As the drawing shows, the length of a guitar string is 0.628 m. The frets are nu
ID: 1452604 • Letter: A
Question
As the drawing shows, the length of a guitar string is 0.628 m. The frets are numbered for convenience. A performer can play a musical scale on a single string, because the spacing between the frets is designed according to the following rule: When the string is pushed against any fret j, the fundamental frequency of the shortened string is larger by a factor of the twelfth root of two (122) than it is when the string is pushed against the fret j - 1.
Assuming that the tension in the string is the same for any note, find the spacing
(a) between fret 3 and fret 2
______ cm and
(b) between fret 8 and fret 7
______cm.
0.628 m 7654 3 21 0Explanation / Answer
The fundamental frequency of a guitar string is given by the formula
f = (1 / (2 * L)) * root (T / u)
where L is the length of the string that is vibrating, T is the tension and u is the density per unit length.
For this problem, we can express the frequency as
f = k / L
where k is a constant.
The frets are arranged to be a semitone apart. (There are 12 semitones in an octave.) When the frequency rises by 1 octave it doubles, which is why the factor of the twelfth root of 2 appears.
This factor = 1.059463. Call this factor b.
To shift up in frequency by 1 semitone when it is fingered to touch the first fret, its length will be
0.628 / 1.059463 = 0.628 / b
For the j th fret the length will be
0.628 / b^j
and for the j+1 fret it will be
0.628 / b^(j + 1)
The distance between the frets will be
(0.628 / b^j) - (0.628 / b^(j+1))
= 0.628 * (1/b^j - 1/b^(j+1))
= 0.628 * (b^(j+1) - b^j) / (b^j * (b^(j+1))
= 0.628 * b^j * (b - 1) / (b^j * (b^(j+1))
= 0.628 * (b - 1) / b^(j+1)
= 0.037343 / 1.059463^(j+1) m
= 3.7343 / 1.059463^(j+1) cm
a)
Between fret 3 and 2 (i.e. j = 3) the spacing is
3.7343 / 1.059463 ^ 4
= 2.96391693 cm
b)
Between fret 8 and 7 (j = 7) the spacing is
3.7343 / 1.059463^8
= 2.352463264 cm
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