mass-spring sys + harmonic mass-spring sys + harmonic mass-spring sys + harmonic
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06-92-315
Mechanical Vibrations
Lecture 2
Mass Spring System
and Harmonic Motion
Dr. Ahmed Deif
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Spring Mass System
From strength of
materials recall:
A plot of force versus displacement:
FBD:
linear
nonlinear
experiment fk kx
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Free-body diagram and
equation of motion
Newtons Law:
Again a 2nd
order ordinary differential equation
(1.2)
m&&x(t) kx(t) m&&x(t) kx(t) 0
x(0) x0 , &x(0) v0
)(tXm..
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Solution to 2nd order DEs
x(t) Asin(nt)
Lets assume a solution:
Differentiating twice gives:
x(t) nAcos(nt)&&x(t) n
2Asin(nt) -n
2x(t)
Substituting back into the equations of motion gives:
mn2Asin(nt) kAsin(nt) 0
mn2 k 0 or n
k
mNatural
frequency
t
x(t)
rad/s
)(tX..
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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).
m
k
x
Displacement
Mass Spring
Stiffness (k)
Mass (m)
fk kx(t)
fm ma(t) m&&x(t)
Proportional to displacement
Proportional to acceleration
statics
dynamics
)(tXm..
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Examples of Single-Degree-of-
Freedom Systems
m
ql =length
Pendulum
q(t) g
lq(t) 0
Gravity g
Shaft and Disk
q
J&&q(t) kq(t) 0
TorsionalStiffness
kMoment
of inertia
J
)(tJ..
q
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Initial Conditions
If a system is vibrating then we must assume that somethingmust 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 x0and then released at t = 0(i.e. given
Potential energy) or
given an initial velocity v0(i.e. given Kinetic energy) or
Some combination of the two above cases
From our earlier solution we know that:
x0 x(0) Asin(n 0 ) Asin()
v0 &x(0) nA cos(n 0 ) nA cos())0(.
X
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Initial ConditionsSolving these equation gives:
v0
n
x0
1
nn
2x02 v0
2
t
x(t)
x0
Slope
here is v0
n
A 1
nn
2x0
2 v02
Amplitude
1 24 4 4 34 4 4
, tan1nx0
v0
Phase
1 24 4 34 4
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The total solution for the spring
mass system is
(1.10)
Called the solution to a simple harmonic oscillator
and describes oscillatory motion, or simple harmonic motion.
Note (Example 1.1.2)
x(0) n
2x0
2 v02
n
nx0
n2x
0
2 v02
x0
as it should
x(t) n
2x02 v0
2
nsin nt tan
1 nx0
v0
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A note on arctangents
Note that calculating arctangent from acalculator requires some attention. First, allmachines work in radians.
The argument atan(-/+) is in a different quadrantthen atan(+/-), and usual machine calculationswill return an arctangent in between -p/2 and+p/2, reading only the atan(-) for both of theabove two cases.
In MATLAB, use the atan2(x,y) function to get
the correct phase.
+
+
-
_+
+
-
-
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Example 1.1.3: wheel, tire suspension m=30
kg,
n= 10 hz, what is k?
k mn2 30 kg 10 cylce
secg2p rad
cylce
1.184 105 N/m
There are of course more complex models of suspension systems
and these appear latter in the course as our tools develop
2
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Summary of simple harmonic
motion
t
x(t)
x0 Slopehere is v0
n
Period
T 2p
n
Amplitude
A
Maximum
Velocity
An
fn n rad/s
2p rad/cyclen cycles
2p sn
2pHz
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Harmonic Motion
T 2p rad
n rad/s
2p
ns (1.11)
The natural frequency in the commonly used units of hertz:
The period is the time elapsed to complete one complete cylce
fn n
2p
n rad/s
2p rad/cyclen cycles
2p sn
2pHz (1.12)
For the pendulum:
n
g
l rad/s, T 2pl
g s
n k
Jrad/s, T 2p
J
ks
For the disk and shaft:
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Relationship between Displacement,
Velocity and Acceleration
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
-1
0
1
x
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-20
0
20
v
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1-200
0
200
Time (sec)
a
A=1, n=12
x(t) Asin(nt)
x(t) nAcos(nt)
x(t) n2Asin(nt)
Note how the relative magnitude of each increases forn>1
Displacement
Velocity
Acceleration
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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 n
and the max amplitude of vibration.
n k
m
857.8 N/m
49.210-3 kg132 rad/s
fn
n
2p 21 Hz
T2p
n
1
fn
1
21cyles
sec
0.0476 s
x(t)max A 1
nn
2x0
2 v02 x0 10 mm
0
Note: common
Units are Hertz
To avoid Costly errors use fnwhen working in Hertz and nwhen in rad/s
Units depend on system
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Compute the solution and max velocity and
acceleration
v(t)max nA 1320 mm/s = 1.32 m/s
a(t)max n2A 174.24 103 mm/s2
=174.24 m/s2 17.8g!
tan1nx0
0
p
2
rad
x(t) 10sin(132tp/ 2) 10cos(132t) mm
g= 9.8 m/s2
2.92 mph
90
~0.4 in max
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Does gravity matter in spring
problems?
Let be the deflection caused byhanging a mass on a spring
( = x1-x0 in the figure)Then from static equilibrium: mg kNext sum the forces in the vertical for some point x> x1 measured
from m&&x k x mg kx mg k
01 24 34 m&&x(t) kx(t) 0
So no, gravity does not have an effect on the vibration
(note that this is not the case if the spring is nonlinear)
X..
X..
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Example Pendulums & measuring g
A 2 m pendulum
swings with a period
of 2.893 s
What is the
acceleration due to
gravity at that
location?
g4p
2
T
2l
4p2
2.893
2
s
22 m
g= 9.434 m/s2
This is gin Denver, CO USA, at 1638m
and a latitude of 40
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Review of Complex Numbers and
Complex Exponential
b
a
A
c a jb Aejq
A complex number can be written with a real and imaginary
part or as a complex exponential
Where
a Acosq,b Asinq
Multiplying two complex numbers:
c1c2 A1A2ej(q1 q2 )
Dividing two complex numbers:
c1
c2
A1
A2ej(q1 q2 )
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Equivalent Solutions to 2nd
order Differential Equations
x(t) Asin(nt)x(t) A1 sinnt A2 cosnt
x(t) a1ejn t a2e
jn t
All of the following solutions are equivalent:
The relationships between A and , A 1and A 2, and a1and a2can be found in Window 1.4 of the course text, page 17.
Sometimes called the Cartesian form
Called the magnitude and phase form
Called the polar form
Each is useful in different situations
Each represents the same information
Each solves the equation of motion
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Derivation of the solution
This approach will be used again for more complicated problems
(1.18)
Substitute x(t) aet
into m&&x kx 0 m
2ae
t kaet 0
m2 k 0
k
m
k
mj n j
x(t) a1en jt and x(t) a2e
n jt
x(t) a1en jt a2e
n jt
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Is frequency always positive?
From the preceding analysis, = + n thenx(t) a1e
n jt a2en jt
Using the Euler relations for trigonometric functions, the
above solution can be written as
x(t) Asin nt (1.19)It is in this form that we identify as the natural frequency nand this is positive, the + sign being used up in the transformation
from exponentials to the sine function.
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Calculating RMS
A peak value
x limT 1Tx(t)dt = average value
0
T
x 2 limT
1
Tx2 (t)dt
0
T
= mean-square value
xrms x2 = root mean square value
Proportionalto energy
Not very useful since for
a sine function the
average value is zero
May need to be limited dueto physical constraints
Also useful when the vibration is random
(1.21)
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The Decibel or dB scale
It is often useful to use a logarithmic scale to plot vibrationlevels (or noise levels). One such scale is called the decibelor
dB scale. The dB scale is always relative to some reference
value x0. It is define as:
For example: if an acceleration value was 19.6m/s2 then relative
to 1g(or 9.8m/s2) the level would be 6dB,
10log1019.6
9.8
2
20log10 2 6dB
Or for Example 1.2.1: The Acceleration Magnitude
is 20log10(17.8)=25dBrelative to 1g.
dB 10log10 xx0
2
20log10 xx0
(1.22)
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