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Chapter 10
Simple Harmonic
Motion and Elasticity
10.1 The Ideal Spring and Simple Harmonic Motion
xkF Applied
x =
spring constant
Units: N/m
10.1 The Ideal Spring and Simple Harmonic Motion
Example 1 A Tire Pressure Gauge
The spring constant of the spring
is 320 N/m and the bar indicator
extends 2.0 cm. What force does the
air in the tire apply to the spring?
xkF Applied
x =
( )( ) N4.6m020.0mN320 ==
10.1 The Ideal Spring and Simple Harmonic Motion
Conceptual Example 2 Are Shorter Springs Stiffer?
A 10-coil spring has a spring constant k. If the spring is
cut in half, so there are two 5-coil springs, what is the spring
constant of each of the smaller springs?
10.1 The Ideal Spring and Simple Harmonic Motion
HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING
10.2 Simple Harmonic Motion and the Reference Circle
tAAx ωθ coscos ==
DISPLACEMENT
10.2 Simple Harmonic Motion and the Reference Circle
tAAx ωθ coscos ==
10.2 Simple Harmonic Motion and the Reference Circle
period T: the time required to complete one cycle
frequency f: the number of cycles per second (measured in Hz)
T
f
1
=
T
f
π
πω
2
2 ==
amplitude A: the maximum displacement
10.2 Simple Harmonic Motion and the Reference Circle
VELOCITY
10.2 Simple Harmonic Motion and the Reference Circle
Example 3 The Maximum Speed of a Loudspeaker Diaphragm
The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm.
(a)What is the maximum speed of the diaphragm?
(b)Where in the motion does this maximum speed occur?
10.2 Simple Harmonic Motion and the Reference Circle
ACCELERATION
 tAaa
a
cx ωωθ coscos
max
2
−=−=
10.2 Simple Harmonic Motion and the Reference Circle
FREQUENCY OF VIBRATION
m
k
=ω
tAax ωω cos2
−=tAx ωcos=
xmakxF =−=∑
2
ωmAkA −=−
10.2 Simple Harmonic Motion and the Reference Circle
Example 6 A Body Mass Measurement Device
The device consists of a spring-mounted chair in which the astronaut
sits. The spring has a spring constant of 606 N/m and the mass of
the chair is 12.0 kg. The measured
period is 2.41 s. Find the mass of the
astronaut.
10.2 Simple Harmonic Motion and the Reference Circle
totalm
k
=ω 2
total ωkm =
T
f
π
πω
2
2 ==
( ) astrochair2total
2
mm
T
k
m +==
π
( )
( )( ) kg77.2kg0.12
4
s41.2mN606
2
2
2
chair2astro
=−=
−=
π
π
m
T
k
m
10.3 Energy and Simple Harmonic Motion
A compressed spring can do work.
10.3 Energy and Simple Harmonic Motion
10.3 Energy and Simple Harmonic Motion
DEFINITION OF ELASTIC POTENTIAL ENERGY
The elastic potential energy is the energy that a spring
has by virtue of being stretched or compressed. For an
ideal spring, the elastic potential energy is
2
2
1
elasticPE kx=
SI Unit of Elastic Potential Energy: joule (J)
10.3 Energy and Simple Harmonic Motion
Conceptual Example 8 Changing the Mass of a Simple
Harmonic Oscilator
The box rests on a horizontal, frictionless
surface. The spring is stretched to x=A
and released. When the box is passing
through x=0, a second box of the same
mass is attached to it. Discuss what
happens to the (a) maximum speed
(b) amplitude (c) angular frequency.
10.3 Energy and Simple Harmonic Motion
Example 8 Changing the Mass of a Simple Harmonic Oscilator
A 0.20-kg ball is attached to a vertical spring. The spring constant
is 28 N/m. When released from rest, how far does the ball fall
before being brought to a momentary stop by the spring?
10.3 Energy and Simple Harmonic Motion
of EE =
2
2
12
2
12
2
12
2
12
2
12
2
1
ooooffff kymghImvkymghImv +++=+++ ωω
oo mghkh =2
2
1
( )( ) m14.0
mN28
sm8.9kg20.02
2
2
==
=
k
mg
ho
10.4 The Pendulum
A simple pendulum consists of
a particle attached to a frictionless
pivot by a cable of negligible mass.
only)angles(small
L
g
=ω
only)angles(small
I
mgL
=ω
10.4 The Pendulum
Example 10 Keeping Time
Determine the length of a simple pendulum that will
swing back and forth in simple harmonic motion with
a period of 1.00 s.
2
2
L
g
T
f ===
π
πω
( ) ( ) m248.0
4
sm80.9s00.1
4 2
22
2
2
===
ππ
gT
L
2
2
4π
gT
L =
• A spring returns to its original shape when the force compressing or
stretching it is removed.
• In fact, all materials become distorted in some way when they are
squeezed or stretched, and many of them, such as rubber, return to
their original shape when the squeezing or stretching is removed.
• Such materials are said to be “elastic.”
• The interatomic forces that hold the atoms of a solid together are
particularly strong, so considerable force must be applied to stretch
or compress a solid object.
• Experiments have shown that the magnitude of the force can be
expressed by the following relation, provided that the amount of stretch
or compression is small compared to the original length of the object:
10.7 Elastic Deformation
Young Modulus
• The amount of elongation of an object, such as the rod depends not
only on the force applied to it, but also on the material of which is
made and its dimensions.
• Y is the proportionality constant called Young’s Modulus or Elastic
Modulus sometimes represented by E.
Young’s modulus has the units of pressure: N/m2
10.7 Elastic Deformation
10.7 Elastic Deformation
Example 12 Bone Compression
In a circus act, a performer supports the combined weight (1080 N) of
a number of colleagues. Each thighbone of this performer has a length
of 0.55 m and an effective cross sectional area of 7.7×10-4
m2
. Determine
the amount that each thighbone compresses under the extra weight.
10.7 Elastic Deformation
A
L
L
YF
o





 ∆
=
( )( )
( )( ) m101.4
m107.7mN104.9
m55.0N540 5
2429
−
−
×=
××
==∆
YA
FL
L o
Strain and Stress
ooo LxLLVV ∆∆∆
Young Modulus
10.7 Elastic Deformation
SHEAR DEFORMATION AND THE SHEAR MODULUS
A
L
x
SF
o





 ∆
=
The shear modulus has the units of pressure: N/m2
10.7 Elastic Deformation
10.7 Elastic Deformation
Example 14 J-E-L-L-O
You push tangentially across the top
surface with a force of 0.45 N. The
top surface moves a distance of 6.0 mm
relative to the bottom surface. What is
the shear modulus of Jell-O?
A
L
x
SF
o





 ∆
=
xA
FL
S o
∆
=
10.7 Elastic Deformation
( )( )
( ) ( )
2
32
mN460
m100.6m070.0
m030.0N45.0
=
×
= −
S
xA
FL
S o
∆
=
10.7 Elastic Deformation
VOLUME DEFORMATION AND THE BULK MODULUS





 ∆
−=∆
oV
V
BP
The Bulk modulus has the units of pressure: N/m2
10.7 Elastic Deformation
10.8 Hooke’s Law
HOOKE’S LAW FOR STRESS AND STRAIN
Stress is directly proportional to strain.
Strain is a unitless quantitiy.
SI Unit of Stress: N/m2
Tutorial Exercises 4 March 2013
Section 10.4 no 14
Problems 13, 25, 27, 28, 43 to 47, 52, 60, 76, 81

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Ch10 simple harmonic motion 25 02 2013

  • 2. 10.1 The Ideal Spring and Simple Harmonic Motion xkF Applied x = spring constant Units: N/m
  • 3. 10.1 The Ideal Spring and Simple Harmonic Motion Example 1 A Tire Pressure Gauge The spring constant of the spring is 320 N/m and the bar indicator extends 2.0 cm. What force does the air in the tire apply to the spring? xkF Applied x = ( )( ) N4.6m020.0mN320 ==
  • 4. 10.1 The Ideal Spring and Simple Harmonic Motion Conceptual Example 2 Are Shorter Springs Stiffer? A 10-coil spring has a spring constant k. If the spring is cut in half, so there are two 5-coil springs, what is the spring constant of each of the smaller springs?
  • 5.
  • 6. 10.1 The Ideal Spring and Simple Harmonic Motion HOOKE’S LAW: RESTORING FORCE OF AN IDEAL SPRING
  • 7.
  • 8. 10.2 Simple Harmonic Motion and the Reference Circle tAAx ωθ coscos == DISPLACEMENT
  • 9. 10.2 Simple Harmonic Motion and the Reference Circle tAAx ωθ coscos ==
  • 10. 10.2 Simple Harmonic Motion and the Reference Circle period T: the time required to complete one cycle frequency f: the number of cycles per second (measured in Hz) T f 1 = T f π πω 2 2 == amplitude A: the maximum displacement
  • 11. 10.2 Simple Harmonic Motion and the Reference Circle VELOCITY
  • 12. 10.2 Simple Harmonic Motion and the Reference Circle Example 3 The Maximum Speed of a Loudspeaker Diaphragm The frequency of motion is 1.0 KHz and the amplitude is 0.20 mm. (a)What is the maximum speed of the diaphragm? (b)Where in the motion does this maximum speed occur?
  • 13. 10.2 Simple Harmonic Motion and the Reference Circle ACCELERATION  tAaa a cx ωωθ coscos max 2 −=−=
  • 14. 10.2 Simple Harmonic Motion and the Reference Circle FREQUENCY OF VIBRATION m k =ω tAax ωω cos2 −=tAx ωcos= xmakxF =−=∑ 2 ωmAkA −=−
  • 15. 10.2 Simple Harmonic Motion and the Reference Circle Example 6 A Body Mass Measurement Device The device consists of a spring-mounted chair in which the astronaut sits. The spring has a spring constant of 606 N/m and the mass of the chair is 12.0 kg. The measured period is 2.41 s. Find the mass of the astronaut.
  • 16. 10.2 Simple Harmonic Motion and the Reference Circle totalm k =ω 2 total ωkm = T f π πω 2 2 == ( ) astrochair2total 2 mm T k m +== π ( ) ( )( ) kg77.2kg0.12 4 s41.2mN606 2 2 2 chair2astro =−= −= π π m T k m
  • 17. 10.3 Energy and Simple Harmonic Motion A compressed spring can do work.
  • 18. 10.3 Energy and Simple Harmonic Motion
  • 19. 10.3 Energy and Simple Harmonic Motion DEFINITION OF ELASTIC POTENTIAL ENERGY The elastic potential energy is the energy that a spring has by virtue of being stretched or compressed. For an ideal spring, the elastic potential energy is 2 2 1 elasticPE kx= SI Unit of Elastic Potential Energy: joule (J)
  • 20. 10.3 Energy and Simple Harmonic Motion Conceptual Example 8 Changing the Mass of a Simple Harmonic Oscilator The box rests on a horizontal, frictionless surface. The spring is stretched to x=A and released. When the box is passing through x=0, a second box of the same mass is attached to it. Discuss what happens to the (a) maximum speed (b) amplitude (c) angular frequency.
  • 21. 10.3 Energy and Simple Harmonic Motion Example 8 Changing the Mass of a Simple Harmonic Oscilator A 0.20-kg ball is attached to a vertical spring. The spring constant is 28 N/m. When released from rest, how far does the ball fall before being brought to a momentary stop by the spring?
  • 22. 10.3 Energy and Simple Harmonic Motion of EE = 2 2 12 2 12 2 12 2 12 2 12 2 1 ooooffff kymghImvkymghImv +++=+++ ωω oo mghkh =2 2 1 ( )( ) m14.0 mN28 sm8.9kg20.02 2 2 == = k mg ho
  • 23. 10.4 The Pendulum A simple pendulum consists of a particle attached to a frictionless pivot by a cable of negligible mass. only)angles(small L g =ω only)angles(small I mgL =ω
  • 24. 10.4 The Pendulum Example 10 Keeping Time Determine the length of a simple pendulum that will swing back and forth in simple harmonic motion with a period of 1.00 s. 2 2 L g T f === π πω ( ) ( ) m248.0 4 sm80.9s00.1 4 2 22 2 2 === ππ gT L 2 2 4π gT L =
  • 25. • A spring returns to its original shape when the force compressing or stretching it is removed. • In fact, all materials become distorted in some way when they are squeezed or stretched, and many of them, such as rubber, return to their original shape when the squeezing or stretching is removed. • Such materials are said to be “elastic.” • The interatomic forces that hold the atoms of a solid together are particularly strong, so considerable force must be applied to stretch or compress a solid object. • Experiments have shown that the magnitude of the force can be expressed by the following relation, provided that the amount of stretch or compression is small compared to the original length of the object: 10.7 Elastic Deformation
  • 26. Young Modulus • The amount of elongation of an object, such as the rod depends not only on the force applied to it, but also on the material of which is made and its dimensions. • Y is the proportionality constant called Young’s Modulus or Elastic Modulus sometimes represented by E. Young’s modulus has the units of pressure: N/m2
  • 28. 10.7 Elastic Deformation Example 12 Bone Compression In a circus act, a performer supports the combined weight (1080 N) of a number of colleagues. Each thighbone of this performer has a length of 0.55 m and an effective cross sectional area of 7.7×10-4 m2 . Determine the amount that each thighbone compresses under the extra weight.
  • 29. 10.7 Elastic Deformation A L L YF o       ∆ = ( )( ) ( )( ) m101.4 m107.7mN104.9 m55.0N540 5 2429 − − ×= ×× ==∆ YA FL L o
  • 30. Strain and Stress ooo LxLLVV ∆∆∆
  • 32. 10.7 Elastic Deformation SHEAR DEFORMATION AND THE SHEAR MODULUS A L x SF o       ∆ = The shear modulus has the units of pressure: N/m2
  • 34. 10.7 Elastic Deformation Example 14 J-E-L-L-O You push tangentially across the top surface with a force of 0.45 N. The top surface moves a distance of 6.0 mm relative to the bottom surface. What is the shear modulus of Jell-O? A L x SF o       ∆ = xA FL S o ∆ =
  • 35. 10.7 Elastic Deformation ( )( ) ( ) ( ) 2 32 mN460 m100.6m070.0 m030.0N45.0 = × = − S xA FL S o ∆ =
  • 36. 10.7 Elastic Deformation VOLUME DEFORMATION AND THE BULK MODULUS       ∆ −=∆ oV V BP The Bulk modulus has the units of pressure: N/m2
  • 38. 10.8 Hooke’s Law HOOKE’S LAW FOR STRESS AND STRAIN Stress is directly proportional to strain. Strain is a unitless quantitiy. SI Unit of Stress: N/m2
  • 39. Tutorial Exercises 4 March 2013 Section 10.4 no 14 Problems 13, 25, 27, 28, 43 to 47, 52, 60, 76, 81