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
§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.
§10.2 Hooke’s Law
∆L
F
L
F
Apply a force to both ends of a long wire. These forces will
stretch the wire from length L to L+∆L.
∆L ∝ F
∆L ∝ L (e.g. two springs in series)
1
1
∆L ∝
≈
number of rods Area
FL
∆L ∝
A
Define:
∆L
strain =
L
The fractional
change in length
F
stress =
A
Force per unit crosssectional area
Hooke’s Law (F∝x) can be written in terms of stress and
strain (stress ∝ strain).
F
∆L
=Y
A
L
YA
The spring constant k is now 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.
The Young'
s modulus (also known as modulus of elasticity, elastic modulus
or tensile modulus) allows the behavior of a material under load to be
calculated. For instance, it can be used to predict the amount a wire will
extend under tension, or to predict the load at which a thin column will buckle
under compression
tensile stress
Y=
tensile strain
Material
Rubber (small strain)
Young's modulus
(E) in GPa
Young's modulus (E) in
lbf/in² (psi)
0.01-0.1
1,500-15,000
3-3.5
435,000-505,000
Nylon
3-7
290,000-580,000
Oak wood (along grain)
11
1,600,000
High-strength concrete (under
compression)
30
4,350,000
Magnesium metal (Mg)
45
6,500,000
Aluminium alloy
69
10,000,000
Glass (all types)
72
10,400,000
Brass and bronze
103-124
17,000,000
Titanium (Ti)
105-120
15,000,000-17,500,000
150
21,800,000
190-210
30,000,000
450
65,000,000
Tungsten carbide (WC)
450-650
65,000,000-94,000,000
Single Carbon nanotube [1]
1,000+
145,000,000
1,050-1,200
150,000,000175,000,000
Polystyrene
Carbon fiber reinforced plastic
(unidirectional, along grain)
Wrought iron and steel
Silicon carbide (SiC)
Diamond (C)
F1
F2
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 5.8×104 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.
F
∆L
=Y
L
A
F L
∆L =
A Y
Force of
ceiling
on beam
Force of
floor on
beam
F
∆L =
A
For steel Y=200×109 Pa.
5.8 ×10 4 N
L
=
7.5 ×10 −3 m 2
Y
2.5 m
−4
=
1
.
0
×
10
m
9
2
200 × 10 N/m
Example (text problem 10.6): 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 20.0 N?
F
∆L
=Y
A
L
F L
20.0 N
=
∆L =
A Y
1.0 ×10 −6 m 2
= 5.0 ×10 −3 m = 5.0 mm
0.5 m
2.0 ×109 N/m 2
§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 crosssectional area is called tensile strength.
The ultimate strength of a material is the maximum stress
that it can withstand before breaking.
F
A
∆L
L
How to get this graphs?
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 2.5×108 Pa. What is the minimum
diameter the wire should have to support her?
F
Want stress =
< elastic limit
A
F
mg
A>
=
elastic limit elastic limit
D
π
2
2
mg
>
elastic limit
4mg
D>
= 1.7 ×10 −3 m = 1.7 mm
π ∗ elastic limit
§10.4 Shear and Volume Deformations
A shear deformation
occurs when two forces
are applied on opposite
surfaces of an object.
Shear Force F
=
Shear Stress =
Surface Area A
Define:
displacement of surfaces ∆x
Shear Strain =
=
L
separation of surfaces
Hooke’s law (stress∝strain) for shear deformations is
∆x
F
=S
L
A
where S is the
shear modulus
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
∆x
=S
A
L
F
From Hooke’s Law:
∆x
F = SA
L
(
)(
= 940 N/m 0.0025 m
2
2
)
0.64 cm
= 0.30 N
5.0 cm
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
For a volume deformation, Hooke’s Law is (stress∝strain):
∆V
∆P = − B
V
where B is called the bulk modulus. The bulk modulus (B) of a
substance essentially measures the substance'
s resistance to uniform
compression. It is defined as the pressure increase needed to effect a given
relative decrease in volume.
Water: 2.2×109 Pa (value increases at higher pressures)
Air: 1.42×105 Pa
Steel: 1.6×1011 Pa
Solid helium (approximate): 5×107 Pa
Example (text problem 10.24): 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
0.23 m 3 1.75 ×106 Pa
∆V = −
=−
B
60.0 ×109 Pa
= −6.7 ×10 −6 m 3
(
)(
)
Deformations summary table
Tensile or
compressive
Shear
Volume
Stress
Force per unit crosssectional 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)
Oscillations about equilibrium
(Synonym:Vibration)
What means Oscillation?
Oscillation is the periodic variation, typically in time, of some measure
as seen, for example, in a swinging pendulum.
Many things oscillate/vibrate: Periodic motion
(a motion that repeats itself over and over)
Pulse
Oscillations are the origin of the
sensation of musical tone
……….. in Aerospace: Orbits
Electrical/Computer:
LRC resonance in circuits
Physics: Atomic Vibrations, String Theory, Electromagnetic Waves
……more examples
….heart beat, breathing, sleeping, taking shower, eating, chewing, blinking, drinking
….motion of planets, stars, motion of electrons, atoms
….wind (Tacoma Bridge)
….vocal cords, ear drums
Why does something vibrate/oscillate?
Whenever the system is displaced from equilibrium, a restoring force pulls it
back, but it overshoots the equilibrium position.
Parameter used to describe vibrations
Period T Time taken to complete one
cycle of the vibration. Units: s
Frequency f = 1/T
Number of vibration cycles per
second. Units: 1/s (Hz, Hertz)
Amplitude A Maximum
displacement from equilibrium
position
One cycle is time take for a pendulum:
center – right – center – left –
center
Pendulum
Special cases of periodic 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.
For instance when the restoring
force is F = - k x.(Hook’s Law)
Simple harmonic motion
The displacement from equilibrium can be describes as a
cosinusoidal function
A
Period T
Simple harmonic motion is the projection of circular
motion on the x-axis
=
t
Angular velocity
is NOT necessarily
the same as
Angular frequency
Displacement, Velocity and Acceleration of SHM
x(t ) = A cos(ωt + φ )
dx(t )
= − Aω sin(ωt + φ )
dt
dv(t )
= − Aω 2 cos(ωt + φ ) = −ω 2 x(t )
a(t ) =
dt
v(t ) =
A is the amplitude of the
motion, the maximum
displacement from
equilibrium, Aω=vmax, and
Aω2 =amax.
Mass-Spring Java applet
Mass-Spring-System
A Spring always pushes or pulls mass back towards equilibrium
position. The time period can be calculated from Hooke´s Law:
F = −kx F = ma
ma = −kx
m[− Aω 2 cos(ωt )] = −k[ A cos(ωt )]
ω2 =
k
m
or T = 2π
m
k
Independent from amplitude!
(Application: measure the mass of astronauts in
space) Heavier mass – slower oscillations
Stiffer spring (greater k) – rapid oscillations
The period of oscillation is T =
1 2π
=
ω
f
Mass Spring-System in vertical setup
When a mass-spring system is oriented
vertically, it will exhibit SHM with the same
period and frequency as a horizontally
placed system.
Same formulae as for the horizontal setup
but the system oscillates around a new
equilibrium position y0.
y0 = mg/k
Energy Conservation in Oscillatory Motion
Potential Energy of simple harmonic
motion
Energy: E = U + K
U: Potential Energy
K: Kinetic Energy
Spring
Epot = U = ½ k x2
Ekin = K = ½ m v2
Turning points:
E = Umax + 0
(Displacement and U at maximum)
Minimum:
E = Kmax + 0
(Velocity and K at maximum)
Total energy of system
E = U + K = ½ k (A cos( t))2 + ½ m (A sin( t))2
= ½ k A2 cos2( t) + ½ m A2 2 sin2( t)
E = ½ k A2 cos2( t) + ½ m A2 k/m sin2( t) = ½ k A2 (cos2( t)+
sin2( t))
And therefore:
E = ½ k A2
Exercise 1: 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?
1 2 1 2 1 2 1 2
At equilibrium x=0: E = K + U = mv + kx = kA = mv
2
2
2
2
Since E=constant, at equilibrium (x = 0) the KE must be
a maximum. Here v = vmax = Aω.
The amplitude A is given, but ω is not.
2π
2π
=
= 12.6 rads/sec
ω=
T
0.50 s
and v = A = (5.0 cm )(12.6 rads/sec) = 62.8 cm/sec
Exercise 2: 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?
(
)
F = Fmax = mamax = m Aω = mA(2πf ) = 4π 2 mAf 2
2
2
The value is Fmax=1400 N.
(b) What is the mechanical energy of the diaphragm?
Since mechanical energy is conserved, E = KEmax = Umax.
U max =
KEmax
1 2
kA
2
1 2
= mvmax
2
The value of k is unknown so use KEmax.
KEmax =
1 2
1
1
2
2
mvmax = m( Aω ) = mA2 (2πf )
2
2
2
The value is KEmax= 0.13 J.
Exercise 3: 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 sinωt gives A = 8.00 cm and ω = 1.57 rads/sec.
The frequency is:
ω 1.57 rads/sec
f =
2π
=
= 0.250 Hz
2π
Other quantities can also be determined:
2π
2π
=
= 4.00 sec
The period of the motion is T =
ω 1.57 rads/sec
xmax = A = 8.00 cm
vmax = Aω = (8.00 cm )(1.57 rads/sec) = 12.6 cm/sec
amax = Aω 2 = (8.00 cm )(1.57 rads/sec) = 19.7 cm/sec2
2
Pendulum
A mass, called a bob, suspended from a fixed point so that it can
swing in an arc determined by its momentum and the force of gravity.
The length of a pendulum is the distance from the point of
suspension to the center of gravity of the bob. Chance observation of
a swinging church lamp led Galileo to find that a pendulum made
every swing in the same time, independent of the size of the arc. He
used this discovery in measuring time in his astronomical studies. His
experiments showed that the longer the pendulum, the longer is the
time of its swing.
If we assume the angle q is small, for then we can approximate sin θ
with θ (expressed in radian measure). (As an example, if θ = 5.00°=
0.0873 rad, then sin θ = 0.0872, a difference of only about 0.1%.)
With that approximation and some rearranging, we then have
∂ 2θ mgL
−
θ =0
2
∂t
I
Physical Pendulum,
Small amplitude
UPendulum = mgh = mgL (1-cos )
For smaller displacements, the movement of
an ideal pendulum can be described
mathematically as simple harmonic motion
(like the mass-spring), as the change in
potential energy at the bottom of a circular arc
is nearly proportional to the square of the
displacement. Real pendulums do not have
infinitesimal displacements, so their behaviour
is actually of a non-linear kind.
The Physical Pendulum
A "physical" pendulum has
extended size and is a
generalization of the bob pendulum.
An example would be a bar rotating
around a fixed axle. A simple
pendulum can be treated as a
special case of a physical pendulum
with moment of inertia I. ( I = miri2)
Period of a physical pendulum
(Note: l is now the length from the
suspension point to the center of
mass CM instead of L)
Example:
Simple Pendulum: I = mL2
Leg: I = 1/3 mL2
Simple Pendulum
All the mass of a simple
pendulum is concentrated in the
mass m of the particle-like bob,
which is at radius L from the
pivot point. Thus, we can
substitute I = mL2 for the
rotational inertia of the
pendulum.
L
T = 2π
g
for small amplitudes!!
Exercise 4: A clock has a pendulum that performs one full
swing every 1.0 sec. The object at the end of the string
weights 10.0 N. What is the length of the pendulum?
L
T = 2π
g
Solving for L:
(
)
gT 2
9.8 m/s 2 (1.0 s )
L=
=
= 0.25 m
2
2
4π
4π
2
Pivot
Length: L
Mass: M
ICM= 1/12 ML2
CM
Parallel-Axis Theorem
IPivot= 1/12 ML2 + M (½ L)2 = 1/3 ML2
mg
The period is
1
ML2
2L
I
3
T = 2π
= 2π
= 2π
1
3
gMl
g
gM ( L)
2
Ring
Disc
r
r
CM
IPivot= Mr2 + Mr2 = 2Mr2
I
2 Mr 2
T = 2π
= 2π
gMl
gMr
2r
T = 2π
g
CM
IPivot= ½ Mr2 + Mr2 = 3/2 Mr2
I
T = 2π
= 2π
gMl
3r
T = 2π
2g
3
Mr 2
2
gMr
Natural Frequency for different species. To calculate the
moment of inertia we assume the a leg can be treated as rod
of length L
2L
T = 2π
3g
Human: Length of leg L = 1m
T = 2π
Dachshund L=0.2m
Tdachshund
2 1m
= 1.6s
2
3 10m / s
= 0.7 s
If we assume that our legs swings with an max angle of 10degrees or 0.174 rad
v(Human) = s/t = 0.174rad * 1m / T/2 = 0.21 m/s = 0.5mi/h
The Foucault pendulum (pronounced "foo-KOH"), or Foucault's
pendulum, named after the French physicist Léon Foucault, was
conceived as an experiment to demonstrate the rotation of the Earth.
It is a tall pendulum free to oscillate in any vertical plane. The first
public exhibition of a Foucault pendulum took place in February 1851
in the Meridian Room of the Paris Observatory. A few weeks later,
Foucault made his most famous pendulum when he suspended a 28kg bob with a 67-metre wire from the dome of the Panthéon in Paris.
In 1851 it was well known that the earth rotated: observational
evidence included earth'
s measured polar flattening and equatorial
bulge. However, Foucault'
s pendulum was the first dynamical proof of
the rotation in an easy-to-see experiment, and it created a sensation
in both the learned and everyday worlds.
http://www.youtube.com/watch?v=jtkr70fHF08
Foucault'
s Pendulum in
the Panthéon, Paris.
Damped Oscillations
We know that in reality, a spring won'
t oscillate for ever. Frictional forces will
diminish the amplitude of oscillation until eventually the system is at rest.
A mass in air oscillates many times before it comes to rest. A mass in a liquid like
molasses is hardly to oscillate at all.
When dissipative forces such as friction are not negligible, the amplitude of
oscillations will decrease with time. The oscillations are damped.
ρ
ρ
F = −bv
To incorporate friction, we can just say that there is a frictional force that'
s
proportional to the velocity of the mass.
b: damping constant
This is a pretty good approximation for a body moving at a low velocity in air, or in
a liquid.
What we expect is that the amplitude of oscillation decays with time. It is
described with an exponential decay of the amplitude with time, instead of the
amplitude being constant.
The solution is
(http://www.abdn.ac.uk/physics/vpl/pendulum/damped.html)
A
with
Mechanical energy decreases with time
A = A0 e
− bt / 2 m
critically damped: The damping force is such that the system returns to equilibrium
as quickly as possible and stops at that point. ( ζ=1)
overdamped. The damping force is greater than the minimum needed to prevent
oscillations. The system returns to equilibrium without oscillating, but it takes longer
to do so than a critically damped system. . ( ζ>1)
underdamped: It oscillates about the equilibrium point, with ever diminishing
amplitude. ( ζ<1)
Damping factor
zeta
For a (mass spring system)
http://lectureonline.cl.msu.edu/%7Emmp/applist/damped/d.htm
http://www.abdn.ac.uk/physics/vpl/pendulum/applet/applet.html
1.
2.
3.
4.
=1
N
kg
b = 1* 4 ⋅ 6kg ⋅ 234 = 12 39
m
s
Driven Oscillations and Resonance
steady state x(t ) = AD cos(ω D t + φ )
AD =
F /m
(ω02 − ω 2 ) 2 + ωγ 2
When the driving frequency is close to natural frequency,
its amplitude of motion can be quite large.
Resonance curve
A plot of amplitude A versus driving frequency
Small damping
The amplitude can become very large for frequencies
close to natural frequency
Large damping
The amplitude has low, broad peak near the natural
frequencies.
Applet resonance