Chapter 10: Elasticity and Oscillations

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Fisica Generale - Alan Giambattista, Betty McCarty Richardson

Copyright © 2008 – The McGraw-Hill Companies s.r.l.

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 nowL

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 7.510-3 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 42923

4

Y

L

A

FL

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Example (text problem 10.6): A 0.50 m long guitar string, of cross- sectional area 1.010-6 m2, has a Young’s modulus of 2.0109 Pa. By how much must you stretch a guitar string to obtain a tension of 20.0 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

Fpressurestress 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 107.6

Pa 100.60

Pa 1075.1m 23.0

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 10.28): 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 8.62rads/sec 6.12cm 5.0 and

rads/sec 6.12s 50.0

22

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 1.810-4 m at that frequency.

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

2222maxmax 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 = KEmax = Umax.

2maxmax

2max

2

12

1

mvKE

kAU

The value of k is unknown so use KEmax.

2222maxmax 2

2

1

2

1

2

1fmAAmmvKE

The value is KEmax= 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 7.19rads/sec 57.1cm 00.8

cm/sec 6.12rads/sec 57.1cm 00.8

cm 00.8

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:

Assume <<1 radian

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 then the angular frequency of oscillations is found to be:

L

g

The period of oscillations isg

LT 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 25.0

4

s 0.1m/s 8.9

4L

2

2

22

2

2

gT

g

LT

Solving for L:

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Example (text problem 10.84): 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

Chapter 10: Elasticity and Oscillations

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Transcript of Chapter 10: Elasticity and Oscillations