt maF =∑
• Consider tangential components of acceleration and
force for a simple pendulum,
0sin
sin
=+
=−
θθ
θθ
l
g
mlW



Edition
Simple Pendulum (Exact Solution)
19 - 8
0sin =+ θθ
l
gAn exact solution for
leads to
( )
∫
−
=
2
0
22
sin2sin1
4
π
φθ
φ
τ
m
n
d
g
l
which requires numerical
solution.






=
g
lK
n π
π
τ 2
2

Edition
Sample Problem 19.1
19 - 9
A 50-kg block moves between vertical
guides as shown. The block is pulled
40mm down from its equilibrium
position and released.
For each spring arrangement, determine
a) the period of the vibration, b) the
maximum velocity of the block, and c)
the maximum acceleration of the block.
SOLUTION:
• For each spring arrangement, determine
the spring constant for a single
equivalent spring.
• Apply the approximate relations for the
harmonic motion of a spring-mass
system.

Edition
Sample Problem 19.1
19 - 10
mkN6mkN4 21 == kk SOLUTION:
• Springs in parallel:
- determine the spring constant for equivalent spring
mN10mkN10 4
21
21
==
+==
+=
kk
P
k
kkP
δ
δδ
- apply the approximate relations for the harmonic motion
of a spring-mass system
n
n
n
m
k
ω
π
τ
ω
2
srad14.14
kg20
N/m104
=
===
s444.0=nτ
( )( )srad4.141m040.0=
= nmm xv ω
sm566.0=mv
2
sm00.8=ma( )( )2
2
srad4.141m040.0=
= nmm axa

Edition
Sample Problem 19.1
19 - 11
mkN6mkN4 21 == kk
• Springs in series:
- determine the spring constant for equivalent spring
- apply the approximate relations for the harmonic motion
of a spring-mass system
n
n
n
m
k
ω
π
τ
ω
2
srad93.6
kg20
400N/m2
=
===
s907.0=nτ
( )( )srad.936m040.0=
= nmm xv ω
sm277.0=mv
2
sm920.1=ma( )( )2
2
srad.936m040.0=
= nmm axa
mN10mkN10 4
21
21
==
+==
+=
kk
P
k
kkP
δ
δδ

Edition
Free Vibrations of Rigid Bodies
19 - 12
• If an equation of motion takes the form
0or0 22
=+=+ θωθω nn xx 
the corresponding motion may be considered
as simple harmonic motion.
• Analysis objective is to determine ωn.
( ) ( )[ ] mgWmbbbmI ==+= ,22but 2
3
222
12
1
0
5
3
sin
5
3
=+≅+ θθθθ
b
g
b
g 
g
b
b
g
n
nn
3
5
2
2
,
5
3
then π
ω
π
τω ===
• For an equivalent simple pendulum,
35bl =
• Consider the oscillations of a square plate
( ) ( ) θθθ  ImbbW +=− sin

Edition
Sample Problem 19.2
19 - 13
k
A cylinder of weight W is suspended
as shown.
Determine the period and natural
frequency of vibrations of the
cylinder.
SOLUTION:
• From the kinematics of the system, relate
the linear displacement and acceleration
to the rotation of the cylinder.
• Based on a free-body-diagram equation
for the equivalence of the external and
effective forces, write the equation of
motion.
• Substitute the kinematic relations to arrive
at an equation involving only the angular
displacement and acceleration.

Edition
Sample Problem 19.2
19 - 14
SOLUTION:
• From the kinematics of the system, relate the linear
displacement and acceleration to the rotation of the cylinder.
θrx = θδ rx 22 ==
θα rra ==θα 
= θ
ra =
• Based on a free-body-diagram equation for the equivalence of
the external and effective forces, write the equation of motion.
( ) :∑∑ = effAA MM ( ) αIramrTWr +=− 22
( )θδ rkWkTT 2but 2
1
02 +=+=
• Substitute the kinematic relations to arrive at an equation
involving only the angular displacement and acceleration.
( )( ) ( )
0
3
8
22 2
2
1
2
1
=+
+=+−
θθ
θθθ
m
k
mrrrmrkrWWr


m
k
n
3
8
=ω
k
m
n
n
8
3
2
2
π
ω
π
τ ==
m
k
f n
n
3
8
2
1
2 ππ
ω
==

Edition
Sample Problem 19.3
19 - 15
s13.1
lb20
n =
=
τ
W
s93.1=nτ
The disk and gear undergo torsional
vibration with the periods shown.
Assume that the moment exerted by the
wire is proportional to the twist angle.
Determine a) the wire torsional spring
constant, b) the centroidal moment of
inertia of the gear, and c) the maximum
angular velocity of the gear if rotated
through 90o
and released.
SOLUTION:
• Using the free-body-diagram equation for
the equivalence of the external and
effective moments, write the equation of
motion for the disk/gear and wire.
• With the natural frequency and moment
of inertia for the disk known, calculate
the torsional spring constant.
• With natural frequency and spring
constant known, calculate the moment of
inertia for the gear.
• Apply the relations for simple harmonic
motion to calculate the maximum gear
velocity.

Edition
Sample Problem 19.3
19 - 16
s13.1
lb20
n =
=
τ
W
s93.1=nτ
SOLUTION:
• Using the free-body-diagram equation for the equivalence
of the external and effective moments, write the equation
of motion for the disk/gear and wire.
( ) :∑∑ = effOO MM
0=+
−=+
θθ
θθ
I
K
IK


K
I
I
K
n
nn π
ω
π
τω 2
2
===
• With the natural frequency and moment of inertia for the
disk known, calculate the torsional spring constant.
2
2
2
2
1 sftlb138.0
12
8
2.32
20
2
1
⋅⋅=











== mrI
K
138.0
213.1 π= radftlb27.4 ⋅=K

Edition
Sample Problem 19.3
19 - 17
s13.1
lb20
n =
=
τ
W
s93.1=nτ
radftlb27.4 ⋅=K
K
I
I
K
n
nn π
ω
π
τω 2
2
===
• With natural frequency and spring constant known,
calculate the moment of inertia for the gear.
27.4
293.1
I
π= 2
sftlb403.0 ⋅⋅=I
• Apply the relations for simple harmonic motion to
calculate the maximum gear velocity.
nmmnnmnm tt ωθωωωθωωθθ === sinsin
rad571.190 =°=mθ
( ) 





=





=
s93.1
2
rad571.1
2 π
τ
π
θω
n
mm
srad11.5=mω

Edition
Principle of Conservation of Energy
19 - 18
• Resultant force on a mass in simple harmonic motion
is conservative - total energy is conserved.
constant=+VT
=+
=+
222
2
2
12
2
1 constant
xx
kxxm
nω

• Consider simple harmonic motion of the square plate,
01 =T ( ) ( )[ ]
2
2
1
2
1 2sin2cos1
m
m
Wb
WbWbV
θ
θθ
≅
=−=
( ) ( )
( ) 22
3
5
2
1
22
3
2
2
12
2
1
2
2
12
2
1
2
m
mm
mm
mb
mbbm
IvmT
θ
ωθ
ω


=
+=
+= 02 =V
( ) 00 222
3
5
2
12
2
1
2211
+=+
+=+
nmm mbWb
VTVT
ωθθ bgn 53=ω

Edition
Sample Problem 19.4
19 - 19
Determine the period of small
oscillations of a cylinder which rolls
without slipping inside a curved
surface.
SOLUTION:
• Apply the principle of conservation of
energy between the positions of maximum
and minimum potential energy.
• Solve the energy equation for the natural
frequency of the oscillations.

Edition
Sample Problem 19.4
19 - 20
01 =T ( )( )
( )( )2
cos1
2
1
mrRW
rRWWhV
θ
θ
−≅
−−==
( ) ( )
( ) 22
4
3
2
2
2
2
1
2
12
2
1
2
2
12
2
1
2
m
mm
mm
rRm
r
rR
mrrRm
IvmT
θ
θθ
ω


−=





 −
+−=
+= 02 =V
SOLUTION:
• Apply the principle of conservation of energy between the
positions of maximum and minimum potential energy.
2211 VTVT +=+

Edition
Sample Problem 19.4
19 - 21
2211 VTVT +=+
( ) ( ) 0
2
0 22
4
3
2
+−=−+ m
m rRmrRW θ
θ 
( )( ) ( ) ( )22
4
3
2
2 mnm
m rRmrRmg ωθ
θ
−=−
rR
g
n
−
=
3
22
ω
g
rR
n
n
−
==
2
3
2
2
π
ω
π
τ
01 =T ( )( )22
1 mrRWV θ−≅
( ) 22
4
3
2 mrRmT θ−= 02 =V
• Solve the energy equation for the natural frequency of
the oscillations.

Edition
Forced Vibrations
19 - 22
:maF =∑
( ) xmxkWtP stfm =+−+ δωsin
tPkxxm fm ωsin=+
( ) xmtxkW fmst =−+− ωδδ sin
tkkxxm fm ωδ sin=+
Forced vibrations - Occur
when a system is subjected to
a periodic force or a periodic
displacement of a support.
=fω forced frequency

Edition
Forced Vibrations
19 - 23
[ ] txtCtC
xxx
fmnn
particulararycomplement
ωωω sincossin 21 ++=
+=
( ) ( )222
11 nf
m
nf
m
f
m
m
kP
mk
P
x
ωω
δ
ωωω −
=
−
=
−
=
tkkxxm fm ωδ sin=+
tPkxxm fm ωsin=+
At ωf = ωn, forcing input is in
resonance with the system.
tPtkxtxm fmfmfmf ωωωω sinsinsin2
=+−
Substituting particular solution into governing equation,

Edition
Sample Problem 19.5
19 - 24
A motor weighing 350 lb is supported
by four springs, each having a constant
750 lb/in. The unbalance of the motor
is equivalent to a weight of 1 oz located
6 in. from the axis of rotation.
Determine a) speed in rpm at which
resonance will occur, and b) amplitude
of the vibration at 1200 rpm.
SOLUTION:
• The resonant frequency is equal to the
natural frequency of the system.
• Evaluate the magnitude of the periodic
force due to the motor unbalance.
Determine the vibration amplitude
from the frequency ratio at 1200 rpm.

Edition
Sample Problem 19.5
19 - 25
SOLUTION:
• The resonant frequency is equal to the natural frequency of
the system.
ftslb87.10
2.32
350 2
⋅==m
( )
ftlb000,36
inlb30007504
=
==k
W = 350 lb
k = 4(350 lb/in)
rpm549rad/s5.57
87.10
000,36
==
==
m
k
nω
Resonance speed = 549 rpm

Edition
Sample Problem 19.5
19 - 26
W = 350 lb
k = 4(350 lb/in)
rad/s5.57=nω
• Evaluate the magnitude of the periodic force due to the
motor unbalance. Determine the vibration amplitude from
the frequency ratio at 1200 rpm.
( ) ftslb001941.0
sft2.32
1
oz16
lb1
oz1
rad/s125.7rpm1200
2
2
⋅=













=
===
m
f ωω
( )( )( ) lb33.157.125001941.0 2
12
6
2
==
== ωmrmaP nm
( ) ( )
in001352.0
5.577.1251
300033.15
1 22
−=
−
=
−
=
nf
m
m
kP
x
ωω
xm = 0.001352 in. (out of phase)

Edition
19 - 27
• With viscous damping due to fluid friction,
:maF =∑ ( )
0=++
=−+−
kxxcxm
xmxcxkW st

δ
• Substituting x = eλt
and dividing through by eλt
yields the characteristic equation,
m
k
m
c
m
c
kcm −





±−==++
2
2
22
0 λλλ
• Define the critical damping coefficient such that
nc
c m
m
k
mc
m
k
m
c
ω220
2
2
===−





• All vibrations are damped to some degree by
forces due to dry friction, fluid friction, or
internal friction.

Edition
19 - 28
• Characteristic equation,
m
k
m
c
m
c
kcm −





±−==++
2
2
22
0 λλλ
== nc mc ω2 critical damping coefficient
• Heavy damping: c > cc
tt
eCeCx 21
21
λλ
+= - negative roots
- nonvibratory motion
• Critical damping: c = cc
( ) tnetCCx ω−
+= 21 - double roots
- nonvibratory motion
• Light damping: c < cc
( ) ( )tCtCex dd
tmc
ωω cossin 21
2
+= −
=





−=
2
1
c
nd
c
c
ωω damped frequency

Edition
Damped Forced Vibrations
19 - 29
( )[ ] ( )( )[ ]
( )( )
( )
=
−
=
=
+−
==
2
222
1
2
tan
21
1
nf
nfc
nfcnf
m
m
m
cc
cc
x
kP
x
ωω
ωω
φ
ωωωω
δ
magnification
factor
phase difference between forcing and steady
state response
tPkxxcxm fm ωsin=++  particulararycomplement xxx +=

Edition
19 - 30
• Consider an electrical circuit consisting of an inductor,
resistor and capacitor with a source of alternating voltage
0sin =−−−
C
q
Ri
dt
di
LtE fm ω
• Oscillations of the electrical system are analogous to
damped forced vibrations of a mechanical system.
tEq
C
qRqL fm ωsin
1
=++ 

Edition
19 - 31
• The analogy between electrical and mechanical
systems also applies to transient as well as steady-
state oscillations.
• With a charge q = q0 on the capacitor, closing the
switch is analogous to releasing the mass of the
mechanical system with no initial velocity at x = x0.
• If the circuit includes a battery with constant voltage
E, closing the switch is analogous to suddenly
applying a force of constant magnitude P to the
mass of the mechanical system.

Edition
19 - 32
• The electrical system analogy provides a means of
experimentally determining the characteristics of a given
mechanical system.
• For the mechanical system,
( ) ( ) 0212112121111 =−++−++ xxkxkxxcxcxm 
( ) ( ) tPxxkxxcxm fm ωsin12212222 =−+−+ 
• For the electrical system,
( ) 0
2
21
1
1
21111 =
−
++−+
C
qq
C
q
qqRqL 
( ) tE
C
qq
qqRqL fm ωsin
2
12
12222 =
−
+−+ 
• The governing equations are equivalent. The characteristics
of the vibrations of the mechanical system may be inferred
from the oscillations of the electrical system.

Ch19

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Ch19