Chapter 10
Elasticity & Oscillations
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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Chap 10d - Elas & Vibrations - Revised 7-12-10
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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-PHY1401
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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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Stress and Strain
Define:
L
strain 
L
The fractional
change in length
F
stress 
A
Force per unit crosssectional area
Stretching ==> Tensile Stress
Squeezing ==> Compressive Stress
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Hooke’s Law
Hooke’s Law (Fx) can be written in terms of stress
and strain (stress  strain).
F
L
Y
A
L
The spring constant k is now
YA
k
L
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.
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Compressive Stress
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104 N and the length
of the beam is 2.5 m, and the cross-sectional area of the beam is 7.5103 m2.
Find the vertical compression of the beam.
Force of
ceiling
on beam
Force of
floor on
beam
F
L
Y
A
L
 F  L 
L    
 A  Y 
For steel Y = 200109 Pa.
4
2.5 m
 F  L   5.8 10 N 

4

L      

1
.
0

10
m


3
2 
9
2
 A  Y   7.5 10 m  200 10 N/m 
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Tensile Stress
Example (text problem 10.7): A 0.50 m long guitar string, of cross-sectional
area 1.0106 m2, has a Young’s modulus of 2.0109 Pa.
By how much must you stretch a guitar string to obtain a tension of 20 N?
F
L
Y
A
L
0.5 m
 F  L   20.0 N 

L      
6
2 
9
2 
 A  Y   1.0 10 m  2.0 10 N/m 
 5.0 10 3 m  5.0 mm
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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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Ultimate Strength
The ultimate strength of a material is the
maximum stress that it can withstand before
breaking.
Materials support compressive stress better than
tensile stress
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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108 Pa.
What is the minimum diameter the wire should have to support her?
F
stress   elastic limit
A
Want
F
mg
A

elastic limit
elastic limit
2
mg
D
 
elastic limit
2

D
MFMcGraw-PHY1401
4mg
 1.7 10 3 m  1.7 mm
  elastic limit
Chap 10d - Elas & Vibrations - Revised 7-12-10
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Different Representations
F
ΔL
=Y
A
L0
ΔL 1 F
=
L0
Y A
Original equation
Fractional change
1 F
ΔL = L0 

Y A 
1 F

L - L0 = L0 

Y
A


1 F
L = L0 + L0 

Y A 
1 F

L = L0  1+

Y
A


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Change in length
New length
13
Shear Deformations
A shear deformation
occurs when two forces
are applied on opposite
surfaces of an object.
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Stress and Strain
Define:
Shear Force F
Shear Stress 

Surface Area A
displaceme nt of surfaces x
Shear Strain 

separation of surfaces
L
Hooke’s law (stressstrain) for shear deformations is
F
x
S
A
L
MFMcGraw-PHY1401
where S is the shear
modulus
Chap 10d - Elas & Vibrations - Revised 7-12-10
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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
x
S
A
L
F
From Hooke’s Law:
x
F  SA
L



 0.64 cm 
 940 N/m 0.0025 m 
  0.30 N
 5.0 cm 
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2
2
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Volume Deformations
An object completely submerged in a fluid will be squeezed
on all sides.
F
volume stress  pressure 
A
V
The result is a volume strain; volume strain 
V
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Volume Deformations
For a volume deformation, Hooke’s Law is
(stressstrain):
V
P   B
V
where B is called the bulk modulus. The bulk
modulus is a measure of how easy a material is to
compress.
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An anchor, made of cast iron of bulk modulus 60.0109 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 1.75106 Pa.
Find the change in the volume of the anchor.
V
P   B
V
VP
0.230 m 3 1.75  106 Pa
V  

B
60.0 109 Pa
 6.7110 6 m 3

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
Chap 10d - Elas & Vibrations - Revised 7-12-10
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Examples
• I-beam
• Arch - Keystone
• Flying Buttress
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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
Young’s modulus Shear modulus (S)
proportionality (Y)
MFMcGraw-PHY1401
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Bulk Modulus
(B)
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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 = - ω2 x Displacement
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The Spring
The motion of a mass on a spring is an example of
SHM.
Equilibrium
position
y
x
x
The restoring force is F = kx.
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Equation of Motion & Energy
Assuming the table is frictionless:
F
Classic form for SHM
Also,
MFMcGraw-PHY1401
x
= - kx = ma x
k
ax t  = - x t  = - ω2 x t 
m
1 2
1 2
E  t   K  t   U  t   mv  t   kx  t 
2
2
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Spring Potential Energy
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Simple Harmonic Motion
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.
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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:
xt   A cos t
x


vt 
  A sin t
t
v
at  
  A 2 cos t
t
Or by:
xt   A sin t
x
vt  
 A cos t
t
v
at  
  A 2 sin t
t
MFMcGraw-PHY1401
where A is the amplitude of
the motion, the maximum
displacement from
equilibrium, A = vmax, and
A2 = amax.
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Linear Motion - Circular Functions
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Projection of Circular Motion
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The Period and the Angular Frequency
The period of oscillation is
where  is the angular frequency of
the oscillations, k is the spring
constant and m is the mass of the
block.
MFMcGraw-PHY1401
Chap 10d - Elas & Vibrations - Revised 7-12-10
T
2

.

k
m
35
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:
1 2 1 2 1 2
E  K  U  mv  kx  mv
2
2
2
Since E = constant, at equilibrium (x = 0) the
KE must be a maximum. Here v = vmax = A.
MFMcGraw-PHY1401
Chap 10d - Elas & Vibrations - Revised 7-12-10
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Example continued:
The amplitude A is given, but  is not.
2
2


 12.6 rads/sec
T
0.50 s
and v  Aω  5.0 cm 12.6 rads/sec   62.8 cm/sec
MFMcGraw-PHY1401
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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


F

F

ma

m
A


mA
2

f

4

mAf

max
max
2
2
The value is Fmax=1400 N.
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Example continued:
(b) What is the mechanical energy of the diaphragm?
Since mechanical energy is conserved, E = Kmax = Umax.
U max
K max
1 2
 kA
2
1 2
 mvmax
2
The value of k is unknown so use Kmax.
K max
1 2
1
1
2
2
 mvmax  m A   mA2 2f 
2
2
2
The value is Kmax= 0.13 J.
MFMcGraw-PHY1401
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Example (text problem 10.47): The displacement of an
object in SHM is given by:
yt   8.00 cmsin 1.57 rads/sec  t 
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:
 1.57 rads/sec
f 

 0.250 Hz
2
2
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Chap 10d - Elas & Vibrations - Revised 7-12-10
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Example continued:
Other quantities can also be determined:
The period of the motion is
2
2
T

 4.00 sec
 1.57 rads/sec
xmax  A  8.00 cm
vmax  A  8.00 cm 1.57 rads/sec   12.6 cm/sec
amax  A 2  8.00 cm 1.57 rads/sec   19.7 cm/sec 2
2
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Chap 10d - Elas & Vibrations - Revised 7-12-10
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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.
MFMcGraw-PHY1401
Chap 10d - Elas & Vibrations - Revised 7-12-10
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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 φcoordinate is perpendicualr
to the rod and is positive in
the counterclock wise
direction.
MFMcGraw-PHY1401
Chap 10d - Elas & Vibrations - Revised 7-12-10
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2nd
Apply Newton’s
Law to the pendulum
bob.
 F  mg sin   ma
2
v
 Fr  T  mg cos   m r  0
If we assume that φ <<1 rad, then sin φ  φ and cos φ 1, the angular
frequency of oscillations is then:
 F  mg sin   ma  mL
mg sin   mL
  ( g / L)sin 
  ( g / L)

The period of oscillations is
MFMcGraw-PHY1401
T
Chap 10d - Elas & Vibrations - Revised 7-12-10
g
L
2

 2
L
g
44
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?
L
T  2
g
Solving for L:
MFMcGraw-PHY1401


gT 2 9.8 m/s 2 1.0 s 
L

 0.25 m
2
2
4
4
Chap 10d - Elas & Vibrations - Revised 7-12-10
2
45
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).

Lcos
L
L
y  L(1  cos  )
y=0
MFMcGraw-PHY1401
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The Physical Pendulum
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.
MFMcGraw-PHY1401
Chap 10d - Elas & Vibrations - Revised 7-12-10
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The Physical Pendulum
http://hyperphysics.phy-astr.gsu.edu/HBASE/pendp.html
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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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Graphical representations of damped oscillations:
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Damped Oscillations
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
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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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Tacoma Narrows Bridge
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
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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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Extra
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Aeroelasticity
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."
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Aeroelastic Flutter
Flutter
Flutter is a self-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-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 - including wings, aerofoils, but also chimneys and bridges - 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.
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