Chapter 15 Oscillations and Waves - austincc.edu · MFMcGraw-PHY 2425 Chap...
Transcript of Chapter 15 Oscillations and Waves - austincc.edu · MFMcGraw-PHY 2425 Chap...
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Chapter 15
Oscillations and Waves
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 2
Oscillations and Waves
• Simple Harmonic Motion
• Energy in SHM
• Some Oscillating Systems
• Damped Oscillations
• Driven Oscillations
• Resonance
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 3
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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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 4
Characteristics of SHM
• Repetitive motion through a central equilibrium point.
• Symmetry of maximum displacement.
• Period of each cycle is constant.
• Force causing the motion is directed toward the
equilibrium point (minus sign).
• F directly proportional to the displacement from
equilibrium.
Acceleration = - ω2 x Displacement
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 5
A Simple Harmonic Oscillator (SHO)
Frictionless surface
The restoring force is F = −kx.
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 6
Frictionless surface
Two Springs with Different Amplitudes
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 7
SHO Period is Independent of the Amplitude
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 8
The period of oscillation is .2
ω
π=T
where ω is the angular frequency of
the oscillations, k is the spring
constant and m is the mass of the
block.
m
k=ω
The Period and the Angular Frequency
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 9
At the equilibrium point x = 0 so, a = 0 also.
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.
Simple Harmonic Motion
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 10
Representing Simple Harmonic Motion
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 11
Representing Simple Harmonic Motion
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 12
Representing Simple Harmonic Motion
Position - xmax = A
Velocity - vmax = ωA
Acceleration - amax = ω2A
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 13
A simple harmonic oscillator can be described
mathematically by:
( )
( )
( ) 2
x t = Acosωt
dxv t = = -Aωsinωt
dt
dva t = = -Aω cosωt
dt
Or by:
( )
( )
( ) 2
x t = Asinωt
dxv t = = Aωcosωt
dt
dva t = = -Aω sinωt
dt
where A is the amplitude of
the motion, the maximum
displacement from
equilibrium, Aω = vmax, and
Aω2 = amax.
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 14
Linear Motion - Circular Functions
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 15
Projection of Circular Motion
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 16
Circular Motion is the superposition of two linear SHO that are
900 out of phase with each other
sin( )y A tω=
cos( )x A tω=
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 17
sinx = A t -φ
cosω
{ } ϕ
sin tx = A 2π -
cos T
( )( )
π
2
π π
2 2
x = Asin t -
x = A sinωt cos - sin cosωt
x = A sinωt (0) - (1)cosωt
x = -Acosωt
ω
The minus sign means that the
phase is shifted to the right.
A plus sign indicated the phase
is shifted to the left
Shifting Trig Functions
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 18
Shifting Trig Functions
πωt - = 0
2
πωt =
2
π 1 1 Tt = ; =
2ω ω 2π
π T Tt = =
2 2π 4
πsin t - = 0
2ω
Shifted Trig Functions
-1.50
-1.00
-0.50
0.00
0.50
1.00
1.50
-3.00 -2.00 -1.00 0.00 1.00 2.00 3.00
Time
sin(ωt)
sin(ωt-δ)
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 19
Energy
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 20
Assuming the table is frictionless:
( ) ( ) ( )
∑ xx
2
x
F = - kx = ma
ka t = - x t = - ω x t
m
Also, ( ) ( ) ( ) ( )2 21 1
2 2= + = +E K t U t mv t kx t
Equation of Motion & Energy
Classic form for SHM
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 21
Spring Potential Energy
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 22
Spring Total Energy
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 23
Approximating Simple Harmonic Motion
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 24
Approximating Simple Harmonic Motion
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 25
Potential and Kinetic Energy
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 26
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 = Aω.
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 27
( )( ) cm/sec 862rads/sec 612cm 05 and
rads/sec 612s 500
22
...Aωv
..T
===
===ππ
ω
The amplitude A is given, but ω is not.
Example continued:
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 28
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
1.8×10−4 m at that frequency.
(a) What is the maximum force acting on the
diaphragm?
( ) ( ) 2222
maxmax 42 mAffmAAmmaFF ππω =====∑
The value is Fmax=1400 N.
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 29
(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 πω ===
The value is Kmax= 0.13 J.
Example continued:
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 30
Example: 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 sinωt gives A = 8.00 cm
and ω = 1.57 rads/sec. The frequency is:
Hz 250.02
rads/sec 57.1
2===
ππ
ωf
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 31
( )( )
( )( ) 222
max
max
max
cm/sec 719rads/sec 571cm 008
cm/sec 612rads/sec 571cm 008
cm008
...Aa
...Av
.Ax
===
===
==
ω
ω
Other quantities can also be determined:
The period of the motion is sec 00.4rads/sec 57.1
22===
π
ω
πT
Example continued:
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 32
What About Gravity?
When a mass-spring system is oriented vertically,
it will exhibit SHM with the same period and
frequency as a horizontally placed system.
The effect of gravity is canceled out.
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 33
Why We Ignore Gravity with Vertical Springs
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 34
The Simple 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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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 35
The pendulum is best
described using polar
coordinates.
The origin is at the pivot
point. The coordinates are
(r, φ). The r-coordinate
points from the origin
along the rod. The φ-
coordinate is perpendicular
to the rod and is positive in
the counter clockwise
direction.
The Simple Pendulum
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 36
Apply Newton’s 2nd
Law to the pendulum
bob.
2
sin
cosr
F mg ma
vF T mg m
r
φφ φ
φ
= − =
= − =
∑
∑
If we assume that φ <<1 rad, then sin φ ≈ φ and cos φ ≈1, the angular
frequency of oscillations is then:
L
g=ω
The period of oscillations isg
LT π
ω
π2
2==
sin
sin
( / )sin
( / )
F mg ma mL
mg mL
g L
g L
φφ φ α
φ α
α φ
α φ
= − = =
− =
= −
= −
∑
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 37
Example: 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
===
=
ππ
π
Solving for L:
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 38
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-cosθ).
θ
L
y=0
L
Lcosθ
)cos1( θ−= Ly
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 39
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.
The Physical Pendulum
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 40
The Physical Pendulum
IT = 2π
MgD
I is the moment of inertia about
the given axis. The Icm from the
table will need to be modified
using the parallel axis theorem.
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 41
Compound Pendulum
I = Irod + Idisk
M = mrod + Mdisk
D = distance from the axis to
the center of mass of the rod
and disk.
IT = 2π
MgD
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 42
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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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 43
Damped Oscillations Equations2
02
d x dxm + b + mω x = 0
dt dt
ωτ
'
0
-tx(t) = A exp cos( t +δ)
2
2
'
0
0
bω = ω 1 -
2mω0
kω =
mc 0
mτ = ; b = 2mω
b
For b > bc the system is overdamped. For b = bc the system is
critically damped. The object doesn’t oscillate and returns to its
equilibrium posion very rapidly.
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 44
Damped Oscillations
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 45
Graphical representations of damped oscillations:
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 46
•Overdamped: The system returns to equilibrium without
oscillating. Larger values of the damping the return to
equilibrium slower.
•Critically damped : The system returns to equilibrium as
quickly as possible without oscillating. This is often
desired for the damping of systems such as doors.
•Underdamped : The system oscillates (with a slightly
different frequency than the undamped case) with the
amplitude gradually decreasing to zero.
Source: Damping @ Wikipedia
Damped Oscillations
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 47
Damped Oscillations
The larger the damping the more difficult it is to assign
a frequency to the oscillation.
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Damped Oscillations
2E α A
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Forced Oscillations
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 50
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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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 51
Forced Oscillations Equations2
0 02
d x dxm + b + mω x = F cosωt
dt dt
ω δx = Acos( t - )
0
2 2 2 2 2 2
0
FA =
m (ω - ω ) + b ω2 2
0
bωtanδ =
m(ω - ω )
ma friction spring applied force
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 52
2 2
0
-tEnergy α A = A exp
τ
x
dxv = = -ωAsin(ωt - δ)
dt
πω ωxv = -ωAsin(ωt - ) = + Acos t
2
At resonance v and Fo are in phase
2 2102
2 210 02
-tE = mω A = E exp
τ
E = mω A ; τ = m b
0
0
ω mQ = ω τ =
b
Energy and Resonance2
0 02
d x dxm + b + mω x = F cosωt
dt dt
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 53
Power Transfer
ω
ω∆0Q =
The dissipation in the system, represented by “b” keeps the
amplitude from going to infinity.
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 54
Tacoma Narrows Bridge
Nov. 7, 1940
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Tacoma Narrows Bridge
Nov. 7, 1940
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 56
The first Tacoma Narrows Bridge opened to traffic on July 1, 1940. It
collapsed four months later on November 7, 1940, at 11:00 AM (Pacific
time) due to a physical phenomenon known as aeroelastic flutter caused
by a 67 kilometres per hour (42 mph) wind.
The bridge collapse had lasting effects on science and engineering. In
many undergraduate physics texts the event is presented as an example of
elementary forced resonance with the wind providing an external periodic
frequency that matched the natural structural frequency (even though the
real cause of the bridge's failure was aeroelastic flutter[1]).
Its failure also boosted research in the field of bridge aerodynamics/
aeroelastics which have themselves influenced the designs of all the
world's great long-span bridges built since 1940. - Wikipedia
http://www.youtube.com/watch?v=3mclp9QmCGs
Tacoma Narrows Bridge
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 57
Normal Mode Vibrations
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End of Chapter Problems
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Chap 14 - #92
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Extra Slides
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MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 61
∂ ∂
∂ ∂
2 2
2 2 2
y 1 y- = 0
x v t
y(t) = Asin(kx -ωt)
2π 2πk = ; ω= 2πf =
λ T
x ty(t) = Asin 2π -
λ T
x t- = Constantλ T
dx dt- = 0
λ T
dx = f = v
dt T
λλ=
The Full Wave Equation
Traveling with the wave the phase is constant
Wave velocity