1 oscillations_03 Damped Oscillations Forced Oscillations and Resonance SHM shm_v.avi
Dec 19, 2015
1oscillations_03
Damped Oscillations
Forced Oscillations and ResonanceSHM
shm_v.avi
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
Review: SHM mass/spring system
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
0 2 4 6 8 10 12-10
0
10e
xte
nsi
on
y (
m)
SHM (zero damping)
0 2 4 6 8 10 12-20
0
20
velo
city
v (
m/s
)
0 2 4 6 8 10 120
2000
4000
6000
en
erg
y K
Ue E
(J)
time t (s)
bungee3.avi
5
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
6oscillations_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 for answering examination questions, ISEE
1
2
kf
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 (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
9
CP 463
Damped Oscillations
10
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
11
-15 -10 -5 0 5 10 15-10
-8
-6
-4
-2
0
2
4
6
8
10
ext
en
sio
n y
(m
)
velocity v (m/s)
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.
-15 -10 -5 0 5 10 15-10
-8
-6
-4
-2
0
2
4
6
8
10
ext
en
sio
n y
(m
)
velocity v (m/s)
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
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
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 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
15
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
b is the damping factor
16Sinusoidal 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
17
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
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
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
20
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
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 ???
241 Read and answer the question2 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
25Problem 1
Why do some tall building collapse during an earthquake ?
26I S E E
Vibration motion can be resolved into vertical and horizontal motions
27Vertical 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
28
Driver frequency fd
nodes
antinodes
2nd floor disappeared – driving frequency matches natural frequency (3rd harmonic)
Horizontal Motion
ResonanceStanding Waves setup in building
2nd floor
29
ground vibrates horizontally setting up a standing wave
antinode at top of building
node
antinode this floor will collapse
node at ground level
standing_2.avi
30
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 2.0104 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?
31
x = 4.2 m
k = 2x104 N.m-1
v = 30 km.h-1
m = (50 + 70) kg = 120 kg
Solution 2 I S E E
32Tractor 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.