Chapter 10 Elasticity & Oscillations - Austin Community College

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Chapter 10


Elasticity & Oscillations

MFMcGraw
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PHY1401

Chap 10d
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Elas & Vibrations
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Revised 7
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12
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10

2

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


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3

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.

MFMcGraw
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Chap 10d
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4

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.

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5

Define:

L
L


strain
The fractional
change in length

A
F

stress
Force per unit cross
-
sectional area

Stress and Strain

Stretching ==> Tensile Stress

Squeezing ==> Compressive Stress

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6

Hooke’s Law (F

x) can be written in terms of stress
and strain (stress


strain).

L
L
Y
A
F


The spring constant
k

is now

L
YA
k

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
.

Hooke’s Law

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Chap 10d
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7

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 5.8

10
4

N and the length
of the beam is 2.5 m, and the cross
-
sectional area of the beam is
7.5

10

3

m
2
.
Find the vertical compression of the beam.


Force of
floor on
beam

Force of
ceiling
on beam

















Y
L
A
F
L
L
L
Y
A
F
For steel Y = 200

10
9

Pa.

m

10
0
.
1
N/m

10
200
m

5
.
2
m

10
5
.
7
N

10
8
.
5
4
2
9
2
3
4




































Y
L
A
F
L
Compressive Stress

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8

Example (text problem 10.7): A 0.50 m long guitar string, of cross
-
sectional
area 1.0

10

6

m
2
, has a Young’s modulus of 2.0

10
9

Pa.

By how much must you stretch a guitar string to obtain a tension of 20 N?


mm

5.0
m

10
0
.
5
N/m

10
0
.
2
m

5
.
0
m

10
0
.
1
N

0
.
20
3
2
9
2
6




































Y
L
A
F
L
L
L
Y
A
F
Tensile Stress

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Beyond Hooke’s Law

Elastic Limit

If the stress on an object exceeds the
elastic limit
, then
the object will not return to its original length.


Breaking Point

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.

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The
ultimate strength

of a material is the
maximum stress that it can withstand before
breaking.

Ultimate Strength

Materials support compressive stress better than
tensile stress

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12

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 2.5

10
8

Pa.

What is the minimum diameter the wire should have to support her?

Want

limit

elastic

stress


A
F
limit

elastic
limit

elastic
mg
F
A


mm

1.7
m

10
7
.
1
limit

elastic
4
limit

elastic
2
3
2















mg
D
mg
D
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Different Representations

 
 
 
 
 
 
 
 
 
 
 
 
0
0
0
0 0
0 0
0
F
ΔL
=Y
A L
ΔL 1 F
=
L Y A
1 F
ΔL = L
Y A
1 F
L - L = L
Y A
1 F
L = L +L
Y A
1 F
L = L 1+
Y A
Original equation

Fractional change

Change in length

New length

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Shear Deformations

A
shear deformation

occurs when two forces
are applied on opposite
surfaces of an object.


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15

A
F


Area

Surface
Force
Shear


Stress
Shear
L
x



surfaces

of

separation
surfaces

of
nt
displaceme

Strain
Shear
Hooke’s law (stress

strain)
for shear deformations is


L
x
S
A
F


Define:

where S is the shear
modulus

Stress and Strain

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16

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. The shear modulus of the gelatin is 940 Pa.

What is the magnitude of the tangential force?

F

F





N

30
.
0
cm

5.0
cm

64
.
0
m

0025
.
0
N/m

940
2
2










L
x
SA
F
From Hooke’s Law:

L
x
S
A
F


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17

A
F


pressure
stress

volume
An object completely submerged in a fluid will be squeezed
on all sides.

The result is a volume strain;

V
V


strain

volume
Volume Deformations

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For a
volume deformation,

Hooke’s Law is
(stress

strain):

V
V
B
P




where B is called the bulk modulus. The bulk
modulus is a measure of how easy a material is to
compress.

Volume Deformations

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19

An anchor, made of cast iron of bulk modulus 60.0

10
9

Pa and a volume of
0.230 m
3
, is lowered over the side of a ship to the bottom of the harbor
where the pressure is greater than sea level pressure by 1.75

10
6

Pa.

Find the change in the volume of the anchor.






3
6
9
6
3
m

10
71
.
6
Pa

10
0
.
60
Pa

10
75
.
1
m

230
.
0
















B
P
V
V
V
V
B
P
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20

Examples


I
-
beam


Arch
-

Keystone


Flying Buttress

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21

Deformations Summary Table

Bulk Modulus
(B)

Shear modulus (S)

Young’s modulus
(Y)

Constant of
proportionality

Fractional
change in
volume

Ratio of the relative
displacement to the
separation of the two
parallel surfaces

Fractional
change in length

Strain

Pressure

Shear force divided by
the area of the surface
on which it acts

Force per unit
cross
-
sectional
area

Stress


Volume


Shear

Tensile or
compressive

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22

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.

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Characteristics of SHM


Repetitive motion through a central equilibrium point.


Symmetry of maximum displacement.


Period of each cycle is constant.


Force causing the motion is directed toward the
equilibrium point (minus sign).


F directly proportional to the displacement from
equilibrium.

Acceleration =
-

ω
2

x Displacement


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The motion of a mass on a spring is an example of
SHM.

The restoring force is F =

kx.

x

Equilibrium
position


x

y

The Spring

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25

Assuming the table is frictionless:








x
x
2
x
F = - kx = ma
k
a t = - x t = -
ω x t
m
Also,











2 2
1 1
2 2
E t K t U t mv t kx t
   
Equation of Motion & Energy

Classic form for SHM

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26


Spring Potential Energy

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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.

Simple Harmonic Motion


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What About Gravity?

When a mass
-
spring system is oriented vertically,
it will exhibit SHM with the same period and
frequency as a horizontally placed system.



The effect of gravity is cancelled out.

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Spring Compensates for Gravity

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Representing Simple Harmonic Motion


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A simple harmonic oscillator can be described
mathematically by:







t
A
t
v
t
a
t
A
t
x
t
v
t
A
t
x





cos
sin
cos
2











Or by:







t
A
t
v
t
a
t
A
t
x
t
v
t
A
t
x





sin
cos
sin
2










where A is the amplitude of
the motion, the maximum
displacement from
equilibrium, A


= v
max
, and
A

2

= a
max.

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33


Linear Motion
-

Circular Functions

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Projection of Circular Motion

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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


The Period and the Angular Frequency

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36

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:

2
2
2
2
1
2
1
2
1
mv
kx
mv
U
K
E





Since E = constant, at equilibrium (x = 0) the
KE must be a maximum. Here v = v
max
= A

.

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cm/sec

8
62
rads/sec

6
12
cm

0
5

and
rads/sec

6
12
s

50
0
2
2
.
.
.
A
ω
v
.
.
T









The amplitude A is given, but


is not.

Example continued:

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38

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
1.8

10

4

m at that frequency.

(a) What is the maximum force acting on the
diaphragm?





2
2
2
2
max
max
4
2
mAf
f
mA
A
m
ma
F
F









The value is F
max
=1400 N.

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(b) What is the mechanical energy of the diaphragm?

Since mechanical energy is conserved, E = K
max
= U
max.

2
max
max
2
max
2
1
2
1
mv
K
kA
U


The value of k is unknown so use K
max
.





2
2
2
2
max
max
2
2
1
2
1
2
1
f
mA
A
m
mv
K





The value is K
max
= 0.13 J.

Example continued:

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Example (text problem 10.47): The displacement of an
object in SHM is given by:









t
t
y

rads/sec

57
.
1
sin
cm

00
.
8

What is the frequency of the oscillations?

Comparing to y(t) = A sin

t gives A = 8.00 cm
and


= 1.57 rads/sec. The frequency is:


Hz

250
.
0
2
rads/sec

57
.
1
2






f
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2
2
2
max
max
max
cm/sec

7
19
rads/sec

57
1
cm

00
8
cm/sec

6
12
rads/sec

57
1
cm

00
8
cm
00
8
.
.
.
A
a
.
.
.
A
v

.
A
x










Other quantities can also be determined:

The period of the motion is

sec

00
.
4
rads/sec

57
.
1
2
2






T
Example continued:

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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.

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The pendulum is best
described using polar
coordinates.

The origin is at the pivot
point. The coordinates are
(r, φ). The r
-
coordinate
points from the origin
along the rod. The φ
-
coordinate is perpendicualr
to the rod and is positive in
the counterclock wise
direction.

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44

Apply Newton’s 2
nd

Law to the pendulum
bob.

2
sin
cos 0
r
F mg ma
v
F T mg m
r




  
   


If we assume that
φ

<<1 rad, then sin
φ



φ

and cos
φ


1,
the angular
frequency of oscillations is then:


L
g


The period of oscillations is

g
L
T



2
2


sin
sin
(/)sin
(/)
F mg ma mL
mg mL
g L
g L


 
 
 
 
   
 
 
 

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45

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

25
0
4
s

0
1
m/s

8
9
4
L
2
2
2
2
2
2
.
.
.
gT
g
L
T







Solving for L:

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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


)
cos
1
(



L
y
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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.


The Physical Pendulum

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48


http://hyperphysics.phy
-
astr.gsu.edu/HBASE/pendp.html

The Physical Pendulum

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Damped Oscillations

When dissipative forces such as friction are not
negligible, the amplitude of oscillations will decrease
with time. The oscillations are damped.

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50

Graphical representations of damped oscillations:

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Overdamped
: The system returns to equilibrium without
oscillating. Larger values of the damping the return to
equilibrium slower.



Critically damped

: The system returns to equilibrium as
quickly as possible without oscillating. This is often
desired for the damping of systems such as doors.




Underdamped

: The system oscillates (with a slightly
different frequency than the undamped case) with the
amplitude gradually decreasing to zero.


Source: Damping @ Wikipedia

Damped Oscillations

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Damped Oscillations

The larger the damping the more difficult it is to assign
a frequency to the oscillation.

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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
.

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Tacoma Narrows Bridge

Nov. 7, 1940

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Tacoma Narrows Bridge

Nov. 7, 1940

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The first Tacoma Narrows Bridge opened to traffic on July 1, 1940. It
collapsed four months later on November 7, 1940, at 11:00 AM (Pacific
time) due to a physical phenomenon known as aeroelastic flutter caused
by a 67 kilometres per hour (42 mph) wind.

The bridge collapse had lasting effects on science and engineering. In
many undergraduate physics texts the event is presented as an example of
elementary forced resonance with the wind providing an external periodic
frequency that matched the natural structural frequency (even though the
real cause of the bridge's failure was aeroelastic flutter[1]).

Its failure also boosted research in the field of bridge aerodynamics/
aeroelastics which have themselves influenced the designs of all the
world's great long
-
span bridges built since 1940.
-

Wikipedia


http://www.youtube.com/watch?v=3mclp9QmCGs

Tacoma Narrows Bridge

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Chapter 10: Elasticity & Oscillations




Elastic deformations of solids


Hooke's law for tensile and compressive forces


Beyond Hooke's law


Shear and volume deformations


Simple harmonic motion


The period and frequency for SHM


Graphical analysis of SHM


The pendulum


Damped oscillations


Forced oscillations and resonance


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Chap 10d
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Elas & Vibrations
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Extra

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Elas & Vibrations
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Aeroelasticity is the science which studies the
interactions among inertial, elastic, and aerodynamic
forces. It was defined by Arthur Collar in 1947 as "the
study of the mutual interaction that takes place within
the triangle of the inertial, elastic, and aerodynamic
forces acting on structural members exposed to an
airstream, and the influence of this study on design."

Aeroelasticity

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Elas & Vibrations
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Flutter


Flutter is a self
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feeding and potentially destructive vibration where aerodynamic forces on an object couple with a
structure's natural mode of vibration to produce rapid periodic motion. Flutter can occur in any object within a strong
fluid flow, under the conditions that a positive feedback occurs between the structure's natural vibration and the
aerodynamic forces. That is, that the vibrational movement of the object increases an aerodynamic load which in turn
drives the object to move further. If the energy during the period of aerodynamic excitation is larger than the natural
damping of the system, the level of vibration will increase, resulting in self
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exciting oscillation. The vibration levels can
thus build up and are only limited when the aerodynamic or mechanical damping of the object match the energy input,
this often results in large amplitudes and can lead to rapid failure. Because of this, structures exposed to aerodynamic
forces
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including wings, aerofoils, but also chimneys and bridges
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are designed carefully within known parameters to
avoid flutter. It is however not always a destructive force; recent progress has been made in small scale (table top) wind
generators for underserved communities in developing countries, designed specifically to take advantage of this effect.


In complex structures where both the aerodynamics and the mechanical properties of the structure are not fully
understood flutter can only be discounted through detailed testing. Even changing the mass distribution of an aircraft or
the stiffness of one component can induce flutter in an apparently unrelated aerodynamic component. At its mildest this
can appear as a "buzz" in the aircraft structure, but at its most violent it can develop uncontrollably with great speed and
cause serious damage to or the destruction of the aircraft.


In some cases, automatic control systems have been demonstrated to help prevent or limit flutter related structural
vibration.


Flutter can also occur on structures other than aircraft. One famous example of flutter phenomena is the collapse of the
original Tacoma Narrows Bridge.

Aeroelastic Flutter