Applied Natural Sciences - Applied Physicsphys.tue.nl/nfcmr/natuur/Ch14.pdfCopyright © 2012 Pearson...
Transcript of Applied Natural Sciences - Applied Physicsphys.tue.nl/nfcmr/natuur/Ch14.pdfCopyright © 2012 Pearson...
Applied Natural Sciences
Leo Pel
e‐mail: [email protected]
http://tiny.cc/3NAB0
Het basisvak Toegepaste Natuurwetenschappen
http://www.phys.tue.nl/nfcmr/natuur/collegenatuur.html
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
13
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
30
)()( 22
2
txdt
xddtdvta
)cos()( tAtx
)sin()( tAdtdxtv
Remember, simple harmonic motion is not uniformly accelerated motion. 31
32
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
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
At which of the following times does the object have the most negative velocity vx?
Question
39
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
At which of the following times does the object have the most negative acceleration ax?
Question
40
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
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.
Energy in Simple Harmonic Motion
53
This diagram shows how the energy transforms from potential to kinetic and back, while the total energy
remains the same.
Energy in Simple Harmonic Motion
54
55
Energy in Simple Harmonic Motion
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
Energy in Simple Harmonic Motion
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
68
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.
Damped Harmonic Motion
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
Damped Harmonic Motion
/( ) 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
Damped Harmonic Motion
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
Example – Bad Shocks
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
Example – Bad Shocks
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
103
104
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
109
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
115
116
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