Transcript Slide 1
oscillations_01
PHYS1002 Physics 1 (Fundamentals) Oscillations and Waves Semester 1, 2011
Ian Cooper cooper
@physics.usyd.edu.au
Textbook:
College Physics
Chapters
8.3 10.4
(Knight, Jones, Field) 14 15 16 17.1 17.4 1
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 2
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 3
What happens to our skin when we become old ?
4 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? 5 Queenstown NZ – World Home of Bungy What is the physics?
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)?
6
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 7
v
dr dt a
dv dt W
r
1
r
2
F
cos
dr
F
ma U e
1 2
k x
2
F
k x U G
m g h F e
k x E U e
U G
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
. 8 1
Elastic and plastic behaviour
equilibrium length equilibrium position:
x
= 0 restoring force
F e
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 +
x
CP239-245 9
Hooke’s Law (simplest model for restoring force)
Extension or compression is proportional to restoring & applied forces 10
F
e
k x F
k x
restoring force [N] applied force [N] extension or compression [m] spring or elastic constant [N.m
-1 ]
F
(0,0) linearly elastic region
x
plastic region spring does not recover elastic limit as
F
is increased beyond the
elastic limit
the extension will become permanent
F e
(
F sp
)
x
CP239-245
equilibrium compressed Extended (stretched) 11
x F e F x
Hooke’s Law
F
=
k x F
e
= - k x
F F e
CP239-245
F rise F run
O
k
= slope of
F
vs
x
graph
x
0 Stiffest or most rigid spring 12
k
1
k
2
“pliant”
materials: large deformation – small forces
k
3
x k
1 >
k
2 >
k
3 CP239-245
Elastic potential energy
F
linearly elastic region work done in extending wire = area under curve [J N.m] = potential energy
U e
wire [J] stored in extended
x U
e
1 2
x F
1 2
k x
2
Examples:
Pogo stick, longbow, crossbow, pole vaulting, … ..
CP303-304 13
Work done
W
by an applied force
F
through a displacement
x
in extending a spring
W
0
x
F dx
0
x
1 2
k x
2 The work done
W
increases the potential energy of the mass/spring system
U e U e
1 2
k x
2 Reference point
U e
= 0 when
x
= 0 14 2
Problem solving strategy: I S E E I
dentity: What is the question asking (target variables) ?
What type of problem, relevant concepts, approach ?
S
et up: Diagrams Equations Data (units) Physical principals
PRACTICE ONLY MAKES PERMANENT
15
E
xecute: Answer question Rearrange equations then substitute numbers
E
valuate: Check your answer – look at limiting cases sensible ?
units ?
significant figures ?
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. 16 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
Solution 1
Identify / Setup
DNA molecule
F
= 1.55 nN = 1.55
10 -9 N 17
k
= ? N.m
-1 Hooke’s Law
F = k x Execute x
= 5.2 nm = 5.2
k
=
F
/
x
= (1.55
10 -9 / 5.2
10 -9 ) N.m
-1
k
= 0.30 N.m
-1 10 -9 m
Evaluate
OK
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. 18 use the ISEE method
Solution 2
Identify / Setup
nasal strip
F
= 0.25 N
x
= 3.7 mm = 3.7
10 -3 m
k
= ? N.m
Hooke’s Law
F = k x x
= 4.2 mm = 4.2
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 10 -3 m
F
= ? N -1 19
Evaluate
OK
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?
20 use the ISEE method
Solution 3
Identify / Setup
Hooke’s Law
F = k x Execute k
=
F
/
x
= 450 / 0.7 N.m
-1
k
= 6.43
10 2 N.m
-1
Evaluate
OK Bow and arrow
F
= 450 N
x
= 0.700 m
k
= ? N.m
-1 21
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 22
Solution 4 1
Identify / Setup m k
= 100 kg
L
= 1000 N.cm
-1 = 50.0 m 1 cm = 10 -2 m 1 cm
k g
= 10 5 N.m
-1 = 9.80 m.s
-2 -1 = 10 2 m -1 2
x max
= ? m
K
= ½
m v
2
U G
=
m g h U e
= ½
k x
2 Conservation of energy
E
=
U
+
K
= constant
L x max K
1
U G
1
U e
1 = 0 =
m g
= 0 (
L
+
x max
)
K
2
U G
2
U e
2 = 0 = 0 = ½
k x max
2 23
Execute
energy conserved
E
1 =
E
2
m g
(
L
+
x max
) = ½
k x max
2 ½
k x max
2 –
m g x max
–
m g L
=0 Quadratic equation
a x
2 +
b x
+
c
= 0
x
= {-
b
(
b
2 – 4
a c
)} / (2
a
)
a
=
k
/2
b
= -
m g a =
5 10 4
b c
= - 980 = -
c m g L
= -4.9
10 4
x max
= 1.00 m or
x max
= - 0.98 m max stretch
x max
= 1.00 m
Evaluate
OK 24
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. (d) What might happen to this person?
(e) How fast does the person hit the ground?
use the ISEE method 25
Answers to bungee jump Problem 5 (a)
k
= 98 N.m
-1 (b)
F emax
= 1764 N (c)
a
= 19.6 m.s
-2 = 2g (d)
h drop
= 40 m > 30 m (e)
v
= 15 m.s
-1 = 54 km.h
-1 jumper could be killed or seriously injured 26