Applied Natural Sciences - Applied Physicsphys.tue.nl/nfcmr/natuur/Ch14.pdfCopyright © 2012 Pearson...

Applied Natural Sciences

Leo Pel

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Copyright © 2012 Pearson Education Inc.

PowerPoint® Lectures forUniversity Physics, Thirteenth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by Wayne Anderson

Chapter 14

Periodic Motion

LEARNING GOALS

• How to describe oscillations in terms of amplitude, period, frequency,

and angular frequency.

• How to do calculations with simple harmonic motion, an important type of

oscillation.

• How to use energy concepts to analyze simple harmonic motion.

• How to apply the ideas of simple harmonic motion to different physical

situations.

• How to analyze the motions of a simple pendulum.

• What a physical pendulum is, and how to calculate the properties of its motion.

• What determines how rapidly an oscillation dies out.

• How a driving force applied to an oscillator at the right frequency can cause a

very large response, or resonance 4

Oscillating Ruler

All these oscillators have two things in common:

1. The oscillation takes place about an equilibrium position

2. The motion is periodic.

5

Describing Oscillatory Motion

Characteristics of oscillatory motion:

• Amplitude A = max displacement

from equilibrium.

• Period T = time for the motion

to repeat itself.

• Frequency f = # of oscillations per unit time.

1fT

[ f ] = hertz (Hz)

= 1 cycle / s

same period T

same amplitude A

Heinrich Hertz

(1857-1894)6

Frequency and Period

1/ and 1/f T T f is the frequency

(units: Hz oscillations per second)f

is the period (units: s)T

7

Oscillating Ruler

An oscillating ruler completes 28 cycles in 10 s & moves a total distance of 8.0 cm.

What are the amplitude, period, & frequency of this oscillatory motion?

Amplitude = 8.0 cm / 2 = 4.0 cm.

1028

sTcycles

1fT

0.36 /s cycle

2810cycles

s 2.8 Hz

A, T, f do not specify an oscillation completely.

What is the physics; the motion equation?

8

Oscillations

• An object with a stable equilibrium tends to oscillate about that equilibrium

• This oscillation involves at least two types of energy: kinetic and a potential energy

• Once the motion has been started, it will repeat

When energy traded back and forth between kinetic and potential energy: “resonance”

9

• Hooke’s law:

• The force always acts toward the equilibrium position: restoring force

• The mass is initially pulled to a distance A and released from rest

• As the object moves toward the equilibrium position, F and a decrease, but v increases

kxF

Motion of the spring-mass system

10

• At x = 0, F and a are zero, but v is a maximum

• The object’s momentum causes it to overshoot the equilibrium position

• The force and acceleration start to increase in the opposite direction and velocity decreases

• The motion momentarily comes to a stop at x = - A

11

• It then accelerates back toward the equilibrium position

• The motion continues indefinitely

• The motion of a spring mass system is an example of Simple Harmonic Motion (SHM)

12

Acceleration

• When the block is displaced from the equilibrium point and released, it is a particle under a net force and therefore has an acceleration.• The force described by Hooke’s Law is the net force in Newton’s Second Law.

kxdt

xdm 2

2

1. The acceleration is proportional to the displacement of the block

2. Therefore, the kinematic equations cannot be applied.

3. The force is conservative. In the absence of friction, the motion will continue forever.

Real systems are generally subject to friction, so they do not actually oscillate forever.

14

Experimental test

15

1

23

4

5

86

7

SHM Dynamics

x R cos

x

y

1

1

0

12

3

45

6

8

7

1

2 2

16

Kinematics of rotation

Define position by θ

17

Circle 2πr

Angle radian:

2π =3600

Defenition radian:

Θ = s/r

Calculate in radians!

radian is dimensionless

Kinematics of rotation

18

Be aware

19

Quiz

A hard disk is spinning at 6000 rpm (revolutions per minute)Angular velocity is:

Answer: 3. 6000 rpm=6000x2π rad/60 sec=630 rad/s

2. 732 rad/s

1. 704 rad/s

4. don’t know

3. 630 rad/s

(neem π = 3.15)

20

Rotation

/d dt Angular velocity

rad/s

Angular velocity ω: is 2 π times the frequency.

2 f 2T

Period T [s]: is the time for one cycle

21

1

23

4

5

86

7

SHM Dynamics

x(t) R cos R cost

x

y

1

1

0

12

3

45

6

8

7

1

2 2

22

1

2

3

4

5

86

7

SHM Dynamics

x(t) R cos R cost

x

y

1

1

0

12

3

45

6

8

7

1

2 2

cosine= ‘projection on x-axis’Projection will oscillate around zero as function time

sine= ‘projection on y-axis’23

Circular motion and the Equations of Simple Harmonic Motion

Simple harmonic motion is the projection of uniform

circular motion onto a diameter

x Acos24

Cosine

Projection x-axis

Sine

Projection y-axis

cosine

sine

Projections ‘move’ like oscillator 25

SHM Solution

Relation Acost Asint

2

Phase shift

26

T 2

A

A

Drawing of Acost

SHM Solution

2 f 2T

27

Drawing of Acost

SHM Solution

28

Question math

x(t) = A cos(t + )

Which parameter varies in

Figures A, B, and C to the

left?

A

B

C

1. A, ,

2. , , A

3. , A,

4. , A,

29

)()( 22

2

txdt

xddtdvta

)cos()( tAtx

)sin()( tAdtdxtv

Remember, simple harmonic motion is not uniformly accelerated motion. 31

For spring we had as motion equation

kxdt

xdm 2

2

33

Maximum Values of v and a

Because the sine and cosine functions oscillate between ±1, we can

easily find the maximum values of velocity and acceleration for an object

in SHM.

max

2max

kv A Amka A Am

The graphs show:(a) displacement as a function of time(b) velocity as a function of time(c ) acceleration as a function of time

The velocity is 90o out of phase with the displacement and the acceleration is 180o out of phase with the displacement. 34

The Period of a Mass on a SpringThe Period of a Mass on a Spring

Therefore, the period is:

35

Simple Harmonic Motion equations

km

fT

mkf

mk

22121

2

kxdt

xdm 2

2

36

Oscillating Ruler

f 2

12

km

37

This is an x-t graph for an object in simple harmonic motion.

1. t = T/4

2. t = T/2

3. t = 3T/4

4. t = T

Question

At which of the following times does the object have the most negative velocity vx?

38


1. t = T/4

2. t = T/2

3. t = 3T/4

4. t = T

At which of the following times does the object have the most negative velocity vx?

Question

39


1. t = T/4

2. t = T/2

3. t = 3T/4

4. t = T

At which of the following times does the object have the most negative acceleration ax?

Question

40


1. t = T/4

2. t = T/2

3. t = 3T/4

4. t = T

At which of the following times does the object have the most negative acceleration ax?

Question

41

Two Oscillating Systems

The diagram shows two identical masses attached to two identical springs and resting on a horizontal frictionless surface. Spring 1 is stretched to 5 cm, spring 2 is stretched to 10 cm, and the masses are released at the same time.

Which mass reaches the equilibrium position first?

1) 1

2) 2

3) Same time

Question

42

Two Oscillating Systems

Same springs and mass

Spring 1 is stretched to 5 cm, spring 2 is stretched to 10 cm, and the masses are released at the same time.

Which mass reaches the equilibrium position first?

1) 1

2) 2

3) Same time

Because k and m are the same, the systems have the same period, so they must return to equilibrium at the same time.

The frequency and period of SHM are independent of amplitude.

Question

43

Period and Amplitude in SHM

Period and frequency of simple harmonic motion are

completely determined by:

• The mass m and

• The force constant k

In simple harmonic motion the period and the frequency

do not depend on the amplitude A.

km

fT

mkf

22121

2

mk

tAx

= where

)cos(

44

Displacement, Velocity and Acceleration in SHM

x Acost 0 t

mk

tAx

= where

)cos(

45

Series vs. Parallel Springs

What about these two spring configurations?

For Block 1:keff = 2 k andT1 = 2(m/2k)½

For Block 2:keff = k/2 andT2 = 2(2m/k)½

T1 < T246

Example: A Block on a SpringA 2.00 kg block is attached to a spring as shown.

The force constant of the spring is k = 196 N/m.The block is held a distance of 5.00 cm fromequilibrium and released at t = 0

(a) Find the angular frequency , the frequency f, and the period T.(b) Write an equation for x vs. time.

(196 N/m) 9.90 rad/s(2.00 kg)

km

(9.90 rad/s) 1.58 Hz

2 2f

1/ 0.635 sT f 5.00 cm and 0A

(5.00 cm)cos (9.90 rad/s)x t47

Example: A System in SHM

An air-track glider is attached to a spring,pulled 20 cm to the right, and releasedat t=0. It makes 15 completeoscillations in 10 s.

a. What is the period of oscillation?

b. What is the object’s maximum speed?

c. What is its position and velocity at t=0.80 s?

15 oscillations10 s

1.5 oscillations/s 1.5 Hz

f

1/ 0.667 sT f

max2 2 (0.20 m) 1.88 m/s

(0.667 s)Av

T

2 2 (0.80 s)cos (0.20 m)cos 0.062 m 6.2 cm(0.667 s)

tx AT

max2 2 (0.80 s)sin (1.88 m/s)sin 1.79 m/s

(0.667 s)tv v

T

48

Example: A Vertical OscillationA 200 g block hangs from a

spring with spring constant 10 N/m. The block is pulled down to a point where the spring is 30 cm longer than it’s unstretched length, then

released.

Where is the block and what is its velocity 3.0 s later?

/ 0.196 m =19.6 cmL mg k

0( 0.30 m) ( 0.196 m) 0.104 m; =0A

/ (10 N/m) /(0.2 kg) 7.07 rad/sk m

0( ) cos( )( 0.104 m)cos[(7.07 rad/s)(3.0 s)]0.074 m

= 7.4 cm (above equilibrium position).

y t A t

0sin( ) 0.52 m/s 52 cm/sxv A t

49

Example: Finding the Time

A mass, oscillating in simple harmonic motion, starts at x = A and has period T.

At what time, as a fraction of T, does the mass first pass through x = ½A?

11 162

cos2 2 3T Tt T

12

2cos tx A AT

50

ENERGY

51

In an ideal system with no nonconservativeforces, the total mechanical energy is conserved. For a mass on a spring:

Since we know the position and velocity as functions of time, we can find the maximum kinetic

and potential energies:

Energy in Simple Harmonic Motion

52

As a function of time,

So the total energy is constant;

as the kinetic energy increases, the potential energy decreases, and vice versa.


53

This diagram shows how the energy transforms from potential to kinetic and back, while the total energy

remains the same.


54


56

1 1 1 12 2 2 2max2 2 2 2 constantxE mv kx kA mv

2 2 2 2kv A x A xm

222 20 00 0

mv vA x xk


58

Example:Using Conservation of Energy

A 500 g block on a spring is pulled a distance of 20 cm and released. The subsequent oscillations are measured to have a period of 0.80 s. At what position (or positions) is the speed of the block 1.0 m/s?

2 2 2 2kv A x A xm

222 2 (1.0 m/s)(0.20 m) 0.154 m 15.4 cm

(7.85 rad/s)vx A

2 20.80 s so 7.85 rad/s(0.80 s)

TT

59

To double the total energy of a mass-spring system oscillating in simple harmonic motion, the amplitude must increase by a factor of

1. 4.

2.

3. 2.

4.

5. 4 2 1.189.

2 1.414.

2 2 2.828.

Question

60

1. 4.

2.

3. 2.

4.

5.

To double the total energy of a mass-spring system oscillating in simple harmonic motion, the amplitude must increase by a factor of

2 2 2.828.

2 1.414.

4 2 1.189.

Question

1 1 1 12 2 2 2max2 2 2 2 constantxE mv kx kA mv

61

The Simple Pendulum

63

Looking at the forces on the pendulum bob, we see that the restoring force is proportional to sin , whereas the restoring force for a spring is proportional to the displacement (which is in this case).

sinF mg mg

This is not a Hooke’s Law force.

The Simple Pendulum

t td sF ma mg mdt

2

2sin

64

The Small Angle ApproximationThe Small Angle ApproximationHowever, for small angles, sin θ and θ are approximately

equal.

3 5

sin3! 5!

if 1

65

In the tangential direction,

The length, L, of the pendulum is constant, and for small values of

This confirms the mathematical form of the motion is the same as for SHM.

t td sF ma mg mdt

2

2sin

2

2 sind g gdt L L

gL

fT 21

km

fT 21

Note that T does not depend on m, the mass of the pendulum bob.

kxdt

xdm 2

2

Ls

66

Large Amplitude Oscillations

2

2 40 0 02 2

1 1 31 sin / 2 sin / 22 2 4

T T

0where 2 /T L g

How much of an error in calculating the period Tdo we make by substituting θ for sin θ ?

67

Simple Harmonic Motion: Spring non‐ideal case

Fx kx

69

ExampleA visitor to a lighthouse wishes to determine

the height of the tower. She only has a rope without any any measuremen scale. Can she measure the height?

1) Yes

2) No

2

2

2

222

)141592.3(4)8.9(4.9

44

2

gTlg

lT

heightlglT

PP

P

L = Height = 21.93 m

She makes a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 9.40 s.

70

Development of moderntimekeeping driven by navigation

Before mid-1700sLatitude: use quadrant/sextant/octant to sight North Star or sun

Longitude: lunar time – very poor accuracy

71

1714: British Parliament sets £20,000 prize ($10MEuro) to make clock accurate to 2 minutes (0.5° longitude)

John Harrison – 1764

Development of moderntimekeeping driven by navigation

72

Mechanica watch

Torsion‐spring torque: z = –

With torsion constant

Newton’s Second Law for rotation:

I= moment of inertia

2

2 )(dt

tdI

I 2 IT

• No small-angle restriction is necessary (works on sea) k

mf

T 21

kxdt

xdm 2

2

Quartz oscillators

Typical frequency in watch: 32,768 Hz

(period is 31 s)

Most modern clocks use a quartz oscillator

Crystalline quartz is a harmonic oscillator. Oscillation decay is extremely slow (verypure tone) Quartz is piezoelectric. Mechanically‐electrically coupled motion inducedand measured electrically. The rate of expansion and contraction is the resonancefrequency, and is determined by the cut and size of the crystal.

Can think of bonds between atoms in a crystal as springs. So, the restoring force is

proportional to the distance from equilibrium

74

Quartz Clocks

Nearly insensitive to gravity, temperature, pressure, and acceleration

Slow vibration decay leads to precise period

(loses/gains 0.1 sec in 1 year)

Problem: No two pendula or quartz oscillators

are exactly the same

All atoms are: use Ce

75

NIST-F1 Cesium Fountain Atomic ClockThe Primary Time and Frequency Standard for

the United States

Atomic Clocks

Loses less than one second in 60 million years

76

Every GPS satellitecontains an atomic clock

Receivers: high quality quartz clock which is synchronized to atomic clock

77

x

Example: fluid in U‐tube

km

fT 21

kxdt

xdm 2

2

Here mass, with l length water column:

lAm Restoring force is given by:

gAxF 2

gl

fT

221

Hence:

gl

fT 21

Pendulum:

Length l

78

Example: buoyancy

km

fT 21

kxdt

xdm 2

2

ML

y

Buoyancy gives for the depth L:

gLAgM

Restoring force is given by:

gAyF

gL

gALA

gAM

fT

2221

Hence:

gl

fT 21

79

A simple pendulum consists of a point mass suspended by a massless, unstretchable string.

If the mass is doubled while the length of the string remains the same, the period of the pendulum

1. becomes 4 times greater.

2. becomes twice as great.

3. becomes greater by a factor of .

4. remains unchanged.

5. decreases.

2

Question

80

A simple pendulum consists of a point mass suspended by a massless, unstretchable string.

If the mass is doubled while the length of the string remains the same, the period of the pendulum

1. becomes 4 times greater.

2. becomes twice as great.

3. becomes greater by a factor of .

4. remains unchanged.

5. decreases.

2

Question

81

Two basic modes of oscillation:

kxdt

xdm 2

2

km

fT 21

gl

fT 21

spring

pendulum ?

‘mass <> spring constant’

‘length <> gravity’

82

VERY ‘STRANGE SHM’: Kinetic friction

83

DAMPING

84

Damped simple harmonic motion

85

Damped simple harmonic motion

bvFb Damping

constant

Damping

force

Non conservative force

‘Loose energy’

86

Experiment on damping

87

Damped Harmonic Motion

In many systems, the non-conservative force

(called the damping force) is approximately equal to

where b is a constant giving the damping strength and

v is the velocity. The motion of such a mass-spring

system is described by:

bv

2net

2

F

dxkx bdt

a

d

m

xmdt

88

The solution of the differential equation

is of the form

For simplicity, we take x = A at t = 0, then f = 0.


2

2

dxkx bdt d

m d xt

/( ) cos( )tx t Ae t

89

If one plugs the solution:

into Newton’s 2nd law, one will findthe damping time

And the angular frequency,

Where is the un-damped angular frequency


/( ) cos( )tx t Ae t

0 /k m

bm /2

22

2

mb

o

90

The larger the damping constant b the shorter the damping time t. There are 3 damping regimes:

(a) Underdamped

(b) Critically damped

(c) Overdamped


22

2

mb

o 2 cos( )b tmx Ae t

91

Energy in damped oscillations

The damping force is nonconservative; the mechanical

energy of the system decreases continuously.

E 12

mv2x

12

kx2

Rate of change of E: dEdt

mvxdvx

dt kx dx

dt vx max kx

Using:kx bvx max

bvx max kxdEdt

vx bvx bvx2

Thus the energy change is:

92

An example of a damper: automobile shock absorber

93

Example – Bad Shocks

A car’s suspension can be

modeled as a damped

mass‐spring system with

m = 1200 kg, k = 58 kN/m

and b = 230 kg/s.

How many oscillations does it

take for the amplitude of

the suspension to drop to half its initial value?

http://static.howstuffworks.com/gif/car-suspension-1.gif

94


First find out how long it takes

for the amplitude

to drop to half its initial value:

= 2m/b = 10.43 s

exp(‐t/) = ½

→ t = ln 2 = 7.23 s http://static.howstuffworks.com/gif/car-suspension-1.gif

95


The period of oscillation isT = 2/

= 2/√(k/m – 1/2)= 0.904 s

Therefore, in 7.23 s, the shocks oscillate 7.23/0.904 ~ 8 times!

These are really bad shocks!http://static.howstuffworks.com/gif/car-suspension-1.gif

96

Example: extreme bad shocks

97

DRIVEN

98

Driven Oscillations

When an oscillatory system is acted upon by an external force we

say that the system is driven (instate of free oscillator)

Consider an external oscillatory force:

F = F0 cos(d t).

Newton’s 2nd law for the system becomes:

net

0 d

2

2cos

F

dxkx b F tdt

a

d xdt

m

m

99

Driven Oscillations

Again, we try a solution of the

form x(t) = A cos(d t). When

this is plugged into the 2nd law,

we find that the amplitude has

the resonance form

2 22 20

0

( )( ) /

d

d d

Am b m

F

100

2 22 20

0

( )( ) /

d

d d

Am b m

F

1. A maximum as:

• Maximum energy transfer0d

0km

2. Damping does not change resonance freq

• Natural resonance:

101

Driven oscillator: an example

102

Example – Resonance

November 7, 1940 – Tacoma Narrows Bridge Disaster. At about 11:00 am the

Tacoma Narrows Bridge, near Tacoma, Washington collapsed after hitting

its resonant frequency. The external driving force was the wind.

http://www.enm.bris.ac.uk/anm/tacoma/tacnarr.mpg106

Tacoma bridge

107

Mexico 1985

In the 1985 Mexico City earthquake, buildings between 5and 15 stories tall collapsed because they resonated at thesame frequency as the quake. Taller and shorter buildingssurvived.

Resonance: when the period of the seismic wave matches the period of a structure

30 seconds of shaking put the structure into resonance

108

Application: Swaying skyscraper

110

Application: Swaying skyscraper

Tuned mass damper :

f damper = f building ,

damper building = .

Taipei 101 TMD:

41 steel plates,

730 ton, d = 550 cm,

87th-92nd floor.

Also used in:

• Tall smokestacks, Airport control towers, Power-plant

cooling towers, Bridges, Ski lifts.111

Ground resonance

112

Speed wobble skateboard

113

Be aware: damping does not ‘change’ the frequency

You have to solve it by changing

the resonance frequency

In general: go higher > make it stiffer

118

Summary

119

Summary

120

Depending on the direction of the displacement relative to the direction of propagation, we can define wave motion as:

• Transverse – if the direction of displacement is perpendicular to the direction of propagation

Types of waves : Next week

• Longitudinal – if the direction of displacement is parallel to the direction of propagation

121

Applied Natural Sciences - Applied Physicsphys.tue.nl/nfcmr/natuur/Ch14.pdfCopyright © 2012 Pearson...

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