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Math and Music Part II
Richard W. BeveridgeClatsop Community College
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Pythagorean Ratios
• The Pythagoreans knew that the tones produced by vibrating strings were related to the length of the string.
• They also knew that strings in lengths of small whole number ratios produced pleasing tones when played together –musical fourth, fifth and octave.
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Frequency
• The concept of vibrational frequency was considered to be related to the length of the string.
• Galileo focused on the concept of vibrational frequency as opposed to the ratio of lengths in determining musical pitch.
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Frequency
• “I say that the length of strings is not the direct and immediate reason behind the forms of musical intervals, nor is their tension, nor their thickness, but rather, the ratio of the numbers of vibrations and impact of air waves that go to strike our eardrum, which likewise vibrates according to the same measure of times.”
Galileo (1638)
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Frequency
• How do we determine the frequency of a vibrating string?
• By the early 1600’s Mersenne knew that the frequency and pitch of a vibrating string were related to:
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Frequency
l = length
F = tension
σ = cross sectional area
ρ = density
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Frequency
Using a pendulum analogy,
Joseph Sauveur and
Christiaan Huygens
determined the frequency to
be:
ρσυ Fl2
1≈
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Frequency
• Sauveur used “beats” and ratios to determine absolute frequency.
• Beats are a musical phenomenon well-known to musicians and used often in tuning instruments.
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Frequency
• One definition of beats is that they are “periodic fluctuations of loudness produced by the superposition of tones of close, but not identical frequencies.”
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Frequency
• The number of beats per second is actually equal to the difference in the absolute frequencies of the two tones.
• There are two ways to show that this is true. One uses algebra, the other trigonometry.
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Beats Per Second
• To show that the number of beats per second is equal to the difference between the frequencies of the tones, we must consider what is happening acoustically.
• One of the frequencies is higher than the other.
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Beats Per Second
• That means that the higher frequency will have more wavelengths per second than the lower frequency.
• The beats are a result of the two wavelengths coinciding to produce a momentarily louder tone.
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Beats Per Second
For instance, if the two tones
are 8 hertz (8 wavelengths per
second) and 6 hertz, then the
higher frequency wavelength
will have wave peaks at t=0,
81, 8
2 , 83, 8
4 , 85, 8
6, 87and t=1
second.
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Beats Per Second
The lower frequency
wavelength will have wave
peaks at t=0, 61, 6
2 , 63, 6
4 , 65 and
t=1 second.
So they will coincide at t=0,
t=21 and then t=1 and t=1.5
and so on or
twice each second.
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Beats Per Second
• In the example, when the 8 hertz wavelength had completed 4 waves, the 6 hertz wavelength had only completed 3 waves.
• When the higher frequency completes one more wave than the lower frequency, they will coincide and produce a beat.
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Beats Per Second
• Any mathematician knows that one example doesn’t prove anything, so let’s consider the idea in general.
• What is happening is that the higher frequency “laps” the lower frequency.
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Beats Per Second
So, given two frequencies 1f
and 2f , with 12 ff > . Then we
want to find how many
wavelengths it will take for 2f
to complete one more
wavelength than 1f .
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Beats Per Second
So we set
21
1f
NfN +=
If we cross-multiply we get
12 )1( fNNf +=
112 fNfNf +=
112 fNfNf =−
112 )( fffN =−
So,
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Beats Per Second
12
1
fffN−
=
In our example 1f was 6 and
2f was 8, so this value for N
would come out to
3 26
686 ==−
=N
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Beats Per Second
This is what we saw, the
wavelengths coincided on the
3rd wave of the 6 hertz
frequency and the 4th wave of
the 8 hertz frenquency.
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Beats Per Second
So the time period for the first
beat was 63 seconds or
1fN .
If 12
1
fffN−
= , then ⎟⎠⎞
⎜⎝⎛
⎟⎟⎟⎟⎟
⎠
⎞
⎜⎜⎜⎜⎜
⎝
⎛
−=
1
12
1
1 fff
f
fN
12112
1
1
11*fffff
ffN
−=
−=
⎟⎠⎞
⎜⎝⎛⎟
⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
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Beats Per Second
In the example, the beats
occurred every 21second, so
there were 2 beats per second.
In general, the beats occur
every 12
1ff −
seconds so there
are 12 ff − beats per second.
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Beats Per Second
• Showing this relationship using trigonometry uses the identity or statement of equality that:
sin(u)+sin(v) = 2sin(0.5(u+v))cos(0.5(u-v))
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Beats Per Second
• In the cos(0.5(u-v)), we see that the addition of two sound waves ends up being identical to a sound wave with a frequency equal to the difference of the waves and a variable amplitude.
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Beats Per Second
-0.5 0.5 1 1.5
-4-3-2-1
1234
x
y
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Frequency
• Sauveur used “beats” and ratios to determine absolute frequency.
• We’ve seen that beats can determine the difference of two frequencies.
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Frequency
• If we use the approximation for frequency developed by Sauveur and Huygens, we can see that the ratio of two frequencies is the inverse ratio of their lengths, assuming that they have equal tension, cross section and density.
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Frequency
ρσ
ρσF
F
ff
1
2
1
2
21
21
l
l=
1
2
1
2
21
21
l
l=ff
2
11
21
21
221
lll
l==f
f
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Calculating Frequency
• So, if we know the ratio of two frequencies and we know the difference of two frequencies, then we can calculate what the frequencies themselves are.
• This is one method of hand calculating the frequencies of the notes in the European equal tempered scale.
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Calculating Frequency
If we have two frequencies 1f and
2f , with:
dff =− 12
and
rff =1
2
Then we can calculate 1f and 2f
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Calculating Frequency
rff =1
2
Multiply through by 1f
12 * frf =
and divide by r
12 fr
f =
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Calculating Frequency
Then, isolate 2f from the equation
dff =− 12
So
12 fdf +=
And substitute into the other
equation, giving us:
12 fr
f =
11 fr
fd =+
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Calculating Frequency
Now we solve this equation
11 fr
fd =+
Multiply through by r
11 * frfd =+
And subtract 1f from both sides
11* ffrd −=
Next, we factor 1f out on the right
hand side...
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Calculating Frequency
Next, we factor 1f out on the right
hand side...
)1(*1 −= rfd
And divide through by 1−r , giving
us
11 frd =−
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Calculating Frequency
An example:
What if we had two strings – one
that was 111 cm. long and one that
was 110 cm. long ?
We know that the ratio of their
frequencies is the reciprocal of the
ratio of their length.
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Calculating Frequency
2
1
1
2ll
=ff
So,
110111
1
2 =ff
Imagine that we plucked each
string under equal tension and
counted 2 beats per second in the
sound.
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Calculating Frequency
In this example
110111
1
2 == ffr
and
212 =−= ffd
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Calculating Frequency
Remember that
11 frd =−
so that means
11
110111
2 f=−
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
or
1
110110
110111
2 f=−
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
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Calculating Frequency
Which means that
1
11012
1110*2220 f===
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛
This says that the lower frequency
sound in this example is 220 hertz
or 220 cycles per second.
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Calculating Frequency
This is the A note below middle C.
The higher frequency sound is 222
hertz, which is somewhere
between A and A#.
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The Wave Equation
The Wave Equation can be
derived from a consideration of
the vibrating string as a mass
attached to a spring.
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The Wave Equation
Hooke’s Law for springs states
that the “restoring” force of the
spring will be proportional to the
displacement from equilibrium.
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The Wave Equation
If we imagine the string to be a
series of evenly spaced weights
connected with springs, then
we can use the properties of
Hooke’s Law to determine the
behavior of a vibrating string.
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The Wave Equation
If Hooke’s Law says that
, where does kyF −=
][][ xhxxhx yykyykF −+−= −+
come from?
Isn’t the k supposed to be
negative?
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The Wave Equation
From this point, the derivation
of the wave equation involves
the use of Partial Differential
Equations – so I will be brief,
simplify what is happening
and leave out many steps in
between some of the more
complex transitions.
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The Wave Equation
From Newton, Force is defined
as mass*acceleration: maF = .
In Calculus, acceleration is the
second derivative of the
vertical displacement of the
weight with respect to time, so
2
2
*tymF x
∂∂=
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The Wave Equation
From Hooke’s Law
][][ xhxxhx yykyykF −+−= −+
If we set these two forces equal
to each other, we have
][][2
2
xhxxhxx yykyyk
tym −+−=
∂∂
−+
OR
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The Wave Equation
]2[2
2
hxxhxx yyyk
tym −+ +−=
∂∂
As the number of weights
increases and the distance
between them decreases, the
expression on the right-hand
side
]2[ hxxhx yyyk −+ +−
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The Wave Equation
]2[ hxxhx yyyk −+ +−
becomes the second derivative
of the vertical displacement
with respect to the position of
the weight along the horizontal
OR
2
2
*xyk x
∂∂
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The Wave Equation
So,
2
2
2
2
xyk
tym xx
∂∂=
∂∂
The Wave Equation is often
stated in the form
2
22
2
2
xyc
ty xx
∂∂=
∂∂
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The Wave Equation
The solution of this equation
involves interacting sine and
cosine waves whose behavior
depends on the initial
conditions that set the wave in
motion.
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The Wave Equation
The Scottish physicist
James Maxwell (1831-1879)
used the wave equation to
conclude that visible light is
part of the electromagnetic
spectrum.
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The Wave Equation
• Lasers
• X-Ray, RADAR, Radio Telescopes
• Radio, Television, Cell Phones
• Fluid Dynamics
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Harmonics
• The question of harmony confounded musicians and mathematicians alike for many years.
• Galileo believed that it was the simple integer ratios of the notes that produced harmony.
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Harmonics
• Galileo’s idea was that the sound waves would coincide at their common multiples creating a pleasing tone.
• This is similar to what actually does happen to create “beats.”
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Harmonics
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Harmonics
• While Galileo’s explanation makes sense, it turns out not to be true.
• It is the phenomena of overtones or harmonics that produce harmony.
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Harmonics
• While a tone may be vibrating with a root frequency, there are other frequencies present as well.
• Exactly which frequencies are present and how loud each one is determines the sound or timbre of each musical instrument.
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Harmonics
• Typically, wind and stringed instruments produce harmonic tones at integer multiples of the root or fundamental tone.
• So, in a sense, Galileo had the right idea, but was missing the bigger picture.
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Harmonics
• For instance, the A above middle C has a frequency of 440 cycles per second.
• The musical fifth to A, which is E, would have a frequency 1.5 times that of the A, or 660 cycles per second.
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Harmonics
• When the A is sounded on an instrument, the root tone of 440 hertz is heard, as well as the harmonic frequencies of 880 hertz, 1320 hertz, 1760 hertz, 2200 hertz, 2640 hertz and so on.
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Harmonics
• When the E is sounded on an instrument, the root tone of 660 hertz is heard, as well as the harmonic frequencies of 1320 hertz, 1980 hertz, 2640 hertz and so on.
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Harmonics
A: 440 880 1320 1760 2200 2640E: 660 1320 1980 2640 3300 3960
A: 3080 3520 3960 4400 4840 5280E: 4620 5280 5940 6600 7260 7920
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Harmonics
• So, because of the common multiples shared by the frequencies of the root tones, the harmonic tones resonate pleasingly with each other!
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Harmonics
• The following slide is a graph of the tone produced by a trumpet.
• The different pitches and corresponding volumes can be seen, as well as the dissipation of the tones over time
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Harmonics
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Harmonics
• The next slide shows just the sound wave produced by a clarinet.
• The different harmonic frequencies combining together produce the distinct shape.
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Harmonics
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• http://www.clatsopcc.edu/faculty/rbeveridge/Research/Papers_and_Presentations.htm