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Chapter 10

Lecture

Outline

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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

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§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.

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§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.

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Define:

L

Lstrain

The fractional

change in length

A

Fstress

Force per unit cross-

sectional area

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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.

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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

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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

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§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.

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The ultimate strength of a material is the maximum stress

that it can withstand before breaking.

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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

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§10.4 Shear and Volume

Deformations

A shear deformation

occurs when two forces

are applied on opposite

surfaces of an object.

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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

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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. 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

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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

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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.

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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.

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9

63

m 1071.6

Pa 100.60

Pa 1075.1m 230.0

B

PVV

V

VBP

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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

proportionality

Young’s

modulus (Y)

Shear modulus (S) Bulk

Modulus (B)

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§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.

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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

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Assuming the table is frictionless:

txm

kta

makxF

x

xx

Also, 22

2

1

2

1tkxtmvtUtKtE

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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.

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§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

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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.

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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

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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.

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cm/sec 862rads/sec 612cm 05 and

rads/sec 612s 500

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...Aωv

..T

The amplitude A is given, but is not.

Example continued:

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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.

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(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:

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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

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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:

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§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.

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L

m

An FBD for the

pendulum bob:

A simple pendulum:

T

w x

y

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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

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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:

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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.

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§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.

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Graphical representations of damped oscillations:

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§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.

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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