The Physics
Hypertextbook
Opus in profectus

# Music and Noise

## Practice

### practice problem 1

According to Galileo…

Especially harsh is the dissonance between notes whose frequencies are incommensurable; such a case occurs when one has two strings in unison and sounds one of them open, together with a part of the other which bears the same ratio as the side of a square bears to the diagonal….

Galileo Galilei, 1638

Identify this interval on the equal tempered scale.

#### solution

The ratio of the diagonal of a square to a side is √2:1. (Galileo stated the order of the ratio the other way around, but that's a minor detail.) Each half step (a semitone) up the equal tempered scale multiplies the previous note by the twelfth root of two, two half steps (a whole tone) multiplies the note by the twelfth root of two squared, three half steps by the twelfth root of two cubed, and so on…

The first six semitones of the equal tempered scale
interval name interval size
1 semitone   12√2 =  12√2
2 semitones  = whole tone 12√212√2 =  6√2
3 semitones   12√212√212√2 =  4√2
4 semitones  = ditone 12√212√212√212√2 =  3√2
5 semitones   12√212√212√212√212√2 =  2.4√2
6 semitones  = tritone 12√212√212√212√212√212√2 =  2√2

Six semitones is equal to the twelfth root of two to the sixth power, which is equal to the square root of two. This interval is called a tritone, an augmented fourth, or a diminshed fifth — C and F♯, for example, or F and B. Had I given you a more complete quote from Galileo you would have already known this.

Especially harsh is the dissonance between notes whose frequencies are incommensurable; such a case occurs when one has two strings in unison and sounds one of them open, together with a part of the other which bears the same ratio as the side of a square bears to the diagonal. This yields a dissonance similar to the augmented fourth or diminished fifth.

Galileo Galilei, 1638

### practice problem 2

Determine the beat frequency between C4 and G4 (a perfect fifth) when played on an equal tempered scale where A4 = 440 Hz.

#### solution

On an equal tempered scale, C is 9 semitones below A and G is 2 semitones below A.

 ƒC = ƒA/29/12 = 261.6256 HzƒG = ƒA/22/12 = 391.9954 Hz

A perfect fifth is the ratio 3:2. Therefore three times the tonic should equal twice the fifth, but on an equal tempered scale they don't. The difference in these multiples results in a beat in some overtone pairs.

 ƒbeat = 3ƒC − 2ƒGƒbeat = 3ƒA/29/12 − 2ƒA/22/12ƒbeat = 3(440 Hz)/29/12 − 2(440 Hz)/22/12ƒbeat = 784.877 Hz − 783.991 Hzƒbeat =  0.886 Hz

On a properly tuned piano, when C and A were played, one would expect overtone beats every…

Tbeat = ƒbeat−1 = 1.1 s

### practice problem 3

1. A sawtooth wave "contains odd and even harmonics that fall off at −6 dB/octave."

y = sin x − 12 sin 2x + 13 sin 3x − 14 sin 4x +…

 y = ∑ (− 1)n +1 sin nx n
2. A square wave "contains odd harmonics that fall off at −6 dB/octave."

y = sin x + 13 sin 3x + 15 sin 5x + 17 sin 7x +…

 y = ∑ 1 sin(2n − 1)x 2n − 1
3. A triangle wave "contains odd harmonics that fall off at −12 dB/octave."

y = cos x + 19 cos 3x + 125 cos 5x + 149 cos 7x +…

 y = ∑ 1 cos(2n − 1)x (2n − 1)2

#### solution

The Fourier series equations given above use y and x as variables. When sound waves are received by an ear or a microphone, they're detecting fluctuations in pressure (P) over time (t). This means we should use the pressure level equation in decibels.

 LP = 20 log ⎛⎜⎝ ∆P ⎞⎟⎠ ∆P0
1. The second term in the sawtooth wave series has twice the frequency of the first. Plug the ratio of the two amplitudes into the equation derived above.

 LP = 20 log ⎛⎜⎝ 12 ⎞⎟⎠ = −6.0206 dB 11

We could repeat this for other harmonics, but we'd always wind up with the same answer (rounded to one significant digit) of −6 dB per octave.

 LP = 20 log ⎛⎜⎝ 14 ⎞⎟⎠ = −6.0206 dB 12 LP = 20 log ⎛⎜⎝ 16 ⎞⎟⎠ = −6.0206 dB 13 LP = 20 log ⎛⎜⎝ 18 ⎞⎟⎠ = −6.0206 dB 14 LP = 20 log ⎛⎜⎝ 110 ⎞⎟⎠ = −6.0206 dB 15

We're done with this part.

2. No term in the square wave series has a frequency that is a whole number of octaves above another. Pick the amplitudes on the terms that are slightly larger than an octave apart. For example 3 is slightly greater than 2 × 1 = 2, 7 is slightly greater than 2 × 3 = 6, etc. Check the limit of the level difference for these values.

 LP = 20 log ⎛⎜⎝ 13 ⎞⎟⎠ = −9.54 dB 11 LP = 20 log ⎛⎜⎝ 17 ⎞⎟⎠ = −7.34 dB 13 LP = 20 log ⎛⎜⎝ 111 ⎞⎟⎠ = −6.85 dB 15 LP = 20 log ⎛⎜⎝ 115 ⎞⎟⎠ = −6.62 dB 17 LP = 20 log ⎛⎜⎝ 119 ⎞⎟⎠ = −6.49 dB 19 LP = 20 log ⎛⎜⎝ 123 ⎞⎟⎠ = −6.41 dB 111

This sequence is converging too slowly. Let's try some bigger numbers.

 LP = 20 log ⎛⎜⎝ 1223 ⎞⎟⎠ = −6.0596 dB 1111 LP = 20 log ⎛⎜⎝ 12,223 ⎞⎟⎠ = −6.0245 dB 11,111 LP = 20 log ⎛⎜⎝ 122,223 ⎞⎟⎠ = −6.0210 dB 111111 LP = 20 log ⎛⎜⎝ 1222,223 ⎞⎟⎠ = −6.0206 dB 1111,111

I see our limit approaching. It's the same as the result we got with the sawtooth wave. The harmonics fall off by about −6 dB per octave.

3. The situation for the triangle wave is the same as the square wave. We only have odd frequencies, but this time the amplitudes are squared.

 LP = 20 log ⎛⎜⎝ 19 ⎞⎟⎠ = −19.08 dB 11 LP = 20 log ⎛⎜⎝ 149 ⎞⎟⎠ = −14.72 dB 19 LP = 20 log ⎛⎜⎝ 1121 ⎞⎟⎠ = −13.70 dB 125 LP = 20 log ⎛⎜⎝ 1225 ⎞⎟⎠ = −13.24 dB 149 LP = 20 log ⎛⎜⎝ 1361 ⎞⎟⎠ = −12.98 dB 181 LP = 20 log ⎛⎜⎝ 1529 ⎞⎟⎠ = −12.81 dB 1121

This is also converging too slowly for me. Use larger values — like the square of those we used in part b.

 LP = 20 log ⎛⎜⎝ 149,729 ⎞⎟⎠ = −12.12 dB 112,321 LP = 20 log ⎛⎜⎝ 14,941,729 ⎞⎟⎠ = −12.05 dB 11,234,321 LP = 20 log ⎛⎜⎝ 1493,861,729 ⎞⎟⎠ = −12.04 dB 1123,454,321 LP = 20 log ⎛⎜⎝ 149,383,061,729 ⎞⎟⎠ = −12.04 dB 112,345,654,321

This is twice the power level drop of the sawtooth wave, but of course you already knew this. Squaring a quantity in a logarithm is the same as doubling the logarithm of the unsquared quantity. All of the calculations in this last part are interesting, but unnecessary. I guess I just didn't feel like being clever today.

### practice problem 4

Write something completely different.