Tenth Edition
CHAPTER
19
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Phillip J. Cornwell
Lecture Notes:
Brian P. Self
Mechanical Vibrations
California Polytechnic State University
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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Contents
Introduction
Sample Problem 19.4
Free Vibrations of Particles. Simple
Harmonic Motion
Forced Vibrations
Simple Pendulum (Approximate
Solution)
Simple Pendulum (Exact Solution)
Sample Problem 19.1
Sample Problem 19.5
Damped Free Vibrations
Damped Forced Vibrations
Electrical Analogues
Free Vibrations of Rigid Bodies
Sample Problem 19.2
Sample Problem 19.3
Principle of Conservation of Energy
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19 - 2
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Vector Mechanics for Engineers: Dynamics
Because running in the International Space Station
might cause unwanted vibrations, they have installed a
Treadmill Vibration Isolation System.
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2-3
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Vector Mechanics for Engineers: Dynamics
Introduction
• 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.
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Vector Mechanics for Engineers: Dynamics
Free Vibrations of Particles. Simple Harmonic Motion
• 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,
ma  F  W  k  st  x   kx
mx  kx  0
• General solution is the sum of two particular solutions,
 k 
 k 
x  C1 sin 
t   C 2 cos
t 
 m 
 m 
 C1 sin  n t   C 2 cos n t 
• x is a periodic function and n is the natural circular
frequency of the motion.
• C1 and C2 are determined by the initial conditions:
x  C1 sin  nt   C2 cos nt 
C2  x0
v  x  C1 n cos nt   C2 n sin  nt 
C1  v0  n
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Vector Mechanics for Engineers: Dynamics
Free Vibrations of Particles. Simple Harmonic Motion
x  xm sin  nt   
xm 
v0  n 2  x02  amplitude
  tan 1 v0 x0 n   phase angle
n 
fn 
2
n
 period

 n  natural frequency
 n 2
1
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Vector Mechanics for Engineers: Dynamics
Free Vibrations of Particles. Simple Harmonic Motion
• 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.
x  xm sin  nt   
v  x
 xm n cos n t   
 xm n sin  n t     2
a  x
  xm n2 sin  n t   
 xm n2 sin  n t     
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Vector Mechanics for Engineers: Dynamics
Simple Pendulum (Approximate Solution)
• 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.
• Consider tangential components of acceleration and
force for a simple pendulum,
 Ft  mat :  W sin   ml
g


  sin   0
l
for small angles,
g
l
    0
   m sin  n t   
n 
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2
n
 2
l
g
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Vector Mechanics for Engineers: Dynamics
Simple Pendulum (Exact Solution)
g
l
An exact solution for
  sin   0
leads to
l  2
d
n  4

g 0 1  sin 2  2  sin 2 
m
which requires numerical solution.
n 
2K 
l
 2

 
g
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Vector Mechanics for Engineers: Dynamics
Concept Question
The amplitude of a
vibrating system is
shown to the right.
Which of the following
statements is true
(choose one)?
a) The amplitude of the acceleration equals the amplitude of
the displacement
b) The amplitude of the velocity is always opposite (negative to)
the amplitude of the displacement
c) The maximum displacement occurs when the acceleration
amplitude is a minimum
d) The phase angle of the vibration shown is zero
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.1
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.
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.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.1
k1  4 kN m k2  6 kN m
SOLUTION:
• Springs in parallel:
- determine the spring constant for equivalent spring
- apply the approximate relations for the harmonic motion
of a spring-mass system
k
104 N/m
n 

 14.14 rad s
m
20 kg
n 
P  k1  k2
k
P

2
n
 n  0.444 s
vm  x m  n
 0.040 m 14.14 rad s 
 k1  k2
 10 kN m  10 N m
4
vm  0.566 m s
am  x m an2
 0.040 m 14.14 rad s 2
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am  8.00 m s 2
19 - 12
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.1
k1  4 kN m k2  6 kN m
• 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 
k
2400N/m

 6.93 rad s
m
20 kg
2
n
 n  0.907 s
vm  x m  n
P  k1  k2
k
P

 k1  k2
 10 kN m  104 N m
 0.040 m 6.93 rad s 
vm  0.277 m s
am  x m an2
 0.040 m 6.93 rad s 2
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am  1.920 m s 2
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Vector Mechanics for Engineers: Dynamics
Free Vibrations of Rigid Bodies
• If an equation of motion takes the form
x   n2 x  0 or    n2  0
the corresponding motion may be considered
as simple harmonic motion.
• Analysis objective is to determine n.
• Consider the oscillations of a square plate
 W b sin    mb  I


1 m 2b 2  2b 2  2 mb 2 , W  mg
but I  12
3
3g
3g
sin    
 0
5b
5b
3g
2
5b
then  n 
, n 
 2
5b
n
3g
 
• For an equivalent simple pendulum,
l  5b 3
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.2
SOLUTION:
k
• 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.
A cylinder of weight W is suspended
as shown.
Determine the period and natural
frequency of vibrations of the cylinder.
• Substitute the kinematic relations to arrive
at an equation involving only the angular
displacement and acceleration.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.2
SOLUTION:
• From the kinematics of the system, relate the linear
displacement and acceleration to the rotation of the cylinder.
x  r
  2 x  2r

 


 
a  r  r
a  r
• Based on a free-body-diagram equation for the equivalence of
the external and effective forces, write the equation of motion.
 M A   M A eff :
Wr  T2 2r   ma r  I
but T2  T0  k  12 W  k 2r 
• Substitute the kinematic relations to arrive at an equation
involving only the angular displacement and acceleration.
Wr  1 W  2kr 2r   mrr  1 mr 2
2
 

2
8k
 0
3m
8k
n 
3m
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n 
2
n
 2
3m
8k
fn 
 n 1 8k

2 2 3m
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.3
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.
W  20 lb
 n  1.13 s
 n  1.93 s
• With the natural frequency and moment
of inertia for the disk known, calculate
the torsional spring constant.
The disk and gear undergo torsional
vibration with the periods shown.
• With natural frequency and spring
Assume that the moment exerted by the constant known, calculate the moment of
wire is proportional to the twist angle.
inertia for the gear.
Determine a) the wire torsional spring • Apply the relations for simple harmonic
constant, b) the centroidal moment of
motion to calculate the maximum gear
inertia of the gear, and c) the maximum velocity.
angular velocity of the gear if rotated
through 90o and released.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.3
W  20 lb
 n  1.13 s
 n  1.93 s
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.
 K   I 
 M O   M O eff :
K
    0
I
n 
K
I
n 
2
n
 2
I
K
• With the natural frequency and moment of inertia for the
disk known, calculate the torsional spring constant.
2
1  20  8 
2
I  12 mr 2  
   0.138 lb  ft  s
2  32.2  12 
1.13  2
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0.138
K
K  4.27 lb  ft rad
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.3
• With natural frequency and spring constant known,
calculate the moment of inertia for the gear.
1.93  2
I  0.403 lb  ft  s 2
• Apply the relations for simple harmonic motion to
calculate the maximum gear velocity.
W  20 lb
 n  1.13 s
I
4.27
 n  1.93 s
   m sin nt
   mn sin nt
m   mn
 m  90  1.571 rad
n 
K
I
n 
2
n
 2
I
K
 2 
2 
  1.571 rad


1
.
93
s


 n
m   m 
K  4.27 lb  ft rad
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m  5.11rad s
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
• Using the free-body and kinetic diagrams,
write the equation of motion for the
pendulum.
• Determine the natural frequency and
moment of inertia for the disk (use the
small angle approximation).
• Calculate the period.
A uniform disk of radius 250 mm is
attached at A to a 650-mm rod AB of
negligible mass which can rotate freely
in a vertical plane about B. If the rod is
displaced 2°from the position shown
and released, determine the period of
the resulting oscillation.
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Draw the FBD and KD of the pendulum (mbar ~ 0).
Bn
Bt

l
r
man
mat
I
mg
Determine the equation of motion.
M B  I B


mgl sin   I  ml 2 
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*Note that you could also do
this by using the “moment”
from at, and that at = l
mgl sin   I   lmat
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Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Find I, set up equation of motion
using small angle approximation


mgl sin   I  ml 2 
I 
1 2
mr , sin   
2
1 2
2
mr

ml
2
  mgl  0


Determine the natural frequency
n2 

gl
r2
2
 l2

(9.81)(0.650)

2
2
1
(0.250)

(0.650)
2
 14.053
Calculate the period
n 
2
n
 1.676 s
 n  1.676 s
n  3.7487 rad/s
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Vector Mechanics for Engineers: Dynamics
Concept Question
In the previous problem, what
would be true if the bar was
hinged at A instead of welded at A
(choose one)?
a) The natural frequency of the oscillation would be larger
b) The natural frequency of the oscillation would be larger
c) The natural frequencies of the two systems would be the same
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Vector Mechanics for Engineers: Dynamics
Principle of Conservation of Energy
• Resultant force on a mass in simple harmonic motion
is conservative - total energy is conserved.
1 mx 2  1 kx 2  constant
T V  constant
2
2
x 2   n2 x 2 
• Consider simple harmonic motion of the square plate,
T1  0
V1  Wb 1  cos   Wb 2 sin 2  m 2 


 12 Wb  m2
2
T2  12 mvm2  12 I  m
2
 12 mbm   12
 12
53 mb2 m2
23 mb2  m2
V2  0
T1  V1  T2  V2
0  12 Wb  m2  12
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53 mb2  m2 n2  0
 n  3g 5b
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.4
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.
Determine the period of small
oscillations of a cylinder which rolls
without slipping inside a curved
surface.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.4
SOLUTION:
• Apply the principle of conservation of energy between the
positions of maximum and minimum potential energy.
T1  V1  T2  V2
T1  0
V1  Wh  W R  r 1  cos 

 W R  r   m2 2

V2  0
2
T2  12 mvm2  12 I  m


2
 R  r  2
 12 mR  r m2  12 12 mr 2 
 m
 r 
 3 mR  r 2m2
4
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.4
• Solve the energy equation for the natural frequency of the
oscillations.

T1  0
V1  W R  r   m2 2
T2  34 mR  r 2m2
V2  0

T1  V1  T2  V2
0  W R  r 
mg R  r 
 n2 
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 m2
2
 m2
2
2 g
3 Rr
 34 m R  r 2m2  0
 34 m R  r 2  m n 2m
n 
2
n
 2
3 Rr
2 g
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Vector Mechanics for Engineers: Dynamics
Forced Vibrations
Forced vibrations - Occur
when a system is subjected to
a periodic force or a periodic
displacement of a support.
 f  forced frequency
 F  ma :
Pm sin  f t  W  k  st  x   mx
W  k  st  x   m sin  f t   mx
mx  kx  Pm sin  f t
mx  kx  k m sin  f t
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Vector Mechanics for Engineers: Dynamics
Forced Vibrations
x  xcomplementary  x particular
 C1 sin  n t  C2 cos n t   xm sin  f t
Substituting particular solution into governing equation,
 m 2f xm sin  f t  kxm sin  f t  Pm sin  f t
xm 
Pm
k  m 2f

Pm k
1   f  n 2

m
1   f  n 2
mx  kx  Pm sin  f t
mx  kx  k m sin  f t
At f = n, forcing input is in
resonance with the system.
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19 - 29
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Vector Mechanics for Engineers: Dynamics
Concept Question
A small trailer and its load have
a total mass m. The trailer can
be modeled as a spring with
constant k. It is pulled over a
road, the surface of which can
be approximated by a sine
curve with an amplitude of 40
mm and a wavelength of 5 m.
Maximum vibration amplitude
occur at 35 km/hr. What
happens if the driver speeds up
to 50 km/hr?
a) The vibration amplitude remains the same.
b) The vibration amplitude would increase.
c) The vibration amplitude would decrease.
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Tenth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.5
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.
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.
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19 - 31
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.5
SOLUTION:
• The resonant frequency is equal to the natural frequency of
the system.
m
350
 10.87 lb  s 2 ft
32.2
k  4750  3000 lb in
W = 350 lb
k = 4(350 lb/in)
 36,000 lb ft
k
36,000

m
10.87
 57.5 rad/s  549 rpm
n 
Resonance speed = 549 rpm
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Vector Mechanics for Engineers: Dynamics
Sample Problem 19.5
• Evaluate the magnitude of the periodic force due to the
motor unbalance. Determine the vibration amplitude from
the frequency ratio at 1200 rpm.
 f    1200 rpm  125.7 rad/s

1
 1 lb 
  0.001941 lb  s 2 ft
m  1 oz 

 16 oz  32.2 ft s 2 
W = 350 lb
k = 4(350 lb/in)
n  57.5 rad/s
Pm  man  mr 2
 
6 125.7 2  15.33 lb
 0.001941 12
xm 
Pm k
1   f  n 2

15.33 3000
1  125.7 57.52
 0.001352 in
xm = 0.001352 in. (out of phase)
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Vector Mechanics for Engineers: Dynamics
Damped Free Vibrations
• All vibrations are damped to some degree by
forces due to dry friction, fluid friction, or
internal friction.
• With viscous damping due to fluid friction,
 F  ma :
W  k  st  x   cx  mx
mx  cx  kx  0
• Substituting x = elt and dividing through by elt
yields the characteristic equation,
ml2  cl  k  0
2
c
k
 c 
l
 


2m
m
 2m 
• Define the critical damping coefficient such that
2
k
 cc 

  0
m
 2m 
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cc  2 m
k
 2m n
m
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Vector Mechanics for Engineers: Dynamics
Damped Free Vibrations
• Characteristic equation,
2
c
k
 c 
l
 


2m
m
 2m 
ml2  cl  k  0
cc  2m n  critical damping coefficient
• Heavy damping: c > cc
x  C1e l1t  C 2 e l2t
- negative roots
- nonvibratory motion
• Critical damping: c = cc
x  C1  C 2t  e  nt
- double roots
- nonvibratory motion
• Light damping: c < cc
x  e c 2mt C1 sin  d t  C2 cos d t 
2
d  n
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c
1     damped frequency
 cc 
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Vector Mechanics for Engineers: Dynamics
Concept Question
8
6
4
2
0
-2
-4
-6
-8
0
0.5
1
1.5
2
2.5
3
3.
The graph above represents an oscillation that is…
a) Heavily damped b) critically damped
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c) lightly damped
2 - 36
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Vector Mechanics for Engineers: Dynamics
Concept Question
8
6
4
2
0
-2
-4
-6
-8
0
0.5
1
1.5
2
2.5
3
3.
The period for the oscillation above is approximately…
a) 1.25 seconds
b) 2.5 Hz
c) 0.6 seconds
Estimate the phase shift for the oscillation
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Zero
2 - 37
Tenth
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Vector Mechanics for Engineers: Dynamics
Forced vibrations can be caused by a test machine, by rocks on a trail, by
rotating machinery, and by earthquakes. Suspension systems, shock
absorbers, and other energy-dissipating devices can help to dampen the
resulting vibrations.
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2 - 38
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Vector Mechanics for Engineers: Dynamics
Damped Forced Vibrations
mx  cx  kx  Pm sin  f t
x  xcomplementary  x particular
xm
xm


Pm k 
1
tan  

1  
f
2c cc   f  n

1  f n

2
n 

 magnification
factor
 2c cc   f  n 2
 
2 2


 phase difference between forcing and steady
state response
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19 - 39
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
SOLUTION:
• Determine the system natural
frequency, damping constant, and
the unbalanced force.
• Determine the steady state response
and the magnitude of the motion.
A simplified model of a washing machine is shown. A bundle of wet
clothes forms a mass mb of 10 kg in the machine and causes a rotating
unbalance. The rotating mass is 20 kg (including mb) and the radius of the
washer basket e is 25 cm. Knowing the washer has an equivalent spring
constant k = 1000 N/m and damping ratio z = c/cc = 0.05 and during the
spin cycle the drum rotates at 250 rpm, determine the amplitude of the
motion.
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19 - 40
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Given: m= 20 kg, k= 1000 N/m,
f= 250 rpm, e= 25 cm, mb= 10 kg
Find: xm
Calculate the forced circular
frequency and the natural circular
frequency
f 
(2 )(250)
 26.18 rad/s
60
n 
k
1000

 7.0711 rad/s
m
20
Calculate the critical damping constant cc and the damping constant c
cc  2 km  2 (1000)(20)  282.84 N  s/m
c
c
 cc

 c  (0.05)(141.42)  14.1421 N  s/m

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Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Group Problem Solving
Calculate the unbalanced force
caused by the wet clothes
Pm  mb e 2f
Pm  (10 kg)(0.25 m)(26.18 rad/s)2  1713.48 N
Use Eq 19.52 to determine xm
mx  cx  kx  Pm sin  f t
xm 


Pm
k
 m 2f

2
 (c f )2
1713.48
[1000  (20)(26.18)2 ]2  [(141421)(26.18)]2
1713.48
(12,707.8)2  (370.24)2

xm  134.8 mm
1713.48
 0.13478 m
12,713.2
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19 - 42
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Concept Question
The following parameters were
found in the previous problem:
 f  26.18 rad/s
 =0.05
n  7.0711 rad/s
What would happen to the
amplitude xm if the forcing
frequency f was cut in half?
a) The vibration amplitude remains the same.
b) The vibration amplitude would increase.
c) The vibration amplitude would decrease.
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2 - 43
Tenth
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Vector Mechanics for Engineers: Dynamics
Concept Question
Case 1
 f  26.18 rad/s
n  7.0711 rad/s
Case 2
 f  13.09 rad/s
n  7.0711 rad/s
xm 
Pm

k  m 2f

2
 (c f )2
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2 - 44
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Vector Mechanics for Engineers: Dynamics
Electrical Analogues
• Consider an electrical circuit consisting of an inductor,
resistor and capacitor with a source of alternating voltage
di
q
Em sin  f t  L  Ri   0
dt
C
1
Lq  Rq  q  Em sin  f t
C
• Oscillations of the electrical system are analogous to
damped forced vibrations of a mechanical system.
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19 - 45
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Electrical Analogues
• The analogy between electrical and mechanical
systems also applies to transient as well as steadystate 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.
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19 - 46
Tenth
Edition
Vector Mechanics for Engineers: Dynamics
Electrical Analogues
• The electrical system analogy provides a means of
experimentally determining the characteristics of a given
mechanical system.
• For the mechanical system,
m1x1  c1 x1  c2  x1  x 2   k1 x1  k 2  x1  x2   0
m2 x2  c2  x2  x1   k 2  x2  x1   Pm sin  f t
• For the electrical system,
q q q
L1q1  R1 q1  q 2   1  1 2  0
C1
C2
q2  q1


L2 q2  R2 q 2  q1 
 Em sin  f t
C2
• The governing equations are equivalent. The characteristics
of the vibrations of the mechanical system may be inferred
from the oscillations of the electrical system.
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19 - 47