Applied Physics
Introduction to Vibrations and Waves
(with a focus on elastic waves)
Course Outline
Simple Harmonic Motion
&x&+ω 2 x = 0
ω = k /m
“k = elastic property
of the oscillator
Damped Oscillators
Driven Oscillators
Coupled Oscillators
Normal Modes
Waves and the wave equation
Fourier Analysis
(introduction)
Optical waves
Multiple source interference
diffraction
Elastic properties of materials
Stretching, bending, twisting
Shock absorbers, resonances
in mechanical systems
Natural vibrational
frequencies
Vibrations of a solid
Reflecting waves
Ultrasonic waves/testing
Breakdown of waves
into their components
Thin film interference
techniques
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Week 6 Lecture 3:
problems 41, 42, 43
(French pages 119-127, courseware pg 47-57)
Coupled Oscillations: introduction
→ (this and the next two pages are an introduction)
→ So far we have been concerned with SHM.
→ Characterized by a single natural (resonance)
frequency:
k
ω0 =
m
→ However, most real-life systems resonate at many
frequencies!
WHY?
→ Because most real objects can be viewed as a
number of simple oscillators coupled together.
→ We want to examine coupled oscillators to lead into
a study of wave motion, which is essentially an
infinite number of oscillators (the atoms) coupled
together.
2
The Plan over the next few lectures…
1. Start with a very simple system – 2 coupled simple
pendulums – just to introduce the topic.
2. Move on to a coupled system of 2 masses on a light
string.
3. Expand this to N coupled masses on a light string.
4. Finally look at the string as a continuum.
Development of the wave equation!!
For each case we will look at:
→ general behaviour
→ resonance (normal mode) behaviour
3
Coupled oscillation equations can be
used for some very cool modelling!!
4
Coupled Oscillations
We will examine:
→ General Behaviour
→ Normal Mode Behaviour (a special case of general
behaviour) - this is resonance!!!
→ A normal mode (resonance) occurs when all
particles are oscillating with the same
frequency. This will be a natural or resonance
frequency for the system. There is more than
one resonance frequency in a coupled
oscillating system.
Resonance
Natural
Normal mode
=
= frequency
frequency
frequency
Example: Two Coupled, Simple Pendulums (IP demo 9)
→ General Behaviour:
→ Pull A out and let go – A starts oscillating, then B –
gets larger until A stops, then movement back to A.
→ Energy shuttles back and forth, and in this case the
particles have different ω’s.
B
A
Coupled simple pendulums
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Normal Mode behaviour (a special case of general behaviour)
There are two normal modes (NMs) for this system.
In each of the 2 normal modes both masses have the same
frequency (the resonance frequency) but the frequency of
NM1 ≠ NM2.
(note for this specific case the amplitudes are the same for both
particles for each NM – but this is not generally the case).
Normal Mode Equations for these coupled pendulums
(check out IP9 to see these in action!):
Gravity but no spring
contribution for both
particles:
→ NM1
B
x = A0 cos ωt
A
→ NM2
B
A
g
l
ω=
Spring and gravity
contribution for both
particles:
ω=
x A = A0 cos ωt
x B = − A0 cos ωt
g 2k
+
l m
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The General behaviour equations for the
two coupled pendulums:
get
g
g 2k ⎞
1 ⎛
x A = A0 ⎜⎜ cos t + cos
+ t ⎟⎟
l
l m ⎠
2 ⎝
g 2k ⎞
1 ⎛
g
+ t ⎟⎟
x B = A0 ⎜⎜ cos t − cos
l m ⎠
2 ⎝
l
These look messy initially but when you look
closely they are just basically combinations of the
2 normal modes!
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Deriving the Normal mode frequency:
Case of a massless string with 2 point masses on it:
vibrating and at a snapshot of some instant during motion
This is a good illustration of the method you will use
frequently to solve all coupled oscillation problems. I
m2
θ2
m1 = m2 = m
θ3
T
m1
θ1
T
a
x2
T
θ2
T
x1
a
a
Assumptions:
1. x1 and x2 are small compared to the string length
(allows small angle approximation)
2. T (tension) does not change with these small
displacements
3. ignore gravity
What are the normal mode frequencies
(resonance frequencies) for this system?
We need to develop equations of motion for each of
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the two particles, starting with Newton’s 2nd law for each.
Then we can solve for the normal mode frequencies.
Apply Newton’s 2nd law (laterally)
mass 1:
so
∑ F = T cosθ1 − T cosθ 2
T −T = 0
+ve
(small angle approx:
cos θ = 1)
∴ no lateral acceleration
(same for m2)
Apply Newton’s 2nd law (vertically)
mass 1:
mass 2:
d 2 x1
m 2 = T sin θ 2 − T sin θ1
dt
d 2 x2
m 2 = −T sin θ 2 − T sin θ 3
dt
small angle
approximations
sub these into
1
+ve
1
2
x1
a
x − x1
sin θ 2 ≈ tan θ 2 = 2
a
x
sin θ 3 ≈ tan θ 3 = 2
a
sin θ1 ≈ tan θ1 =
d 2 x1
x1
⎛ x2 − x1 ⎞
m 2 = T⎜
⎟−T
a
dt
⎝ a ⎠
multiply through T
and rearrange
Equation of
motion for
mass 1
d 2 x1
Tx1 Tx2
m 2 +2
−
=0
a
a
dt
1a
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d 2 x2
x2
⎛ x − x1 ⎞
m 2 = −T ⎜ 2
T
−
⎟
a
a
dt
⎠
⎝
subbing small angle
approximations into
2
Equation of
motion for
mass 2
d 2 x2
Tx2 Tx1
−
=0
m 2 +2
a
a
dt
2a
Solving these equations for the normal mode conditions…
*** For normal modes, both m1 and m2 have the same frequency ω.***
assume solutions are:
sub both into 1a
x1 = A cos ωt
x 2 = B cos ωt
- - same ω for both masses
---we need to get A, B and ω
− mAω 2 cos ωt +
2TA cos ωt TB cos ωt
−
=0
a
a
2TA TB
− mω A +
−
=0
a
a
2TB TA
− mω 2 B +
−
=0
a
a
2
sub both into 2a
1b
2b
Solve these simultaneously to find an expression for ω
set each equation =
A/B and then equate in
order to eliminate (A/B)
2
2
⎡ − mω 2 + 2T ⎤ = ⎡ T ⎤
or
⎢⎣
a ⎥⎦
⎣⎢ a ⎥⎦
− mω 2 +
T
2T
=±
a
a
T
3T
ω =
or
ma ma
2
1st normal
mode
ω1
frequency
=
T
ma
2nd normal
mode
ω2
frequency
3T
=
ma
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For Normal Mode 1 what are the A and B Values?
→ subbing ω1 into either 1b or 2b gives A = B
→ ∴ the amplitude is the same for both.
so
T
ω1 =
ma
A=B
so x1 = x2 at any point
For Normal Mode 2 what are the A and B values?
→ subbing ω2 into either 1b or 2b gives A = -B
→ same amplitude but antiphase.
so
3T
ω2 =
ma
A = -B
so x1 = -x2 at any point
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What about three particles?
→ 3 particles exhibit 3 normal modes (resonances)
NM1
NM2
Note here that the
middle particle has
the same frequency
as the other two but
no amplitude
NM3
See IP Demo 8
note similarities between the behaviour of the masses in IP8 and the
masses on a string on the overhead.
In General:
→ For ID motion of a system of N particles we have N
normal modes and N corresponding normal mode
(resonance) frequencies. General behaviour
involves a combination of normal mode solutions.
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→ note: 2D systems are 2N and 3D systems are 3N.
Solving for the General State of Motion
(still two particles on a string)
Develop ∑ F = ma equations Æ already have 1a and 2a
Solve simultaneously:
2T
( x1 + x2 ) − T ( x1 + x2 ) = 0 3
m( &x&1 + &x&2 ) +
add 1a and 2a
a
a
2T
subtract 1a and 2a m ( &x&1 − &x&2 ) +
( x1 − x2 ) − T ( x1 − x2 ) = 0 4
a
a
3 becomes
(&x&1 + &x&2 ) + T ( x1 + x2 ) = 0
ma
4 becomes
(&x&1 − &x&2 ) + 3T ( x1 − x2 ) = 0
ma
→
→
→
→
k
&
&
x
+
x=0
This is simply in the form
m
→ the standard SHM equation with ω1 =
Solutions are:
T
3T
and ω 2 =
ma
ma
( x1 + x2 ) = C cos T t
NM 1
freq
NM 2
freq
ma
( x1 − x2 ) = D cos 3T t
ma
Combining these gives – for general state of motion
to combine just
solve for x1,
plug this into
2nd equation.
Then do same
for x2
x1 =
D
C
T
3T
t
t + cos
cos
ma
ma
2
2
x2 =
D
C
T
3T
t
t − cos
cos
ma
ma
2
2
13
(normally don’t worry about solving for C and D)
Summary: Coupled Oscillations
Two Types of Behaviour:
1. General Behaviour
→ particles shuttle energy back and forth
→ particles have different frequencies
note that you will not be required to solve problems involving general
behaviour but you will be expected to understand how the general
behaviour derivation works
2. Normal Mode Behaviour (special case of general
behaviour) all problems are based on this!
→ resonance!
→ all particles have the same frequency
(characterizing feature!)
→ particles may or may not have the same amplitude
→ The number of normal modes (resonance
frequencies) is equal to the number of particles in a
1D system!
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Methods for Solving Coupled Oscillation Problems
→ Solving for Normal Mode Conditions – 2 or more
particles (need to know for exam)
→ Write down Newton’s 2nd law equation for each
particle
→ Assume solutions
x1 = Acosωt
(same frequency for each particle
because in normal mode)
x2 = Bcosωt
x3 = Ccosωt……
→ Substitute these solutions back into the Newton’s 2nd
law equations and solve simultaneously to ωs for
each normal mode
→ Sub ω values back into earlier equations to get
values for amplitudes A, B, C,…
→ Solving for a Description of General Motion (gets
complicated for > 2 masses) (don’t need to know for
exam)
→ Write down the Newton’s 2nd law equation for each
particle (same as for normal mode procedure)
→ Add and subtract these solutions (solve
simultaneously) – this will give you expressions for
(x1 + x2) and (x1 – x2)
→ Solve for x1 and x2
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Solving for Normal Modes – masses and springs
Tutorial problem 41 – find normal modes and amplitudes
(note the
setup shown here applies to all of these types of problems)
k same for all springs
m1 = 2M
m2 = M
k
k
arbitrary shift in each
mass (I always draw
shift in same direction
with one larger than the
other)
At
equil.
m2
m1
HINT: Draw an
k
x2
x1
moving
m1
m2
+x
Newton’s 2nd Law:
for mass
1
for mass
2
m1 &x&1 = −kx1 + k ( x2 − x1 ) = − x1 (2k ) + kx2
m2 &x&2 = − k ( x 2 ) − k ( x 2 − x1 ) = kx1 − 2kx 2
&x&1 = −
m1= 2M
m2 = M
get
k
k
x1 +
x2
M
2M
k
2k
&x&2 =
x1 −
x2
M
M
both masses have
x1 = A1cos ωt
the same ωs in
x2 = A2cos ωt
Normal Modes
→ plug back into differential equations
→ solve simultaneously
→ get ω1 and ω2 then get A1 and A2 by plugging these
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back in
Then Æ solutions are:
Solving for Normal Modes – masses and springs
Tutorial problem 41 – find normal modes and amplitudes
(note the
setup shown here applies to all of these types of problems)
k same for all springs
m1 = 2M
m2 = M
k
k
arbitrary shift in each
mass (I always draw
shift in same direction
with one larger than the
other)
At
equil.
m2
m1
HINT: Draw an
k
x2
x1
moving
m1
m2
+x
Newton’s 2nd Law:
for mass
1
for mass
2
m1 &x&1 = −kx1 + k ( x2 − x1 ) = − x1 (2k ) + kx2
m2 &x&2 = − k ( x 2 ) − k ( x 2 − x1 ) = kx1 − 2kx 2
&x&1 = −
m1= 2M
m2 = M
get
k
k
x1 +
x2
M
2M
k
2k
&x&2 =
x1 −
x2
M
M
both masses have
x1 = A1cos ωt
the same ωs in
x2 = A2cos ωt
Normal Modes
→ plug back into differential equations
→ solve simultaneously
→ get ω1 and ω2 then get A1 and A2 by plugging these
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back in
Then Æ solutions are:
‘N’ Coupled Oscillators – Normal Modes only
(valid only for situations where all particles and springs
are identical and pinned at each end)
(see proof of these in French pg 139-141)
Terminology:
→ n = normal mode value (e.g. normal mode 10 is n=10)
→ N = total number of coupled oscillators
→ p = the number (along the line) of an individual oscillator
nπ ⎤ = frequency of normal mode “n”
⎥
⎣ 2( N + 1)⎦
⎡
ω n = 2ω 0 sin ⎢
natural
frequency
in SHM
=
=
k
for spring/mass
m
T
for transverse mass/spring vibration
ml
l
Apn
pnπ ⎞
= Cn sin⎛⎜
⎟
N
+
1
⎠
⎝
excitation amplitude of
normal mode n
y pn (t ) = Apn cos ω n t =
= amplitude of the pth oscillator
when the system is oscillating
in the nth normal mode
2
1
Cn
A1n
3
A2n
A3n
the displacement of the pth particle
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when the collection of particles is
oscillating in the nth normal mode
Ex. Look at normal mode frequency for 4 particles in
coupled oscillation.
For N = 4 using
NM1
NM2
NM3
NM4
n=1
n=2
n=3
n=4
⎡ nπ ⎤
ω n = 2ω 0 sin ⎢
⎥
(
)
2
+
1
N
⎦
⎣
ω1 =
ω2 =
ω3 =
ω4
2ω0sin[π/10] = 0.618ω0
2ω0sin[2π/10] = 1.176ω0
2ω0sin[3π/10] = 1.618ω0
= ω1
= 1.9ω1
= 2.6ω1
= 3.07ω1
For those of you that remember high school wave stuff, this might look
strange, since for waves the higher order frequencies are supposed to
be integral multiples of the lowest one…. Except these are not waves!!!
But… can we make them simulate a wave?
Very high N!!
Consider the ωn equation above, but for a huge N value
nπ
⎡ nπ ⎤
sin ⎢
≈
⎥
⎣ 2( N + 1)⎦ 2( N + 1)
Small angle
approx.
So we get:
2ω 0 nπ
nω 0π
ωn =
=
2( N + 1) ( N + 1)
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Now look at the first 4 normal modes again (next page)
∴
→
→
=
(1)ω 0π
(N + 1)
n=1
ω1
n=2
ω2
n=3
ω3
= 3ω 1
n=4
ω4
= 4ω 1
=
2ω 0π
(N + 1)
= 2ω 1
When N is large and you consider only the first few normal
modes (n is small) then the normal mode frequencies are
integral multiples of ω1 Æ approximate wave motion!
Final Note: If N is very large then what stops
from being very small?
well, if one considers
ω0 = ⎛⎜
1
⎞2
T
⎟
⎝ ml ⎠
nω 0π
ωn =
(N + 1)
then as N gets bigger the
“l“ term (distance between particles) gets smaller. So ω0
increases as N increases!
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