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Home , Fret, Fundamental frequency, Node (physics), Nut (string instrument)

April 11, 2008

Harmonics on the Jay P. Fillmore1

Prelude for the guitar are described by the location, in terms of , of a at which to touch the string. One usually starts with “natural” harmonics on an open string; this is followed by “artificial” harmonics on a fretted string. The most common is the having one node exactly twelve frets higher (midway between the or the stopped and the saddle)2. But, one also needs to know the locations of other nodes giving the same (or lower) harmonic, since one must not pluck the string at such a node. Three observations: there are many harmonics that are not covered by such descriptions; the locations of nodes are not always exactly at a fret; and harmonics do not always fall at pitches on the scale of . (We will see that third harmonics are always sharp by two cents and fifth harmonics are always flat by 14 cents.) This is not due to “errors”, but to the physics and mathematics underlying harmonics. Here we discuss how to determine: all harmonics, all nodes, and where harmonics fall on the tempered scale. The reader should look at the hands-on-guitar exercise at the end of this article. To this author it was a fascinating surprise. It should entice further reading. Now, it is important to realize that we need to consider just one string of general length. The location of harmonics does not depend on the length of the string, so our analysis may be applied to an open or a fretted string in any tuning. We begin with a familiar example3. g Example 1. a) On string , open, (sounding A2) the third harmonic has two nodes seven and nineteen frets from the nut - at frets VII and XIX. Touching the string at one of these nodes and plucking (but not at the other node) E4 .

1 Professor of Mathematics, Emeritus University of California at San Diego [email protected] 2 Celedonio Romero. The Celedonio Romero Method for the . Juan Orozo Corp., Woodmere, NY, 1990. p.60 ff 3 The reader is encouraged to carry out these examples on a guitar. - 2 -

h b) On string , depressed at fret V (sounding A2) the third harmonic has two nodes seven and nineteen frets from fret V - at fret XII and where fret XXIV would be. Touching the string at one of these nodes and plucking (but not at the other node) sounds E4 . c) On string g , depressed at fret II (sounding B2), the third harmonic has nodes seven and nineteen frets higher - at frets IX and XXI. Touching the string at one of the nodes and plucking (but not at the other node) sounds F4# . □ Our analysis makes use of two sets of coördinates along a string. The first is the usual distance measurement s from the nut or a designated fret. The second, “fret coördinate” z , is based on the tempered scale and starts with zero at the nut or the designated fret. Fret coördinates make use of the frets to locate positions along the string in a way easily visible to the player. The nodes of a harmonic are located at points equally spaced along a string. The main problems will be to express these location in terms of the frets, and to determine what pitches are heard. Only high-school physics and mathematics will be used here: of , and logarithms. Note that this discussion is for an “ideal string” whose motions are the standing solutions of the elementary wave equation. No account is taken regarding the stretching of a string as it is depressed, or the elastic modulus which gives “inharmonicities” not detectable on a guitar. The only fact that we will need from the physics of the motion is that the of of a string is inversely proportional to its length. (This is a consequence of: a string of length L has a fundamental mode (lowest) wave length 2L and frequency f = V/2L , where V is the transverse velocity of the string obtained as the square root of the tension divided by density.) Knowing how to locate all harmonics will later yield some very interesting facts. This begins with the next example which will be revisited twice later - as our understanding unfolds. Example 2A g . On string (sounding A2) play the fifth (natural) harmonic by touching a node at (just short of) fret IV or IX (there are two more nodes); the pitch will be C5# (but flat by f 14 cents). Depress string (sounding D3) at fret IV (which is F3#) and play the third (artificial) harmonic by touching the - 3 -

node at (just short of) fret XI ; the pitch will again be C5# (but sharp by 2 cents). □

The tempered scale4 – pitches We will use the scale of equal temperament with conventional pitch - A4 has frequency f = 440 . are divided into twelve intervals so that the ratios of successive 12 frequencies are 2 = 1.0594630943593. The frequency of a pitch is then expressible as

12 f = 440.( F 2) (Hz) , where F is an integer indicating the number of the pitch is above (for F positive) or below (for F negative) A4 (for which F = 0 ).

12 . 2 -48 = 27.5 Hz to The notes of a piano go from A0 : 440 ( ) 12 . 2 39 = 4186.009 Hz . This range comprises the C8 : 440 ( ) 39 – (-48) = 87 semitones (or seven and one quarter octaves) 12 between the 88 keys5 . -9 . Middle C is C4 : 440 ( 2) = 261.626 Hz. For the guitar, the pitches of most interest have F equal to D : -31 ; E : -29 ; A : -24 ; D : -19 ; G : -14 ; B : -10 ; E : -5 . 2 2 2 3 3 3 4 See Table 1 for a more extensive list. Each of the tempered scale is divided into 12 F 100 parts called cents. The formula f = 440.( 2) now makes sense for any value of F . The interval, in cents, between two pitches of frequencies f and f’ is given by

. f' . 1200 log2 f = 100 (F'- F) ;

log2 x = log x / log 2 = ln x / ln 2 . A quartertone is 50 cents. Most listeners cannot detect 2 or 3 cents.

4 T.D. Rossing, F.R. Moore, P.A. Wheeler. The Science of , 3rd ed., Addison Wesley, San Fancisco, 2002. p.181 ff 5 Octaves start at C and end at B : … B C D E F G A B C … 3 4 4 4 4 4 4 4 5 - 4 -

Fret coördinates – distance For convenience, we will think of the nut as being a “zero-th fret”. When we refer to the string from a designated fret to the saddle, we include the possibility of an open string. Let s denote the distance along a string of length L from a designated fret to the saddle. When a string is depressed at a fret located at a distance s from the designated fret, we have a string of length L – s . Let s’ be the distance from the designated fret to the fret next after the fret depressed. Since the frequency of the pitch is inversely proportional to string length, the string length L - s’ will be that of L - s shortened by the frequency ratio corresponding to a semitone: L - s L - s' = . Repeating this process starting with z = 0 12 2 and s = 0 , we have L L – s = for z = 0, 1, 2, 3, … and 0 < s < L . 12 ( 2)z This yields 1 s = L 1 - ( 12 ) ( 2)z for the distance of the z-th fret from the designated fret. From z L 12 this we have L - s = 2 , so

. L z = 12 log2 L - s . These formulas are, in fact, valid for any numbers 0 < z and 0 < s < L . Frets correspond to integer values of z . An increase of z by 1 corresponds to a semitone. These will be called the fret coördinates for positions on the string from the designated fret to the saddle. (A typical classical guitar has nineteen frets, but fret coördinates beyond there are still useful for locating position along a string.) These equations are the fundamental equations smoothly relating distance s and fret coördinate z .6

6 On a typical classical guitar of 65 cm scale length, the edges of the are located at 43.00 and 51.50 cm. These points are at fret coördinates z = 18.76 and 27.21. The center of the sound hole is at ½(43.00 + 51.50) = 47.25 cm for which z = 22.47 . Note that ½ (18.76 + 27.21) ≠ 22.47 . - 5 -

Aside. If s is obtained from z in the equation above, and s’ is obtained from z + 1 , then a little algebra gives

12 L - s 2 s'- s = with denominator = 17.817 . 12 12 2 2-1

12 2-1 This is the luthier’s Rule of Eighteen: This distance to the next fret is the distance to the saddle divided by eighteen.

Locating harmonics The N-th harmonic of a pitch having f has frequency N.f . The nodes for the N-th harmonic on a string of length L (from a designated fret to the saddle) divide the string into N equal intervals of length L/N . It will be convenient to number the nodes by n , and start nearest to the saddle. We have 1 < n < N , and n must have no integer factor in common with N . (Otherwise a lower order harmonic is n. obtained.) The distance from the saddle is N L , and the n . distance from the designated fret is s = (1 - N ) L . Then z L N 12 N L - s = n , so 2 = n . Consequently: The fret coördinates of . N the nodes are z = 12 log2 n , where N/n is a fraction greater than 1 with n and N having no common factor. Note that any of these nodes gives the same N-th harmonic. Example 3. Fourth harmonics are obtained by taking N = 4 , and n = 1 and 3 . Since log2(4/1) = 2 and log2(4/3) = 0.41504 , we obtain z = 24 and z = 4.98 , respectively. On an open string, the second is approximately at fret V , the first is exactly at fret XXIV. □ Now, for values of z which are not integers, it suffices to “linearly interpolate” their positions between the frets. So, 98% of the distance from fret IV to fret V will be called fret V . Fret XXIV is the same distance from the saddle as is fret V (4.98) from the nut. In general, for an artificial harmonic, the fourth harmonics are 4.98 and 24 frets higher (toward the saddle) from the designated fret where the string is depressed. - 6 -

The nodes for harmonics up to order ten are given in terms of fret coördinates in Table 2. Note that any of the nodes gives the same N-th harmonic. Integer values for z indicate a position exactly at a fret. For other values of z , one uses linear interpolation7 between adjacent frets.

The pitches that are heard It is important to realize that a harmonic does not always coincide with a pitch of the tempered scale – and can be expected to sound a little sharp or flat. So, we now address the problem of determining the pitch on the tempered scale nearest to a specified harmonic, and the separation. Example 4. One commonly begins tuning a guitar by matching: the h , and fourth harmonic of E2 located at fret V on open string g. the third harmonic of A2 located at fret VII on open string 12 The first has frequency 4 × 440.( 2)-29 = 329.628 Hz , which is × . 12 -24 E4 ; the second has frequency 3 440 ( 2) = 330 Hz , which . is above E4 by an interval of 1200 log2(330/329.628) = 1.96 cents8. □ The fundamental frequency of a string (from the designated 12 fret to the saddle) is given by 440.( 2)F , where F is the integer specifying the number of semitones from A4 of 440 Hz . 12 The N-th harmonic has frequency N.440.( 2)F . This will be of 12 x the form 440.( 2) , where x need not be an integer, but does specify the (fractional) number of semitones from A4 of 440 Hz . Let H be the integer nearest to x : x = H + h , -½ < h < ½ . 12 12 Then we have 440.( 2)F.N = 440.( 2)H+h or . F + 12 log2 N = H + h .

7 One can show that between two adjacent frets, the difference between the interpolated and actual fret coördinate is less than 0.00715 . 8 The pitch that is heard has frequency ½(329.628 + 330) = 329.814 Hz with beats of |329.628 – 330| = 0.372 Hz , or 1/0.372 = 2.69 seconds per . - 7 -

12 The separation of the harmonic 440.( 2)H+h and the nearest pitch 12 on the tempered scale 440.( 2)H is then

. . 12 H+h . 12 H 1200 log2(440 ( 2) /440 ( 2) ) = 100 h (cents). This will be negative when the harmonic is flat, and positive when it is sharp.

Example 5. The third harmonic of E2 is located at fret VII on open string h . Using F = -29 and N = 3 , we have . -29 + 12 log2 3 = -9.9804 = -10 + 0.0196 .

So H = -10 and h = 0.0196 . The pitch heard is B3 , but sharp by 100 × 0.0196 = 1.96 , or two cents. Open string d is B3 ; this is useful for preliminary tuning. □ Example 2B – an extension of Example 2A. From Table 2: Two of the four nodes for the fifth (natural) harmonic on string g are located at z = 3.86 (86% of the way from fret III to fret IV) and at z = 8.84 (84% of the way from fret VIII to fret IX). For g we have F = -24 . The fifth harmonic N = 5 A2 on string gives: . -24 + 12 log2 5 = 3.8631 = 4 – 0.1369 . × So, H = 4 which is C5# - flat by 100 0.1369 = 13.69 , or 14 cents. This one-seventh of a semitone flat is easily heard. □ The pitches of harmonics are relative to the fundamental pitch of the vibrating portion of a string between a designated fret and the saddle. It is useful to know these separations from the fundamental in terms of musical intervals. These are easily obtained from the fact that the N-th harmonic increases pitch . by 12 log2 N . . Example 6. 12 log2 3 = 19.0196 = 12 + 7 + 0.0196 shows third harmonics are always higher by one octave + perfect fifth + 2 cents (sharp) . while 12 log2 5 = 27.8631 = 24 + 4 – 0.1369 shows fifth harmonics are always higher by two octaves + major third - 14 cents (flat) . □ Table 3 gives such musical intervals for all harmonics up to the tenth. - 8 -

Harmonics near a given pitch

12 Finally, we wish to find harmonics N.440.( 2)F near to a 12 given pitch 440.( 2)F0 . Since the N-th harmonic increases the . pitch corresponding to F by 12 log2 N : . F + 12 log2 N = F0 + h , with -½ < h < ½ . For a specific N , one determines F so that the left-hand side is near the integer F0 . In the next example, we illustrate a seventh harmonic. Example 2C – the reverse of Examples 2A and 2B. A pitch near C5# has already appeared twice as a harmonic. In this example, we show how to locate a pitch having a seventh harmonic near . C5# . Begin with 12 log2 7 = 33.6883 , and use F0 = 4 to write this as . -30 + 12 log2 7 = 4 – 0.3117 .

This tells us that the seventh harmonic of D2# is C5# and is 31 cents flat. From -30 = - 31 + 1 , we find that D2# is on string h = D depressed at fret I . Also, . 12 log2 7 = 33.6883 = 24 + 10 – 0.3117 shows that a seventh harmonic is higher than the fundamental by two octaves + minor seventh - 31 cents (flat) . . The nodes are at 12 log2 7/n (n = 6,5,...,2,1) frets higher than the designated fret, fret I. We leave it to the reader to determine all pitches having harmonics near C5# . The complete calculation (with N = 2,3,4,...,10) appears in Table 4 . Beyond the seventh, harmonics near C5# require tunings too low to be of interest. The eighth and tenth harmonics are easy consequences of the fourth and fifth. The ninth harmonic is of B1 and is 4 cents sharp. □

- 9 -

Finale g to be We started with finding the fifth harmonic of A2 on near C5# . Now we have learned that C5# also appears as obvious second and fourth harmonics of C4# and C3# , somewhat less obviously as the third and sixth harmonics of F3# and F2# , and as well as a surprising seventh harmonic of D2# . All of these are readily played. Exercise. The next time you have your guitar tuned to h = D , find the seventh harmonic of D2# and compare it with C5# . There are six nodes: any one (only) of which must be touched, and the others of which must not be plucked. String h is depressed . at fret I . The node nearest to this is at 12 log2 7/6 = 2.67 frets higher, or at about 2/3 of the way from fret III to fret IV of the guitar. (On a guitar of 65 cm scale, these nodes are about 8.76 cm = 3.45 in apart – and very near nodes for other harmonics.) Note that this seventh harmonic sounds flat - as predicted.

Table 1 Values of F and f for various pitches of the tempered scale. Values in bold are used with the usual tunings of a guitar. Note Octave F f (Hz) f = 440*EXP((LN(2)/12)*F) C 0 -57 16.352 Lower threshold of hearing A 0 -48 27.500 Lowest note on a piano D 2 -31 73.416 String h open – tuned to D E 2 -29 82.407 String h open A 2 -24 110.000 g open D 3 -19 146.832 f open G 3 -14 195.998 e open B 3 -10 246.942 d open C 4 -9 261.626 Middle C E 4 -5 329.628 String c open 4 -1 415.305 A flat (approx. Renaissance A ) A 4 0 440.000 C 8 39 4186.009 Highest note on a piano C 10 63 16744.036 Upper threshold of hearing

- 10 -

Table 2 Nodes for harmonics to order ten in terms of fret coördinates. Any of the nodes gives the same N-th harmonic. N = 2 n 1 z 12 N = 3 n 2 1 z 7.02 19.02 N = 4 n 3 1 z 4.98 24 N = 5 n 4 3 2 1 z 3.86 8.84 15.86 27.86 N = 6 n 5 1 z 3.16 31.02 N = 7 n 6 5 4 3 2 1 z 2.67 5.83 9.69 14.67 21.69 33.69 N = 8 n 7 5 3 1 z 2.31 8.14 16.98 36 N = 9 n 8 7 5 4 2 1 z 2.04 4.35 10.18 14.04 26.04 38.04 N = 10 n 9 7 3 1 z 1.82 6.17 20.84 39.86 Integer values for z indicate a node exactly at a fret. For other values, use linear interpolation between adjacent frets.

Table 3 Separation of harmonics from the fundamental in terms of musical intervals. N = 2 . 12 log2 2 = 12 + 0 one octave exact N = 3 . 12 log2 3 = 12 + 7 + 0.0196 one octave + perfect fifth 2 cents sharp N = 4 . 12 log2 4 = 24 + 0 two octaves exact N = 5 . 12 log2 5 = 24 + 4 – 0.1369 two octaves + major third 14 cents flat N = 6 . 12 log2 6 = 24 + 7 + 0.0196 two octaves + perfect fifth 2 cents sharp N = 7 . 12 log2 7 = 24 + 10 – 0.3117 two octaves + minor seventh 31 cents flat N = 8 . 12 log2 8 = 36 + 0 three octaves exact N = 9 . 12 log2 9 = 36 + 2 + 0.0391 three octaves + 4 cents sharp N = 10 . 12 log2 10 = 36 + 4 – 0.1369 three octaves + major third 14 cents flat - 11 -

Table 4 Calculations accompanying Example 2C – the determination of pitches having harmonics near C5# (F0 = 4) . Pitch determined Node location from nut C # F + 12.log N = F + h String and Designated fret 5 2 0 designated fret + smallest from Table 2 Col 1 N = 2 12.log 2 = 12 -8 = -10 + 2 2 + 12 = 14 second 2 C # fret XIV . 4 harmonic -8 + 12 log2 2 = 4 + 0 d fret II of: C # exact 4 N = 3 12.log 3 = 19.0196 -15 = -19 + 4 4 + 7.02 = 11.02 third 2 F # fret XI . 3 harmonic -15 + 12 log2 3 = 4 + 0.0196 f fret IV of: F # 2 cents sharp 3 N = 4 12.log 4 = 24 -20 = -24 + 4 4 + 4.98 = 8.98 fourth 2 C # fret IX . 3 harmonic -20 + 12 log2 4 = 4 + 0 g fret IV of: C # exact 3 N = 5 12.log 5 = 27.8631 -24 = -24 + 0 0 + 3.86 = 3.86 fifth 2 A below fret IV . 2 harmonic -24 + 12 log2 5 = 4 – 0.1369 g open of: A 14 cents flat 2 N = 6 12.log 6 = 31.0196 -27 = -29 + 2 2 + 3.16 = 5.16 sixth 2 F # above fret V . 2 harmonic -27 + 12 log2 6 = 4 + 0.0196 h = E fret II of: F # 2 cents sharp 2 N = 7 12.log 7 = 33.6883 -30 = -31 + 1 1 + 2.76 = 3.76 seventh 2 D # below fret IV . 2 harmonic -30 + 12 log2 7 = 4 – 0.3117 h = D fret I of: D # 31 cents flat 2 N = 8 12.log 8 = 36 -36 = -36 + 0 0 + 2.31 = 2.31 eighth 2 C # above fret II . 2 harmonic -32 + 12 log2 8 = 4 + 0 h = C # open of: C # exact 2 2 N = 9 12.log 9 = 38.0391 -34 = -34 + 0 0 + 2.04 = 2.04 ninth 2 B # fret II . 1 harmonic -34 + 12 log2 9 = 4 + 0.0391 h = B # open of: B 4 cents sharp 1 1 N = 10 12.log 10 = 39.8631 -36 = -36 + 0 0 + 1.82 = 1.82 tenth 2 A # below fret II . 1 harmonic -36 + 12 log2 10 = 4 – 0.1369 h = A # open of: A 14 cents flat 1 1