Math and Music Part II. Richard W. Beveridge Clatsop Community College

Transcription

1 Math and Music Part II Richard W. Beveridge Clatsop Community College

2 Pythagorean Ratios The Pythagoreans knew that the tones produced by vibrating strings were related to the length o the string. They also knew that strings in lengths o small whole number ratios produced pleasing tones when played together musical ourth, ith and octave.

3 Frequency The concept o vibrational requency was considered to be related to the length o the string. Galileo ocused on the concept o vibrational requency as opposed to the ratio o lengths in determining musical pitch.

4 Frequency I say that the length o strings is not the direct and immediate reason behind the orms o musical intervals, nor is their tension, nor their thickness, but rather, the ratio o the numbers o vibrations and impact o air waves that go to strike our eardrum, which likewise vibrates according to the same measure o times. Galileo (638)

5 Frequency How do we determine the requency o a vibrating string? By the early 600 s Mersenne knew that the requency and pitch o a vibrating string were related to:

6 Frequency l = length F = tension σ = cross sectional area ρ = density

7 Frequency Using a pendulum analogy, Joseph Sauveur and Christiaan Huygens determined the requency to be: υ l F ρσ

8 Frequency Sauveur used beats and ratios to determine absolute requency. Beats are a musical phenomenon wellknown to musicians and used oten in tuning instruments.

9 Frequency One deinition o beats is that they are periodic luctuations o loudness produced by the superposition o tones o close, but not identical requencies.

10 Frequency The number o beats per second is actually equal to the dierence in the absolute requencies o the two tones. There are two ways to show that this is true. One uses algebra, the other trigonometry.

11 Beats Per Second To show that the number o beats per second is equal to the dierence between the requencies o the tones, we must consider what is happening acoustically. One o the requencies is higher than the other.

12 Beats Per Second That means that the higher requency will have more wavelengths per second than the lower requency. The beats are a result o the two wavelengths coinciding to produce a momentarily louder tone.

13 Beats Per Second For instance, i the two tones are 8 hertz (8 wavelengths per second) and 6 hertz, then the higher requency wavelength will have wave peaks at t=0, 8,, 3, 4, 5, 6, 7 and t= second.

14 Beats Per Second The lower requency wavelength will have wave peaks at t=0,,, 3, 4, 5 and t= second. So they will coincide at t=0, t= and then t= and t=.5 and so on or twice each second.

15 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 requency completes one more wave than the lower requency, they will coincide and produce a beat.

16 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 requency laps the lower requency.

17 Beats Per Second So, given two requencies and, with >. Then we want to ind how many wavelengths it will take or to complete one more wavelength than.

18 Beats Per Second So we set N = N + I we cross-multiply we get So, N ( + = N ) N + = N N = N N ( = )

19 Beats Per Second N = In our example was 6 and was 8, so this value or N would come out to N = = 6 = 3

20 Beats Per Second This is what we saw, the wavelengths coincided on the 3 rd wave o the 6 hertz requency and the 4 th wave o the 8 hertz renquency.

21 Beats Per Second So the time period or the irst beat was 6 3 seconds or N. I N =, then = N * N = =

22 Beats Per Second In the example, the beats occurred every second, so there were beats per second. In general, the beats occur every seconds so there are beats per second.

23 Beats Per Second Showing this relationship using trigonometry uses the identity or statement o equality that: sin(u)+sin(v) = sin(0.5(u+v))cos(0.5(u-v))

24 Beats Per Second In the cos(0.5(u-v)), we see that the addition o two sound waves ends up being identical to a sound wave with a requency equal to the dierence o the waves and a variable amplitude.

25 Beats Per Second 4 3 y x

26 Frequency Sauveur used beats and ratios to determine absolute requency. We ve seen that beats can determine the dierence o two requencies.

27 Frequency I we use the approximation or requency developed by Sauveur and Huygens, we can see that the ratio o two requencies is the inverse ratio o their lengths, assuming that they have equal tension, cross section and density.

28 Frequency = l l F ρσ F ρσ = l l = l l = l l

29 Calculating Frequency So, i we know the ratio o two requencies and we know the dierence o two requencies, then we can calculate what the requencies themselves are. This is one method o hand calculating the requencies o the notes in the European equal tempered scale.

30 Calculating Frequency I we have two requencies and, with: = d and = r Then we can calculate and

31 Calculating Frequency = r Multiply through by = r* and divide by r r =

32 Calculating Frequency Then, isolate rom the equation So = d = d + And substitute into the other equation, giving us: r d + r = =

33 Calculating Frequency Now we solve this equation d + r = Multiply through by r d + = r* And subtract rom both sides d = r* Next, we actor out on the right hand side...

34 Calculating Frequency Next, we actor out on the right hand side... d = *( r ) And divide through by r, giving us r d =

35 Calculating Frequency An example: What i we had two strings one that was cm. long and one that was 0 cm. long? We know that the ratio o their requencies is the reciprocal o the ratio o their length.

36 Calculating Frequency = So, = l l 0 Imagine that we plucked each string under equal tension and counted beats per second in the sound.

37 Calculating Frequency In this example r = = 0 and d = =

38 Calculating Frequency Remember that r d = so that means = 0 or = 0 0 0

39 Calculating Frequency Which means that 0= * 0 = = 0 This says that the lower requency sound in this example is 0 hertz or 0 cycles per second.

40 Calculating Frequency This is the A note below middle C. The higher requency sound is hertz, which is somewhere between A and A#.

41 The Wave Equation The Wave Equation can be derived rom a consideration o the vibrating string as a mass attached to a spring.

42 The Wave Equation Hooke s Law or springs states that the restoring orce o the spring will be proportional to the displacement rom equilibrium.

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45 The Wave Equation I we imagine the string to be a series o evenly spaced weights connected with springs, then we can use the properties o Hooke s Law to determine the behavior o a vibrating string.

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47 The Wave Equation I Hooke s Law says that F = ky, where does F = x+ h x h k[ y yx] + k[ y yx] come rom? Isn t the k supposed to be negative?

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49 The Wave Equation From this point, the derivation o the wave equation involves the use o Partial Dierential Equations so I will be brie, simpliy what is happening and leave out many steps in between some o the more complex transitions.

50 The Wave Equation From Newton, Force is deined as mass*acceleration: F = ma. In Calculus, acceleration is the second derivative o the vertical displacement o the weight with respect to time, so F = y m* x t

51 The Wave Equation From Hooke s Law ] [ ] [ x h x x h x y y k y y k F + = + I we set these two orces equal to each other, we have ] [ ] [ x h x x h x x y y k y y k t y m + = + OR

52 The Wave Equation y m t x = k[ y x + yx + h y x h ] As the number o weights increases and the distance between them decreases, the expression on the right-hand side k[ yx+ h yx + yx h]

53 The Wave Equation k[ yx+ h yx + yx h] becomes the second derivative o the vertical displacement with respect to the position o the weight along the horizontal OR k y * x x

54 The Wave Equation So, y m t x = k x y x The Wave Equation is oten stated in the orm t y x = c x y x

55 The Wave Equation The solution o this equation involves interacting sine and cosine waves whose behavior depends on the initial conditions that set the wave in motion.

56 The Wave Equation The Scottish physicist James Maxwell (83-879) used the wave equation to conclude that visible light is part o the electromagnetic spectrum.

57 The Wave Equation Lasers X-Ray, RADAR, Radio Telescopes Radio, Television, Cell Phones Fluid Dynamics

58 Harmonics The question o harmony conounded musicians and mathematicians alike or many years. Galileo believed that it was the simple integer ratios o the notes that produced harmony.

59 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.

60 Harmonics

61 Harmonics While Galileo s explanation makes sense, it turns out not to be true. It is the phenomena o overtones or harmonics that produce harmony.

62 Harmonics While a tone may be vibrating with a root requency, there are other requencies present as well. Exactly which requencies are present and how loud each one is determines the sound or timbre o each musical instrument.

63 Harmonics Typically, wind and stringed instruments produce harmonic tones at integer multiples o the root or undamental tone. So, in a sense, Galileo had the right idea, but was missing the bigger picture.

64 Harmonics For instance, the A above middle C has a requency o 440 cycles per second. The musical ith to A, which is E, would have a requency.5 times that o the A, or 660 cycles per second.

65 Harmonics When the A is sounded on an instrument, the root tone o 440 hertz is heard, as well as the harmonic requencies o 880 hertz, 30 hertz, 760 hertz, 00 hertz, 640 hertz and so on.

66 Harmonics When the E is sounded on an instrument, the root tone o 660 hertz is heard, as well as the harmonic requencies o 30 hertz, 980 hertz, 640 hertz and so on.

67 Harmonics A: E: A: E:

68 Harmonics So, because o the common multiples shared by the requencies o the root tones, the harmonic tones resonate pleasingly with each other!

69 Harmonics The ollowing slide is a graph o the tone produced by a trumpet. The dierent pitches and corresponding volumes can be seen, as well as the dissipation o the tones over time

70 Harmonics

71 Harmonics The next slide shows just the sound wave produced by a clarinet. The dierent harmonic requencies combining together produce the distinct shape.

72 Harmonics

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