As the drawing shows, the length of a guitar string is 0.628 m. The frets are nu

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... (I need help with A please)

(a) between fret 3 and fret 2.----> _______ cm and

(b) between fret 8 and fret 7. --> 2.352cm (I know this answer is correct for b)

Explanation / Answer

Here,
Length if string, l = 0.628 m

Solving for Fundamental frequency,
f = k / L ----------------------------(1)

where k is a constant.

Since 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, b = 1.059463 --------------------(2)

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 -------------------(3)

For the j th fret the length will be
0.628 / b^j ------------------------------(4)

and for the j+1 fret it will be
0.628 / b^(j + 1) --------------------------(5)

so distance between the frets will be , subtracting 3 and 4

(0.628 / b^j) - (0.628 / b^(j+1)) = s
s = 0.628 * (1/b^j - 1/b^(j+1))
s = 0.628 * (b - 1) / b^(j+1)

using eqn 2, 3, 4 we get,

s = 0.037343 / 1.059463^(j+1) m
s = 3.7343 / 1.059463^(j+1) cm ------------(6)

Part a ::::::::::::::::::
Between fret 3 and 2 (i.e. j = 3) the spacing is
From eqn 6

s = 3.7343 / 1.059463^(3+1)
s = 2.964 cm

Part b ::::::::::::::::::
Between fret 8 and 7 (j = 7) the spacing is
From eqn 6

s = 3.7343 / 1.059463^(7+1)
s = 2.352 cm

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