### oscillations_03

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oscillations_03
Damped Oscillations
Forced Oscillations and Resonance
SHM
Review: SHM
2
mass/spring system
x  xmax cos  t 
v   xmax sin  t 
a   xmax 2 cos  t 
a   2 x
2
  2 f 
T
k
2
 
m
1 2 1 2
PE  k x  k xmax cos2  t 
2
2
1
1
1 2
2
2
2
2
KE  m v  m xmax  sin t   k xmax sin 2 t 
2
2
2
Etotal  KE  PE 
1 2
1
2
k xmax  m vmax
= constant
2
2
1
1 2 1 2
2
m v  k x  k xmax
2
2
2
2
 v    xmax
 x2
CP 445
3
SHM
position x
10
0
-10
0
10
20
30
40
50
60
70
80
90
100
0
10
20
30
40
50
60
70
80
90
100
0
10
20
30
40
50
time t
60
70
80
90
100
velocity v
5
0
acceleration a
-5
1
0
-1
CP445
energy K Ue E (J)
velocity v (m/s)
extension y (m)
SHM (zero damping)
10
0
-10
0
2
4
6
8
10
12
2
4
6
8
10
12
2
4
6
time t (s)
8
10
12
20
0
-20
0
6000
4000
2000
0
0
5
b=0
0.12
E
energy K U E (J)
0.1
0.08
KE
PE
0.06
0.04
0.02
0
0
2
4
time t (s)
6
8
CP 455
6
oscillations_03:
MINDMAP SUMMARY
Reference frame, restoring force, damping force, driving force (harmonic
driving force), net (resultant) force, Newton’s Second Law, natural
frequency of vibration, free oscillations, damping (underdamped, critical
damping, overdamped), exponential decay, graphical interpretation of
damping, forced oscillations, resonance, driving frequency, resonance
curve, self-excited oscillations, examples of resonant phenomena, strategy
f 
1
2
k
m
7
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
8
Damped Oscillations
Oscillations in real systems die away
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
9
Damped Oscillations
CP 463
10
b=6
0.12
energy K U E (J)
0.1
E
0.08
PE
0.06
0.04
0.02 KE
0
0
2
4
time t (s)
6
8
CP 463
11
10
10
8
8
6
6
4
4
extension y (m)
extension y (m)
The graphs show the vertical motion of an object attached to the end of a
spring. Describe the motion of the object. Up is the positive direction.
2
0
-2
2
0
-2
-4
-4
-6
-6
-8
-8
-10
-15
-10
-5
0
velocity v (m/s)
5
10
15
-10
-15
-10
-5
0
velocity v (m/s)
5
10
15
12
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
1
f 
2
k
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
13
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 energy transfer
which can cause the oscillating system to fail - see wine
glass experiment.
Famous example of resonance: soldiers marching on
bridge
CP 465
14
Resonance phenomena occur widely in natural and in technological
applications:
Emission & absorption of light
Lasers
Tuning of radio and television sets
Mobile phones
Microwave communications
Machine, building and bridge design
Musical instruments
Medicine
nuclear magnetic resonance
(magnetic resonance imaging)
x-rays
hearing
Nuclear magnetic resonance scan
CP 465
Response Curve
b is the damping factor
15
0.4
amplitude A (m)
0.35
b=2
0.3
0.25
0.2
0.15
b=8
0.1
0.05
0
b = 10
0
0.5
1
1.5
fd d/ f/
o o
2
2.5
3
CP 465
16
Sinusoidal driving force fd / fo = 0.1
Sinusoidal driving force fd / fo = 1
b=2
b=2
1
0.5
position x (m)
position x (m)
1
0
-0.5
-1
0
20
40
60
time t (s)
80
100
0.5
0
-0.5
-1
0
20
40
60
time t (s)
80
100
CP 465
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Impulsive force – constant force
applied for a short time interval.
Sinusoidal driving force fd / fo = 2
b=2
1
1
0.5
0.5
position x (m)
position x (m)
b=2
0
-0.5
-1
0
20
40
60
time t (s)
80
100
0
-0.5
-1
0
20
40
60
time t (s)
80
100
CP 465
18
Forced vibrations - resonance
fD = 0.4 fo
fD = 1.1 fo
fD = 1.6 fo
Animation courtesy of Dr. Dan Russell, Kettering University
19
An optical technique called interferometry reveals the oscillations of a wine glass
http://www.acoustics.salford.ac.uk/feschools/waves/wine3video.htm
http://www.acoustics.salford.ac.uk/feschools/waves/shm3.htm
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Resonance and Hearing
Cochlea
staples
Basilar
membrane
Eardrum
Auditory canal
vibrations of small bones
of the middle ear
vibration of eardrum
due to sound waves
1
fo 
2
k
m
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
3000 Hz
30 Hz
CP467
21
Self excited oscillations
Sometimes 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
22
CP 466
23
What is a good strategy for answering
examination questions ???
1
2
Type of problem
Identify the physics – what model can be used?
24
Use exam formula sheet
3
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
Problem 1
Why do some tall building collapse during an earthquake ?
25
26
ISEE
Vibration motion can be resolved into
vertical and horizontal motions
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Vertical motion
natural frequency of vibration
m
1
fo 
2
k
k
m
driving frequency fd
0.4
Resonance
fd  fo
amplitude A (m)
0.35
b=2
0.3
0.25
0.2
0.15
b=8
0.1
0.05
b = 10
large amplitude oscillations – building collapses
0
0
0.5
1
1.5
 d / o
2
2.5
3
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Horizontal Motion
Resonance
Standing Waves
setup in building
antinodes
2nd floor
Driver frequency fd
nodes
2nd floor disappeared – driving
frequency matches natural
frequency (3rd harmonic)
29
antinode at top of building
node
antinode
 this floor will collapse
node at ground level
ground vibrates horizontally
setting up a standing wave
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Problem 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 2.0104 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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Solution 2
ISEE
m = (50 + 70) kg = 120 kg
k = 2x104 N.m-1
v = 30
km.h-1
Dx = 4.2 m
Tractor speed
v = Dx / Dt = 30 km.h-1 = (30)(1000) / (3600) m.s-1 = 8.3 m.s-1
The time interval between hitting the bumps (Dx = 4.2 m)
Dt = Dx / 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 / Dt = 1 / 0.51 = 2.0 Hz.
The natural frequency of vibration of the spring-seat-person is
1
f 
2
k
1

m 2
2  104
 2.1 Hz
120
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-seatperson 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.
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