Text: 18) In the preceding section, the cause of beats due to two simple tones of nearly the same pitch was explained, and it was seen that the number of beats per second was always equal to the difference of the numbers of vibrations per second of each tone; so that as the interval between them increased so would the number of beats increase in a given time. Hence it is obvious that if the interval became sufficiently large, the beats would succeed each other so rapidly as to become undistinguished. For instance, in the case of the fifth whose lower and upper tones are produced by vibrations numbering 264 and 396 per second respectively, the number of beats per second would be 132 and would therefore be undistinguished ­ and still more so supposing the upper tone to have 397 or more vibrations per second; but, on the other hand it is a well-known fact, that if an imperfect fifth, octave, or any other tolerably simple interval is played on a violin or vioncello, the beats are most distinctly heard succeeding each other at perceptible intervals ­ whereas according to what was said above they should occur so rapidly as not to be heard at all. Two explanations of this phenomenon have been given, of which by far the most simple is due to Helmholtz ­ and which here follows. It appears that when the tones are simple and at a sufficiently large interval the beats should occur too rapidly to be heard, wheras when the interval is played on a violin they are easily distinguishable. The reason of this fact is that in the latter case the tones are no longer simple but compound ­ and the beats which are heard are not due to the fundamental tones themselves but arise from two of their harmonic components which are nearly unison. Suppose the ratio of the interval between the fundamental tones to be m/n, that is, let m/n, be the fraction, reduced to its lowest terms, which is formed by putting in the numerator the number of vibrations per second of the upper tone, and in the denominator those of the lower. Then it is plain that the nth harmonic component of the tone m, will be of the same pitch as the mth harmonic component of the tone n; for they will each have exactly mn vibrations per second. Now let M/N be the ratio, expressed in the same way, of another interval, nearly, but not quite, equal to m/n; then the nth harmonic component of M will have Nm vibrations per second, while the mth component of N will have Nm. Now since M/N is nearly equal to m/n, the difference between Mn and Nm will be a small number; and when the two notes are sounded together the number of beats per second will be equal to that difference. For example, let m/n, be the ratio of a fifth, that is the fraction 3/2, and let M/N represent very nearly the same interval, say 397/264; then the difference between Mn and Nm, or 794 and 792, is 2; hence if two strings tuned apart at an interval represented by 397/264 are sounded simultaneously there will be two beats heard per second.

See Also: BEATS, DIFFERENCE TONES, RATIOS

Source: 125