oscillations_01
PHYS1002 Physics 1
(Fundamentals)
Oscillations and Waves
Semester 1, 2011
Ian Cooper
cooper@physics.usyd.edu.au
Textbook: College Physics (Knight, Jones, Field)
Chapters 8.3 10.4 14 15 16 17.1 17.4
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2
OVERVIEW
Mindmaps – A3 summaries
Oscillations
• Elastic materials & Hooke’s law
• Simple harmonic motion
• Damped Oscillations
• Resonance
Waves
• Transverse & longitudinal waves
• Behaviour of waves
• Sound
• Superposition principle and interference
• Standing Waves
• Beats and Doppler effect
• Electromagnetic spectrum
• Refractive index
• Thin film interference
3
Equation Mindmaps
For each equations on the Exam Formula Sheet – you should
construct an Equation Mindmap
Symbols – interpretation, units, signs
Visualization & interpretation
Assumptions
Special constants
Graphical interpretation
Applications, Comments
Numerical Examples
4
What happens to our skin when we become old ?
What Physics is in
the pictures?
What is are the basic
components of a car suspension
system ?
How is the elastic potential energy
related to earthquake damage?
Queenstown NZ – World Home of Bungy
What is the physics?
5
6
Why measure the restoring
force of a DNA molecule ?
What does Hooke’s Law
have to do with a nasal strip
(device for improving air
flow through nasal
passages)?
7
oscillations_01: MINDMAP SUMMARY
Elastic, Plastic, Reference frame (coordinate system, origin, equilibrium position),
displacement (extension, compression, applied force, restoring force, gravitational
force, net (resultant) force, Newton’s Second Law), Hooke’s Law, spring constant
(spring stiffness), equilibrium, velocity, acceleration, work, kinetic energy,
potential energy (reference point), gravitational potential energy, elastic potential
energy, total energy, conservation of energy, ISEE, solve quadratic equations
dv
dt
 F  ma
W   F cos  dr
1 2
Ue  k x
2
v
dr
dt
r2
r1
a
F kx
UG  m g h
Fe   k x
E  K  Ue  UG
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To study any physical phenomena, a physicist starts with a
simple model.
For a vibrating mechanical system, we consider the simplest
model in which an object disturbed from its equilibrium
position is acted upon by a restoring force that acts to return
the object back to its equilibrium position. The restoring force is
proportional to the displacement of the object from equilibrium
and acts in the opposite direction to the displacement. This
restoring force obeys Hooke’s Law.
The motion of the object acted on by this type of restoring force
is periodic and is called Simple Harmonic Motion.
1
9
Elastic and plastic behaviour
equilibrium length
equilibrium position: x = 0
+x
restoring force Fe
applied force F
displacement x y, s, x…
When force F applied, wire (spring) extends a distance x
Elastic behaviour:
Wire returns to original length when force is removed
Plastic behaviour:
Distortion remains when force is removed
CP239-245
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Hooke’s Law (simplest model for restoring force)
Extension or compression is proportional to restoring & applied forces
Fe  k x F  k x
restoring force
[N]
F
applied force
[N]
extension or compression
[m]
spring or elastic constant
[N.m-1]
as F is increased beyond the
elastic limit the extension will
plastic region
spring does not become permanent
linearly
elastic
region
recover
elastic limit
(0,0)
x
Fe  (Fsp)x
CP239-245
equilibrium
compressed
11
Extended (stretched)
Fe
x
F
Fe
x
Hooke’s Law
F
F = k x Fe = - k x
CP239-245
12
F
Stiffest or most rigid
spring
F
k1
k2
rise
“pliant” materials:
large deformation –
small forces
run
k3
O
k = slope of F vs x graph
x
0
x
k1 > k2 > k3
CP239-245
13
Elastic potential energy
work done in extending wire
F
linearly
elastic
region
x
= area under curve [J  N.m]
= potential energy Ue stored in extended
wire
[J]
1
1 2
Ue  x F  k x
2
2
Examples:
Pogo stick, longbow, crossbow, pole vaulting, …..
CP303-304
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Work done W by an applied force F in extending a spring
through a displacement x
W 
x
0
1 2
F dx    kx  dx  k x
0
2
x
The work done W increases the potential energy of the
mass/spring system Ue
1 2
Ue  k x
2
Reference point Ue = 0 when x = 0
2
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Problem solving strategy: I S E E
Identity:
What is the question asking (target variables) ?
What type of problem, relevant concepts, approach ?
PRACTICE ONLY
MAKES PERMANENT
Set up:
Diagrams
Equations
Data (units)
Physical principals
Execute:
Answer question
Rearrange equations then substitute numbers
Evaluate:
Check your answer – look at limiting cases
sensible ?
units ?
significant figures ?
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DNA is a long-chain molecule that is normally tightly coiled. Amazingly it is
possible to grab the two ends of a DNA molecule and gently stretch it while
measuring the restoring force using optical tweezers. Knowing the restoring
force tells how various enzymes act to cut and then reseal coils in the DNA
structure.
Problem 1
A DNA molecule is anchored at one end, then a force of
1.55 nN pulls on the other end, causing the molecule to stretch by
5.2 nm. What is the spring constant of the DNA molecule ?
use the ISEE method
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Solution 1
DNA molecule
Identify / Setup
F = 1.55 nN = 1.5510-9 N
k = ? N.m-1
Hooke’s Law
x = 5.2 nm = 5.210-9 m
F=kx
Execute
k = F / x = (1.5510-9 / 5.210-9 ) N.m-1
k = 0.30 N.m-1
Evaluate
OK
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Problem 2
A nasal strip can improve the air flow through nasal passages.
The nasal strip consists of two flat polyester springs enclosed by
an adhesive tape covering. Measurements show that a nasal
strip can exert a force of 0.25 N on the nose, causing it to
expand by 3.7 mm. Calculate the effective force constant of the
nasal strip and the force required to expand the nose by 4.2 mm.
use the ISEE method
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Solution 2
nasal strip
Identify / Setup
F = 0.25 N
x = 3.7 mm = 3.710-3 m
Hooke’s Law
F=kx
x = 4.2 mm = 4.210-3 m F = ? N
Execute
k = F / x = (0.25 / 3.710-3 ) N.m-1 = 68 N.m-1
F = k x = (68)(4.210-3) N = 0.28 N
Evaluate
OK
k = ? N.m-1
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Problem 3
When a bowstring is pulled back in preparation for
shooting an arrow, the system behaves in a Hookean
fashion. Suppose the string is drawn 0.700 m and
held with a force of 450 N.
What is the elastic constant k of the bow?
use the ISEE method
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Solution 3
Bow and arrow
Identify / Setup
F = 450 N
x = 0.700 m
Hooke’s Law
F=kx
Execute
k = F / x = 450 / 0.7 N.m-1
k = 6.43102 N.m-1
Evaluate
OK
k = ? N.m-1
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Problem 4
During the filming of a movie a 100.0 kg stuntman
steps off the roof of a building and free-falls. He is
attached to a safety line 50.0 m long that has an
elastic constant 1000 N.cm-1.
What is the maximum stretch of the line at the instant
he comes to rest, assuming Hooke’s Law is valid?
Hint: Consider how the various kinds of energy change
use the ISEE method
Solution 4
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1
Identify / Setup
m = 100 kg L = 50.0 m
k = 1000 N.cm-1
1 cm = 10-2 m 1 cm-1 = 102 m-1
k = 105 N.m-1
g = 9.80 m.s-2
2
xmax = ? m
L
xmax
K = ½ m v2 UG = m g h Ue = ½ k x2
Conservation of energy
E = U + K = constant
K1 = 0
UG1 = m g (L + xmax)
Ue1 = 0
K2 = 0
UG2 = 0
Ue2 = ½ k xmax2
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Execute
energy conserved E1 = E2  m g (L + xmax) = ½ k xmax2
½ k xmax2 – m g xmax – m g L =0
Quadratic equation a x2 + b x + c = 0
x = {-b  (b2 – 4 a c)} / (2a)
a = k /2
b=-mg
a = 5104
b = - 980
c=-mg L
c = -4.9104
xmax = 1.00 m or xmax = - 0.98 m
max stretch
Evaluate
OK
xmax = 1.00 m
Problem 5
Consider a person taking a bungee jump. The mass of the
jumper is 60.0 kg. The natural length of the bungee cord is
9.00 m. At the bottom of the jump, the bungee cord has
extended by 18.0 m.
(a) What is the spring constant?
(b) What is the maximum elastic force restoring force
exerted on the jumper?
(c) What is the acceleration experienced by the
jumper at the bottom of the jump?
The person misses the ground by 3.00 m. Another person who
has a mass of 120 kg takes the same cord (without permission)
and takes the plunge.
use the ISEE method
(d) What might happen to this person?
(e) How fast does the person hit the ground?
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Answers to bungee jump Problem 5
(a) k = 98 N.m-1
(b) Femax = 1764 N
(c) a = 19.6 m.s-2 = 2g
(d) hdrop = 40 m > 30 m
(e) v = 15 m.s-1 = 54 km.h-1
jumper could be killed or seriously injured