Engineering Vibration
4th Edition
Prentice Hall
© D. J. Inman
daninman@umich.edu
We make our living in dynamics and
vibration
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Vibrations are Everywhere
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(Show movies)
Modeling and Degrees of Freedom
The examples on the previous slide many degrees of freedom
and many parts, we will start with one degree of freedom and
work towards many.
• Recall from your study of statics and
physics that a degree of freedom is the
independent parameter needed to describe
the configuration of a physical system
• So single degree of freedom system, which
is where we start, is a system whose
position in time and space can be defined
by
one
coordinate,
a displacement
Lack
of consideration
of here
dynamic
loads and vibrationor
caused this to relatively new bridge to vibrate wildly
position.
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3/43
Degrees of Freedom: The Minimum
Number of coordinates to specify a
configuration
• For a single particle confined to a line, one
coordinate suffices so it has one degree of
freedom
• For a single particle in a plane two
coordinates define its location so it has two
degrees of freedom
• A single particle in space requires three
coordinates so it has three degrees of
freedom
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Example 1.1.1 The Pendulum
• Sketch the structure or
part of interest
• Write down all the
forces and make a
“free body diagram”
• Use Newton’s Law
and/or Euler’s Law to
find the equations of
motion
∑ M = J α , J = m
0
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0
2
0
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The problem is one dimensional,
hence a scalar equation results
J 0α (t) = −mgl sin θ (t) ⇒ m θ(t) + mg
θ (t) = 0
sin

2
restoring
force
Here the over dots denote differentiation with respect to time t
This is a second order, nonlinear ordinary differential equation
We can linearize the equation by using the approximation sin θ ≈ θ
g


⇒ m θ (t) + mgθ (t) = 0 ⇒ θ (t) + θ (t) = 0

2
Requires knowledge of θ (0) and θ(0)
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the initial position and velocity.
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Next consider a spring mass system
and perform a static experiment:
• From strength of
materials recall:
FBD:
nonlinear
linear
A plot of force versus displacement:
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experiment ⇒ fk = kx
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Free-body diagram and equation
of motion
•Newton’s Law:
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m
x(t) = −kx(t) ⇒ m
x(t) + kx(t) = 0
 = v0
x(0) = x0 , x(0)
(1.2)
Again a 2nd order ordinary differential equation
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Stiffness and Mass
Vibration is cause by the interaction of two different forces
one related to position (stiffness) and one related to
acceleration (mass).
Proportional to displacement
Stiffness (k)
Displacement
x
fk = −kx(t)
Mass (m)
k
statics
fm = ma(t) = m
x(t)
m
Mass
Spring
dynamics
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Proportional to acceleration
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Examples of Single-Degree-ofFreedom Systems
Pendulum
Shaft and Disk
l =length
Gravity g
θ
Torsional
Stiffness
k
m
θ
g

θ (t) + θ (t) = 0
l
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Moment
of inertia
J
Jθ(t) + kθ (t) = 0
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Solution of 2nd order DEs
x(t)
Lets assume a solution:
x(t) = Asin(ω nt + φ ) (1.3)
t
Differentiating twice gives:
 = ω n A cos(ω nt + φ )
x(t)
(1.4)

x(t) = −ω n2 Asin(ω nt + φ ) = -ω n2 x(t)
(1.5)
Substituting back into the equations of motion gives:
−mω n2 Asin(ω nt + φ ) + kAsin(ω nt + φ ) = 0
−mω + k = 0
2
n
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or
ωn =
k
m
Natural
frequency
rad/s
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Summary of simple harmonic
motion
x(t)
Period
2π
T=
ωn
x0
Amplitude
A
Slope
here is v0
φ
t
Maximum
Velocity
ωn
ωn A
ω n rad/s
ω n cycles ω n
fn =
=
=
Hz
2π rad/cycle
2π s
2π
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12/43
Initial Conditions
If a system is vibrating then we must assume that something
must have (in the past) transferred energy into to the system
and caused it to move. For example the mass could have
been:
•moved a distance x0 and then released at t = 0 (i.e. given
Potential energy) or
•given an initial velocity v0 (i.e. given some kinetic energy) or
•Some combination of the two above cases
From our earlier solution we know that:
x0 = x(0) = Asin(ω n 0 + φ ) = Asin(φ )
 = ω n A cos(ω n 0 + φ ) = ω n A cos(φ )
v0 = x(0)
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Initial Conditions Determine the
Constants of Integration
Solving these two simultaneous equations for A and ϕ gives:
1
2 2
2
−1 ⎛ ω n x0 ⎞
A=
ω n x0 + v0 , φ = tan ⎜
ωn
⎝ v0 ⎟⎠
 
Amplitude
Phase
Slope
here is v0
x(t)
x0
1
ω n2 x02 + v02
ωn
t
φ
ωn
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x0
φ
v0
ωn
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Thus the total solution for the
spring mass system becomes:
ω n2 x02 + v02
⎛
−1 ω n x0 ⎞
x(t) =
sin ⎜ ω nt + tan
ωn
v0 ⎟⎠
⎝
(1.10)
Called the solution to a simple harmonic oscillator
and describes oscillatory motion, or simple harmonic motion.
Note (Example 1.1.2)
ω n2 x02 + v02
x(0) =
ωn
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ω n x0
ω x +v
2 2
n 0
as it should
2
0
= x0
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A note on arctangents
•
Note that calculating arctangent from a
calculator requires some attention. First, all
machines work in radians.
•
The argument atan(-/+) is in a different quadrant
then atan(+/-), and usual machine calculations
will return an arctangent in between -π/2 and
+π/2, reading only the atan(-) for these two
different cases.
+
φ
-
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+
+
_
φ
+
In MATLAB, use the atan2(x,y) function to get
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the correct phase.
Example 1.1.3 wheel, tire suspension
m = 30 kg, fn = 10 hz, what is k?
⎛ cylce 2π rad ⎞
5
k = mω = ( 30 kg ) ⎜ 10
i
= 1.184 × 10 N/m
⎟
⎝
sec cylce ⎠
2
n
There are of course more complex models of suspension systems
and these appear latter in the course as our tools develope
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Section 1.2 Harmonic Motion
The period is the time elapsed to complete one complete cylce
2π rad
2π
T=
=
s
ω n rad/s ω n
(1.11)
The natural frequency in the commonly used units of hertz:
ωn
ω n rad/s
ω n cycles ω n
fn =
=
=
=
Hz
2π 2π rad/cycle
2π s
2π
For the pendulum:
ωn =
(1.12)
g

rad/s, T = 2π
s

g
For the disk and shaft:
ωn =
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k
J
rad/s, T = 2π
s
J
k
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Relationship between Displacement,
Velocity and Acceleration
1
x(t) = Asin(ω nt + φ )
x
Displacement
0
-1
0
20
 = ω n A cos(ω nt + φ )
x(t)
v
Velocity
a

x(t) = −ω n2 Asin(ω nt + φ )
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0.1
0.2
0.3
0.4
0.5 0.6
Time (sec)
0.7
0.8
0.9
1
0
-20
0
200
Acceleration
0
-200
0
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A=1, ωn=12
Note how the relative magnitude of each increases for ωn>1
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Example 1.2.1Hardware store spring, bolt: m= 49.2x10
-3
kg, k=857.8 N/m and x0 =10 mm. Compute ωn and the max amplitude
of vibration.
Note: common
Units are Hertz
k
857.8 N/m
ωn =
=
= 132 rad/s
-3
m
49.2 ×10 kg
To avoid Costly errors use f
ωn
when working in Hertz and ω
fn =
= 21 Hz
when in rad/s
2π
2π 1
1
T=
= =
0.0476 s
ωn fn 21 cyles
sec
0
1
2 2
2
x(t ) max = A =
ωn x0 + v0 = x0 = 10 mm
ωn
n
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Units depend on system 20/43
n
Compute the solution and max velocity and
acceleration:
v(t )max = ωn A = 1320 mm/s= 1.32 m/s
2.92 mph
a(t ) max = ω n A = 174.24 ×10 mm/s
2
3
2
= 174.24 m/s ≈ 17.8g!
g = 9.8 m/s2
π
−1 ⎛ω n x0 ⎞
90°
φ = tan ⎜
⎟ = rad
⎝ 0 ⎠ 2
x(t ) = 10 sin(132t + π / 2) = 10 cos(132t ) mm
2
~0.4 in max
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Does gravity matter in spring
problems?
Δ
Let Δ be the deflection caused by
hanging a mass on a spring
(Δ = x1-x0 in the figure)
Then from static equilibrium:
mg = kΔ
Next sum the forces in the vertical for some point x > x1 measured
from Δ m
x = −k ( x + Δ ) + mg = −kx + mg − kΔ



=0
⇒ m
x(t) + kx(t) = 0
So no, gravity does not have an effect on the vibration
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(note that this is not the case if the spring is nonlinear)
© D. J. Inman
22/43
Example 1.2.2 Pendulums and
measuring g
• A 2 m pendulum
swings with a period
of 2.893 s
• What is the
acceleration due to
gravity at that
location?
2π

T=
= 2π
ωn
g
4π
4π
g= 2 =
2m
2 2
T
2.893 s
⇒ g = 9.796 m/s 2
2
2
This is g in Denver, CO USA, at 1638m
and a latitude of 40°
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23/43
Review of Complex Numbers and Complex
Exponential (See Appendix A)
A complex number can be written with a real and imaginary
part or as a complex exponential
c = a + jb = Ae jθ
Where
a = A cosθ ,b = Asin θ
ℑ
a
Multiplying two complex numbers:
c1c2 = A1 A2 e
j (θ1 + θ 2 )
Dividing two complex numbers:
c1 A1 j (θ1 −θ2 )
=
e
c2 A2
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A
b
ℜ
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Equivalent Solutions to 2nd order
Differential Equations (see Window 1.4)
All of the following solutions are equivalent:
x(t) = Asin(ω nt + φ )
Called the magnitude and phase form
x(t) = A1 sin ω nt + A2 cos ω nt Sometimes called the Cartesian form
x(t) = a1e jω n t + a2 e− jω n t
Called the polar form
The relationships between A and ϕ, A1 and A2, and a1 and a2
can be found in Window 1.4 of the text, page 19..
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•Each is useful in different situations
•Each represents the same information
•Each solves the equation of motion 25/43
Derivation of the solution
Substitute x(t) = aeλt into m
x + kx = 0 ⇒
mλ 2 aeλt + kaeλt = 0 ⇒
mλ 2 + k = 0 ⇒
k
k
λ = ± − =±
j = ±ω n j ⇒
m
m
x(t) = a1eω n jt and x(t) = a2 e−ω n jt ⇒
x(t) = a1eω n jt + a2 e−ω n jt
(1.18)
This approach will be used again for more complicated problems
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26/43
Is frequency always positive?
From the preceding analysis, λ = + ωn then
ω n jt
x(t) = a1e
+ a2 e
− ω n jt
Using the Euler relations* for trigonometric functions, the
above solution can be written as (recall Window 1.4)
x(t) = Asin (ω nt + φ )
(1.19)
It is in this form that we identify as the natural frequency
ωn and this is positive, because the + sign being used up
in the transformation from exponentials to the sine
function.
* http://en.wikipedia.org/wiki/Euler's_formula
e = cos x + i sin x
ix
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27/43
Calculating root mean square
(RMS) values
May need to be limited due
to physical constraints
A = peak value
T
Not very useful since for
a sine function the
average value is zero
1
x = lim ∫ x(t)dt = average value
T →∞ T
0
T
1 2
2
x = lim ∫ x (t)dt = mean-square value
T →∞ T
0
xrms = x 2 = root mean square value
(1.21)
Proportional
to energy
Also useful when the vibration is random
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28/43
The Decibel or dB scale
It is often useful to use a logarithmic scale to plot vibration
levels (or noise levels). One such scale is called the decibel or
dB scale. The dB scale is always relative to some reference
value x0. It is define as:
2
⎛ x⎞
⎛ x⎞
dB = 10 log10 ⎜ ⎟ = 20 log10 ⎜ ⎟
⎝x ⎠
⎝x ⎠
0
(1.22)
0
For example: if an acceleration value was 19.6m/s2 then relative
to 1g (or 9.8m/s2) the level would be 6dB,
2
⎛ 19.6 ⎞
10 log10 ⎜
= 20 log10 ( 2 ) = 6dB
⎟
⎝ 9.8 ⎠
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Or for Example 1.2.1: The Acceleration Magnitude
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is 20log10(17.8)=25dB relative to 1g.
1.3 Viscous Damping
All real systems dissipate energy when they vibrate. To
account for this we must consider damping. The most simple
form of damping (from a mathematical point of view) is called
viscous damping. A viscous damper (or dashpot) produces a
force that is proportional to velocity.
Mostly a mathematically motivated form, allowing
a solution to the resulting equations of motion that predicts
reasonable (observed) amounts of energy dissipation.
Damper (c)

fc = −cv(t) = −cx(t)
x
fc
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30/43
Differential Equation Including
Damping
For this damped single degree of freedom system the force acting
on the mass is due to the spring and the dashpot i.e. fm= - fk - fc.
Displacement
x

m
x(t) = −kx(t) − cx(t)
M
or
 + kx(t) = 0 (1.25)
m
x(t) + cx(t)
k
c
To solve this for of the equation it is useful to assume a
solution of the form (again):
x(t) = ae
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λt
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Solution to DE with damping
included (dates to 1743 by Euler)
The velocity and acceleration can then be calculated as:
x˙(t) = λae λt
x˙˙(t) = λ2 ae λt
If this is substituted into the equation of motion we get:
aeλt (mλ 2 + cλ + k) = 0
(1.26)
Divide equation by m, substitute for natural frequency and
assume a non-trivial solution
λt
ae ≠ 0
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⇒
c
( λ + λ + ω n2 ) = 0
m
2
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Solution to our differential
equation with damping included:
For convenience we will define a term known as the
damping ratio as:
c
ζ=
2 km
(1.30)
Lower case Greek zeta
NOT ξ as some like to use
The equation of motion then becomes:
(λ + 2ζω n λ + ω ) = 0
2
2
n
Solving for λ then gives,
λ1,2 = −ζω n ± ω n ζ 2 − 1
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(1.31)
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Possibility 1. Critically damped motion
Critical damping occurs when ζ =1. The damping coefficient
c in this case is given by:
ζ =1 ⇒ c = ccr = 2 km = 2mω n

Solving for λ then gives,
definition of critical
damping coefficient
λ1,2 = −1ωn ± ωn 1 −1 = −ωn
2
The solution then takes the form
A repeated, real root
x(t) = a1e−ω n t + a2te−ω n t
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Needs two independent solutions, hence the t
in the second term
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Critically damped motion
a1 and a2 can be calculated from initial conditions (t=0),
x = (a1 + a2t)e−ω n t
⇒ a1 = x0
•
•
No oscillation occurs
Useful in door
mechanisms, analog
gauges
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0.6
x0=0.4mm v0=1mm/s
0.5
Displacement (mm)
v = (−ω n a1 − ω n a2t + a2 )e−ω n t
v0 = −ω n a1 + a2
⇒ a2 = v0 + ω n x0
k=225N/m m=100kg and ζ = 1
x0=0.4mm v0=0mm/s
x0=0.4mm v0=-1mm/s
0.4
0.3
0.2
0.1
0
-0.1
0
1
2
Time (sec)
3
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4
Possibility 2: Overdamped motion
An overdamped case occurs when ζ >1. Both of the roots of the
equation are again real.
k=225N/m m=100kg and ζ =2
0.6
λ1,2 = −ζωn ± ωn ζ −1
2
0.5
2
x(t) = e−ζωn t (a1e−ωn t ζ −1 + a2 eω n t ζ −1 )
a1 and a2 can again be calculated
from initial conditions (t=0),
−v0 + ( − ζ + ζ 2 −1)ωn x0
a1 =
2ω n ζ 2 −1
Displacement (mm)
2
x0=0.4mm v0=1mm/s
x0=0.4mm v0=0mm/s
x0=0.4mm v0=-1mm/s
0.4
0.3
0.2
0.1
0
-0.1
0
1
2
Time (sec)
v0 + (ζ + ζ −1)ωn x0
a2 =
2ω n ζ 2 −1
Slower to respond than
2
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critically damped case
3
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4
Possibility 3: Underdamped motion
An underdamped case occurs when ζ <1. The roots of the
equation are complex conjugate pairs. This is the most
common case and the only one that yields oscillation.
λ1,2 = −ζω n ± ωn j 1− ζ 2
x(t) = e
−ζωn t
(a1e
jωn t 1−ζ 2
+ a2 e
− jω n t 1−ζ 2
)
= Ae −ζω n t sin(ω d t + φ)
The frequency of oscillation ωd is called the damped natural
frequency is given by.
ωd = ωn 1 − ζ 2
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(1.37)
37/43
Constants of integration for the
underdamped motion case
As before A and ϕ can be calculated from initial conditions (t=0),
1
(v 0 + ζωn x0 )2 + (x 0ω d ) 2
ωd
⎛ x0ω d ⎞
−1
φ = tan ⎜
⎟
⎝ v0 + ζω n x0 ⎠
1
A=
Gives an oscillating
response with exponential
decay
• Most natural systems vibrate
with and underdamped
response
• See Window 1.5 for details
and other representations
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Displacement
•
0.5
0
-0.5
-1
0
1
2
3
Time (sec)
4
5
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Example 1.3.1: consider the spring of 1.2.1, if c = 0.11
kg/s, determine the damping ratio of the spring-bolt system.
m = 49.2 × 10
−3
kg, k = 857.8 N/m
−3
ccr = 2 km = 2 49.2 × 10 × 857.8
= 12.993 kg/s
c
0.11 kg/s
ζ=
=
= 0.0085 ⇒
ccr 12.993 kg/s
the motion is underdamped
and the bolt will oscillate
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39/43
Example 1.3.2
The human leg has a measured natural
frequency of around 20 Hz when in its rigid (knee
locked) position, in the longitudinal direction (i.e.,
along the length of the bone) with a damping
ratio of ζ = 0.224. Calculate the response of the
tip if the leg bone to an initial velocity of v0 = 0.6
m/s and zero initial displacement (this would
correspond to the vibration induced while landing
on your feet, with your knees locked form a
height of 18 mm) and plot the response. What is
the maximum acceleration experienced by the
leg assuming no damping?
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40/43
Solution:
20 cycles 2π rad
ωn =
= 125.66 rad/s
1
s cycles
ω d = 125.66 1− (.224) = 122.467 rad/s
2
(0.6 + (0.224 )(125.66)( 0)) + (0)(122.467)2
2
A=
122.467
= 0.005 m
⎛ (0)(ω d ) ⎞
⎟=0
φ = tan ⎜
⎝ v0 + ζω n (0 )⎠
-1
⇒ x( t ) = 0.005e −2 8.148t sin(122.467t )
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41/43
Use the undamped formula to get maximum
acceleration:
2
⎛
⎞
v
A = x 02 + ⎜ 0 ⎟ , ω n = 125.66, v 0 = 0.6, x 0 = 0
⎝ωn ⎠
v0
0.6
A=
m=
m
ωn
ωn
⎛ 0.6 ⎞
max( x˙˙) = −ω A = −ω ⎜ ⎟ = (0.6)(125.66 m/s2 ) = 75.396 m/s2
⎝ ωn ⎠
2
n
2
n
2
75.396
m/s
maximum acceleration =
g = 7.68 g's
2
9.81 m/s
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42/43
Here is a plot of the displacement
response versus time
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43/43
Example 1.3.3 Compute the form of the response of an
underdamped system using the Cartesian form of the solution given in
Window 1.5.
sin(x + y) = sin x sin y + cos x cos y ⇒
x(t) = Ae−ζω n t sin(ω d t + φ ) = e−ζω n t (A1 sin ω d t + A2 cos ω d t)
x(0) = x0 = e0 (A1 sin(0) + A2 cos(0)) ⇒ A2 = x0
x = −ζω n e−ζω n t (A1 sin ω d t + A2 cos ω d t)
+ ω d e−ζω n t (A1 cos ω d t − A2 sin ω d t)
v0 = −ζω n (A1 sin 0 + x0 cos 0) + ω d (A1 cos 0 − x0 sin 0)
v0 + ζω n x0
⇒ A1 =
⇒
ωd
⎛ v + ζω n x0
⎞
x(t) = e−ζω n t ⎜ 0
sin ω d t + x0 cos ω d t ⎟
ωd
⎝
⎠
University of Michigan
© D. J. Inman
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