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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Chapter 10
Lecture
Outline
2
Chapter 10: Elasticity and
Oscillations
•Elastic Deformations
•Hooke’s Law
•Stress and Strain
•Shear Deformations
•Volume Deformations
•Simple Harmonic Motion
•The Pendulum
•Damped Oscillations, Forced Oscillations, and Resonance
3
§10.1 Elastic Deformation of Solids
A deformation is the change in size or shape of an object.
An elastic object is one that returns to its original size and
shape after contact forces have been removed. If the forces
acting on the object are too large, the object can be
permanently distorted.
4
§10.2 Hooke’s Law
F F
Apply a force to both ends of a long wire. These forces will
stretch the wire from length L to L+L.
5
Define:
L
Lstrain
The fractional
change in length
A
Fstress
Force per unit cross-
sectional area
6
Hooke’s Law (Fx) can be written in terms of stress and
strain (stress strain).
L
LY
A
F
The spring constant k is now L
YAk
Y is called Young’s modulus and is a measure of an
object’s stiffness. Hooke’s Law holds for an object
to a point called the proportional limit.
7
Example (text problem 10.1): A steel beam is placed vertically in
the basement of a building to keep the floor above from sagging.
The load on the beam is 5.8104 N and the length of the beam is
2.5 m, and the cross-sectional area of the beam is 7.5103 m2.
Find the vertical compression of the beam.
Force of
floor on
beam
Force of
ceiling
on beam
Y
L
A
FL
L
LY
A
F
For steel Y = 200109 Pa.
m 100.1N/m 10200
m 5.2
m 105.7
N 108.5 4
2923
4
Y
L
A
FL
8
Example (text problem 10.7): A 0.50 m long guitar string, of
cross-sectional area 1.0106 m2, has a Young’s modulus of
2.0109 Pa. By how much must you stretch a guitar string to
obtain a tension of 20 N?
mm 5.0m 100.5
N/m 100.2
m 5.0
m 100.1
N 0.20
3
2926
Y
L
A
FL
L
LY
A
F
9
§10.3 Beyond Hooke’s Law
If the stress on an object exceeds the elastic limit, then the
object will not return to its original length.
An object will fracture if the stress exceeds the breaking
point. The ratio of maximum load to the original cross-
sectional area is called tensile strength.
10
The ultimate strength of a material is the maximum stress
that it can withstand before breaking.
11
Example (text problem 10.10): An acrobat of mass 55 kg is
going to hang by her teeth from a steel wire and she does
not want the wire to stretch beyond its elastic limit. The
elastic limit for the wire is 2.5108 Pa. What is the minimum
diameter the wire should have to support her?
Want limit elastic stress A
F
limit elasticlimit elastic
mgFA
mm 1.7m 107.1limit elastic
4
limit elastic2
3
2
mgD
mgD
12
§10.4 Shear and Volume
Deformations
A shear deformation
occurs when two forces
are applied on opposite
surfaces of an object.
13
A
F
Area Surface
ForceShear StressShear
L
x
surfaces of separation
surfaces ofnt displaceme Strain Shear
Hooke’s law (stressstrain) for shear deformations is
L
xS
A
F
Define:
where S is the
shear modulus
14
Example (text problem 10.25): The upper surface of a cube
of gelatin, 5.0 cm on a side, is displaced by 0.64 cm by a
tangential force. If the shear modulus of the gelatin is 940
Pa, what is the magnitude of the tangential force?
F
F
N 30.0cm 5.0
cm 64.0m 0025.0N/m 940 22
L
xSAF
From Hooke’s Law:
L
xS
A
F
15
A
F pressurestress volume
An object completely submerged in a fluid will be squeezed
on all sides.
The result is a volume strain; V
Vstrain volume
16
For a volume deformation, Hooke’s Law is (stressstrain):
V
VBP
where B is called the bulk modulus. The bulk modulus is a
measure of how easy a material is to compress.
17
Example (text problem 10.24): An anchor, made of cast iron
of bulk modulus 60.0109 Pa and a volume of 0.230 m3, is
lowered over the side of a ship to the bottom of the harbor
where the pressure is greater than sea level pressure by
1.75106 Pa. Find the change in the volume of the anchor.
36
9
63
m 1071.6
Pa 100.60
Pa 1075.1m 230.0
B
PVV
V
VBP
18
Deformations summary table
Tensile or
compressive
Shear
Volume
Stress Force per unit
cross-sectional
area
Shear force divided
by the area of the
surface on which it
acts
Pressure
Strain Fractional
change in
length
Ratio of the relative
displacement to the
separation of the two
parallel surfaces
Fractional
change in
volume
Constant of
proportionality
Young’s
modulus (Y)
Shear modulus (S) Bulk
Modulus (B)
19
§10.5 Simple Harmonic Motion
Simple harmonic motion (SHM)
occurs when the restoring force
(the force directed toward a stable
equilibrium point) is proportional to
the displacement from equilibrium.
20
The motion of a mass on a spring is an example of SHM.
The restoring force is F = kx.
x
Equilibrium
position
x
y
21
Assuming the table is frictionless:
txm
kta
makxF
x
xx
Also, 22
2
1
2
1tkxtmvtUtKtE
22
At the equilibrium point x = 0 so a = 0 too.
When the stretch is a maximum, a will be a maximum too.
The velocity at the end points will be zero, and it is a
maximum at the equilibrium point.
23
§10.6-7 Representing Simple
Harmonic Motion
When a mass-spring system is oriented vertically, it will
exhibit SHM with the same period and frequency as a
horizontally placed system.
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SHM
graphically
25
A simple harmonic oscillator can be described mathematically
by:
tAt
vta
tAt
xtv
tAtx
cos
sin
cos
2
Or by:
tAt
vta
tAt
xtv
tAtx
sin
cos
sin
2
where A is the amplitude of the
motion, the maximum
displacement from equilibrium,
A = vmax, and A2 = amax.
26
The period of oscillation is .2
T
where is the angular frequency of the
oscillations, k is the spring constant and
m is the mass of the block. m
k
27
Example (text problem 10.30): The period of oscillation of an
object in an ideal mass-spring system is 0.50 sec and the
amplitude is 5.0 cm. What is the speed at the equilibrium
point?
At equilibrium x = 0:
222
2
1
2
1
2
1mvkxmvUKE
Since E = constant, at equilibrium (x = 0) the
KE must be a maximum. Here v = vmax = A.
28
cm/sec 862rads/sec 612cm 05 and
rads/sec 612s 500
22
...Aωv
..T
The amplitude A is given, but is not.
Example continued:
29
Example (text problem 10.41): The diaphragm of a speaker
has a mass of 50.0 g and responds to a signal of 2.0 kHz by
moving back and forth with an amplitude of 1.8104 m at
that frequency.
(a) What is the maximum force acting on the diaphragm?
2222
maxmax 42 mAffmAAmmaFF
The value is Fmax=1400 N.
30
(b) What is the mechanical energy of the diaphragm?
Since mechanical energy is conserved, E = Kmax = Umax.
2
maxmax
2
max
2
1
2
1
mvK
kAU
The value of k is unknown so use Kmax.
2222
maxmax 22
1
2
1
2
1fmAAmmvK
The value is Kmax= 0.13 J.
Example continued:
31
Example (text problem 10.47): The displacement of an object
in SHM is given by:
tty rads/sec 57.1sincm 00.8
What is the frequency of the oscillations?
Comparing to y(t) = A sint gives A = 8.00 cm and
= 1.57 rads/sec. The frequency is:
Hz 250.02
rads/sec 57.1
2
f
32
222
max
max
max
cm/sec 719rads/sec 571cm 008
cm/sec 612rads/sec 571cm 008
cm008
...Aa
...Av
.Ax
Other quantities can also be determined:
The period of the motion is sec 00.4rads/sec 57.1
22
T
Example continued:
33
§10.8 The Pendulum
A simple pendulum is constructed by attaching a mass to
a thin rod or a light string. We will also assume that the
amplitude of the oscillations is small.
34
L
m
An FBD for the
pendulum bob:
A simple pendulum:
T
w x
y
35
Apply Newton’s 2nd Law
to the pendulum bob.
r
vmmgTF
mamgF
y
tx
2
cos
sin
If we assume that <<1 rad, then sin and cos 1, the
angular frequency of oscillations is then:
L
g
The period of oscillations is g
LT
2
2
36
Example (text problem 10.60): A clock has a pendulum that
performs one full swing every 1.0 sec. The object at the end
of the string weighs 10.0 N. What is the length of the
pendulum?
m 250
4
s 01m/s 89
4L
2
2
22
2
2
...gT
g
LT
Solving for L:
37
Example (text problem 10.94): The gravitational potential
energy of a pendulum is U = mgy. Taking y = 0 at the lowest
point of the swing, show that y = L(1-cos).
L
y=0
L
Lcos
)cos1( Ly
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A physical pendulum is any rigid object that is free to
oscillate about some fixed axis. The period of oscillation of
a physical pendulum is not necessarily the same as that of
a simple pendulum.
39
§10.9 Damped Oscillations
When dissipative forces such as friction are not negligible,
the amplitude of oscillations will decrease with time. The
oscillations are damped.
40
Graphical representations of damped oscillations:
41
§10.10 Forced Oscillations and
Resonance
A force can be applied periodically to a damped oscillator (a
forced oscillation).
When the force is applied at the natural frequency of the
system, the amplitude of the oscillations will be a maximum.
This condition is called resonance.
42
Summary
•Stress and Strain
•Hooke’s Law
•Simple Harmonic Motion
•SHM Examples: Mass-Spring System, Simple Pendulum
and Physical Pendulum
•Energy Conservation Applied to SHM
Recommended