1 LECTURE 4 CP Ch 14 Damped Oscillations Forced Oscillations and Resonance.
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Transcript of 1 LECTURE 4 CP Ch 14 Damped Oscillations Forced Oscillations and Resonance.
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1LECTURE 4 CP Ch 14
Damped Oscillations
Forced Oscillations and Resonance
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2
max ma
2
2 2 2max
2 2 2 2 2 2max
x
a
max
2m x
cos sin
cos
1 1cos
2 21 1 1
sin sin2 2 2
x t v t
a t a x
PE k x k x t
KE m v m x t
x x
x
k x t
2 2max max
1 1= constant
2 2totalE KE PE k x m v
2 2 2 2 2max max
1 1 1
2 2 2m v k x k x v x x
2
22 f
Tk
m
CP 445
SHM
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3
0 10 20 30 40 50 60 70 80 90 100-10
0
10SHM
posi
tion
x
0 10 20 30 40 50 60 70 80 90 100-5
0
5
velo
city
v
0 10 20 30 40 50 60 70 80 90 100-1
0
1
acce
lera
tion
a
time t
CP445
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4
0 2 4 6 80
0.02
0.04
0.06
0.08
0.1
0.12b = 0
ener
gy
K U
E (
J)
time t (s)
KE PE
E
CP 455
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5
Mathematical modelling for harmonic motion
Newton’s Second Law can be applied to the oscillating system
F = restoring force + damping force + driving force
F(t) = - k x(t) - b v(t) + Fd(t)
For a harmonic driving force at a single frequency
Fd(t) = Fmaxcos(t + ).
This differential equation can be solved to give x(t), v(t) and a(t).
CP 463
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6
Damped oscillations
Oscillations in real systems die away (the amplitude steadily decreases) over time - the oscillations are said to be damped
For example:The amplitude of a pendulum will decrease over time due to air resistance
If the oscillating object was in water, the greater resistance would mean the oscillations damp much quicker.
CP 463
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7
Damped oscillations
CP 463
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0 2 4 6 80
0.02
0.04
0.06
0.08
0.1
0.12b = 6
ener
gy
K U
E (
J)
time t (s)
KE
PE
E
CP 463
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9
Forced oscillations Driven Oscillations & Resonance
If we displace a mass suspended by a spring from equilibrium and let it go it oscillates at its natural frequency
f 1
2k
m
If a periodic force at another frequency is applied, the oscillation will be forced to occur at the applied frequency - forced oscillations
CP 465
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Resonance
Forced oscillations are small unless the driving frequency is close to the natural frequency
When the driving frequency is equal to the natural frequency the oscillations can be large - this is called resonance
Away from resonance, energy transfer to the oscillations is inefficient. At resonance there is efficient transfer which can cause the oscillating system to fail - see wine glass experiment.
Famous example of resonance: soldiers marching on bridge
CP 465
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Resonance phenomena occur widely in natural and in technological applications:
Emission & absorption of lightLasersTuning of radio and television setsMobile phonesMicrowave communicationsMachine, building and bridge designMusical instrumentsMedicine – nuclear magnetic resonance magnetic resonance imaging – x-rays – hearing
Nuclear magnetic resonance scanCP 465
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0 0.5 1 1.5 2 2.5 30
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4am
plitu
de A
(m
)
d /o
b = 2
b = 8
b = 10
CP 465
Response Curve
fd / fO
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13Sinusoidal driving force fd / fo = 0.1
0 20 40 60 80 100-1
-0.5
0
0.5
1b = 2
posi
tion
x (
m)
time t (s)
Sinusoidal driving force fd / fo = 1
0 20 40 60 80 100-1
-0.5
0
0.5
1b = 2
pos
ition
x
(m
)time t (s)
CP 465
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14
Sinusoidal driving force fd / fo = 2
0 20 40 60 80 100-1
-0.5
0
0.5
1b = 2
pos
ition
x
(m
)
time t (s)
Impulsive force – constant force applied for a short time interval.
0 20 40 60 80 100-1
-0.5
0
0.5
1b = 2
pos
ition
x
(m
)time t (s)
CP 465
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http://www.acoustics.salford.ac.uk/feschools/waves/wine3video.htm
http://www.acoustics.salford.ac.uk/feschools/waves/shm3.htm
An optical technique called interferometry reveals the oscillations of a wine glass
Great Links to visit
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Eardrum
Auditory canal
Cochlea
Basilar membrane
vibrations of small bones of the middle ear
vibration of eardrumdue to sound waves
Inner air – basilar membrane – as the distance increases from the staples, membrane becomes wider and less stiff – resonance frequency of sensitive hair cells on membrane decreases
staples
1
2o
kf
m
3000 Hz 30 Hz
Resonance and Hearing
CP467
staples
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17
Self excited oscillationsSometimes apparently steady forces can cause large oscillations at the natural frequency
Examples
singing wine glasses (stick-slip friction)
Tacoma Narrows bridge (wind eddies)
CP 466
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18
CP 466
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19
What is a good strategy for answering
examination questions ???
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201 Read and answer the question
2 Type of problem
Identify the physics – what model can be used? Use exam formula sheet
3 Answer in point form
Break the question into small parts, do step by step showing all working and calculations (if can’t get a number in early part of a question, use a “dummy” number. Explicit physics principles (justification, explanation)
Annotated diagrams (collect and information & data – implicit + explicit
Equations Identify Setup Execute Evaluate
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21Problem 4.1
Why do some tall building collapse during an earthquake ?
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22I S E E
Vibration motion can be resolved into vertical and horizontal motions
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23Vertical motion
m
k
1
2o
kf
m
natural frequency of vibration
driving frequency fd
0 0.5 1 1.5 2 2.5 30
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
ampl
itude
A (
m)
d /o
b = 2
b = 8
b = 10
Resonance fd fo large amplitude oscillations – building collapses
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Driver frequency fd
nodes
antinodes
2nd floor disappeared – driving frequency matches natural frequency (3rd harmonic)
Horizontal Motion
ResonanceStanding Waves setup in building
2nd floor
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25
Problem 4.2
Consider a tractor driving across a field that has undulations at
regular intervals. The distance between the bumps is about 4.2
m. Because of safety reasons, the tractor does not have a
suspension system but the driver’s seat is attached to a spring to
absorb some of the shock as the tractor moves over rough
ground. Assume the spring constant to be 2.0104 N.m-1 and
the mass of the seat to be 50 kg and the mass of the driver, 70
kg. The tractor is driven at 30 km.h-1 over the undulations.
Will an accident occur?
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26
x = 4.2 m
k = 2x104 N.m-1
v = 30 km.h-1
m = (50 + 70) kg = 120 kg
Solution 4.2 I S E E
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27Tractor speed v = x / t = 30 km.h-1 = (30)(1000) / (3600) m.s-1 = 8.3 m.s-1
The time interval between hitting the bumps (x = 4.2 m)
t = x / v = (4.2 / 8.3) s = 0.51 s
Therefore, the frequency at which the tractor hits the bumps and energy is
supplied to the oscillating system of spring-seat-person
f = 1 / t = 1 / 0.51 = 2.0 Hz.
The natural frequency of vibration of the spring-seat-person is
42.
1 1 2 10
2 2 1201 Hz
kf
m
This is an example of forced harmonic motion. Since the driving frequency (due to hitting the bumps) is very close to the natural frequency of the spring-seat-person the result will be large amplitude oscillations of the person and which may lead to an unfortunate accident. If the speed of the tractor is reduced, the driving frequency will not match the natural frequency and the amplitude of the vibration will be much reduced.