Math and Music Part II

Richard W. BeveridgeClatsop Community College

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.

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.

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)

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:

Frequency

l = length

F = tension

σ = cross sectional area

ρ = density

Frequency

Using a pendulum analogy,

Joseph Sauveur and

Christiaan Huygens

determined the frequency to

be:

ρσυ Fl2

1≈

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.

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.”

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.

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.

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.

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.

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.

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.

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.

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 .

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,

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

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.

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

−=

−=

⎟⎠⎞

⎜⎝⎛⎟

⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

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.

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))

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.

Beats Per Second

-0.5 0.5 1 1.5

-4-3-2-1

1234

x

y

Frequency

• Sauveur used “beats” and ratios to determine absolute frequency.

• We’ve seen that beats can determine the difference of two frequencies.

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.

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

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.

Calculating Frequency

If we have two frequencies 1f and

2f , with:

dff =− 12

and

rff =1

2

Then we can calculate 1f and 2f

Calculating Frequency

rff =1

2

Multiply through by 1f

12 * frf =

and divide by r

12 fr

f =

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 =+

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...

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 =−

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.

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.

Calculating Frequency

In this example

110111

1

2 == ffr

and

212 =−= ffd

Calculating Frequency

Remember that

11 frd =−

so that means

11

110111

2 f=−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

or

1

110110

110111

2 f=−

⎟⎟⎟

⎠

⎞

⎜⎜⎜

⎝

⎛

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.

Calculating Frequency

This is the A note below middle C.

The higher frequency sound is 222

hertz, which is somewhere

between A and A#.

The Wave Equation

The Wave Equation can be

derived from a consideration of

the vibrating string as a mass

attached to a spring.

The Wave Equation

Hooke’s Law for springs states

that the “restoring” force of the

spring will be proportional to the

displacement from equilibrium.

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.

The Wave Equation

If Hooke’s Law says that

, where does kyF −=

][][ xhxxhx yykyykF −+−= −+

come from?

Isn’t the k supposed to be

negative?

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.

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

∂∂=

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

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 −+ +−

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

∂∂

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

∂∂=

∂∂

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.

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.

The Wave Equation

• Lasers

• X-Ray, RADAR, Radio Telescopes

• Radio, Television, Cell Phones

• Fluid Dynamics

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.

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.”

Harmonics

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.

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.

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.

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.

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.

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.

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

Harmonics

• So, because of the common multiples shared by the frequencies of the root tones, the harmonic tones resonate pleasingly with each other!

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

Harmonics

Harmonics

• The next slide shows just the sound wave produced by a clarinet.

• The different harmonic frequencies combining together produce the distinct shape.

Harmonics

• http://www.clatsopcc.edu/faculty/rbeveridge/Research/Papers_and_Presentations.htm