Chapter 10 -General conclusions and recommendations Chapter 10
Chapter 10
description
Transcript of Chapter 10
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Chapter 10
Sinusoidally Driven Oscillations
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Question of Chapter 10
How do the characteristic frequencies generated in one object (say a piano string) excite vibrations in another object (say a sounding board)?
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A Simple Driving System
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Natural Frequency (o)
If the board is a door, then the natural frequency is around 0.4 Hz.
If the system is driven at 0.4 Hz, large amplitudes result.
Smaller amplitudes result for driver frequency different from 0.4 Hz.
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Actual motions
The door starts with complex motions (transient) that settle down to sinusoidal, no matter the motor rate.
The final frequency is always the driving frequency of the motor ().
The amplitude of the oscillations depends on how far from the natural frequency the motor is.
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Driving at Various Rates
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Amplitude vs. Frequency
Natural Frequency
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<< o
Motor frequency is far below the natural frequency ( << o) door moves almost in
step with motor.
Door moves toward motor when bands are stretched most.
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< o
Door lags behind the motor.
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= o
Door lags by one quarter cycle.
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>> o
Door lags by one-half cycle.
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Summarizing
/o Door Lags
<< 1 0
< 1 Small
= 1 ¼-cycle
>> 1 ½-cycle
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Computer Model
Click on the link and experiment
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Nature of the Transient
Transients are reproducible If crank starts in the same position, we get the
same transient
Damped Harmonic Oscillations Shown by changing the damping Imagine the bottom of the door immersed in an
oil bath The amount of immersion gives the damping
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Small DampingTransient Part
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0.00 0.25 0.50 0.75 1.00
Time
Am
plit
ud
e
Steady State Part
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0 1 2 3 4 5
Time
Am
plit
ud
e
Driven Oscillator - Lightly Damped
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0 1 2 3 4 5
Time
Am
plit
ud
e
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Heavier DampingTransient Part
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.00 0.25 0.50 0.75 1.00
Time
Am
plit
ud
e
Steady State Part
-0.2
-0.1
0.0
0.1
0.2
0.3
0 1 2 3 4 5
Time
Am
plit
ud
e
Driven Oscillator - Heavier Damping
-0.2
-0.1
0
0.1
0.2
0.3
0 1 2 3 4 5
TimeA
mp
litu
de
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Two Part Motion
Damped harmonic oscillation (transient) is at the natural frequency
Driven (steady state) oscillation is at the driver frequency
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Driver Frequency = Natural Frequency
Transient Part
-0.2
-0.1
0.0
0.1
0.2
0.3
0.00 0.25 0.50 0.75 1.00
Time
Am
plit
ud
e
Steady State Part
-0.4
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
0 0.5 1 1.5 2 2.5 3
Time
Am
plit
ud
e
Driving Force = Natural Frequency
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0 0.5 1 1.5 2 2.5 3
TimeA
mp
litu
de
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Damping and the Steady State
As long as we are far from natural frequency, damping doesn’t affect the steady state.
Near the natural frequency, damping does have an effect.
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Damping and the Steady State
As damping is increased the height of the peak decreases
Small damping
Large damping
Am
plit
ude
Frequency
W½
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Trends with Damping
As damping increases we expect the halving time to decrease ( ) Oscillations die out quicker for larger damping.
12
1T
D
max
1A
D
As damping increases the maximum amplitude decreases ( )
Also notice W½ D. Larger damping
means a broader curve.
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Percentage Bandwidth (PBW)
Range of frequencies for which the response is it least half the maximum amplitude.
Let N be the number of oscillations that the pendulum makes in T½.
Direct measurement yields
PBW = 38.2/N measured in %
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Example of PBW
Imagine tuning an instrument by using a tuning fork (A 440) while playing A.
If you are not matching pitch, the tuning fork is not being driven at its natural frequency and the amplitude will be small.
Only at a frequency of 440 Hz will the amplitude of the tuning fork be large
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Example of PBW - continued
T½ = 5 sec (it takes about 5 seconds for the tuning fork to decay to half amplitude)N = (440 Hz) (5 sec) = 2200 cyclesSo when you get a good response from the tuning fork, you have found pitch to better than
PBW = 38.2/2200 = 0.017%or 0.076 Hz!
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Caution!
You must play long, sustained tones
Short “toots” will stimulate the transient which recall is at the natural frequency of the tuning fork (440 Hz) Without the sustained driving force of the instrument,
we will never get to the steady state and the tuning fork will ring due to the transient.
You will think the instrument is in pitch when it is not.
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Systems with Two Natural Modes
Each mode has its own frequency, decay time, and shape.
The modes are always damped sinusoidal.
Superposition applies.
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Simple Two Mass Model
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Normal Modes of Two Mass Model (Chapter 6)
Let Mode 1 have a natural frequency of 10 Hz and Mode 2 a natural frequency of 17.32 Hz.
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Driving Point Response Function or Resonance Curve
Am
plit
ude
Frequency
10 Hz 17.32 Hz
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Frequencies Between Peaks
Mass one has a mode one component and should lag a half-cycle behind the driver( > o1)
Mass one also has a mode two component to its motion, and here the driving frequency is less than the natural frequency ( << o2) Mass one keeps in step with the driver
These conflicting tendencies account for the small amplitude here
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New Terms
Driving Point Response Curve – measure the response at the mass being driven
Transfer Point Response Curve – measure the response at another mass in the system (not a driven mass)
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Properties of a Sinusoidally Driven System
At startup there is a transient that is made up of the damped sinusoids of all of the natural frequencies.
Once the transient is gone the steady state is at the driving frequency. When the driving frequency is close to one of the natural frequencies, the amplitude is a maximum and resembles that natural mode.
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A Tin Tray
The tray is clamped at three places. Sensors ( )and drivers ( )are used as pairs in the locations indicated.
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Response Curves
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General Principles
Sensor cannot pickup any mode whose nodal line runs through it. Notice that Sensor 2 is on the centerline It cannot pick up modes with nodal lines
through the center, such as…
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Sensor 2 is blind to…
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Sensor 2 is very sensitive to…
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Drivers Ability to Excite Modes
If a driver falls on the nodal line of a mode, that mode will not be excited
If a driver falls between nodal lines of a mode, that mode will be excited
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Steady State Response
Superposition of all the modes excited and their amplitudes at the detector positions. Some modes may reinforce or cancel other modes.
Example – consider the modes on the next screen Colored sections are deflected up at this time and the
uncolored sections are deflected down The vertical lines show where in the pattern of each we
are for a particular position on the plate
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Superposing Two Modes
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Summary
Altering the location of either the driver or the detector will greatly alter what the transfer response curve will be.
Altering the driver frequency will also change the response.
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Three Cases Presented
Deflections of the same sign (giving a larger deflection) Add
Deflections of opposite sign (canceling each other out) Subtract
Deflection of one mode lined up with the node of the other (deflection due to one mode only) Single
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The G4 Phantom at G3
392, 784, 1176, 1568, 1960, 2352, …
196, 392, 588, 784, 980, 1176, …
Depress G3 slowly
Press & release G4