Question

When a player's finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string's fundamental frequency. The string's tension and mass per unit length remain unchanged.

If the unfingered length of the string is l=65cm, determine the positions x of the first six frets, if each fret raises the pitch of the fundamental by one musical note in comparison to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is 21/12
x1=
x2=
x3=
x4=
x5=
x6=

Answer 1

Step-by-step explanation:

For frequencies n generated in a string , the expression is as follows

n = 1 /2L√ ( T/m )

n is fundamental frequency , T is tension in string , m is mass per unit length and L is length of string.

If T and m are constant , then

n x L = constant , hence n is inversely proportional to L or length of string.

Frequencies increase by 21/12 = 1.75 , length must decrease by 1 / 1.75 times

Initial length of string is 65 cm .

x1 = 65 x 1 / 1.75 = 37.14 cm

x2 = 37.14 x 1/ 1.75 = 21.22 cm

x3 = 21.22 x 1 / 1.75 = 12.12 cm

x4= 12.12 x 1 / 1.75 = 6.92 cm

x5 = 6.92 x 1 / 1.75 = 3.95 cm

x6 = 3.95 x 1 / 1.75 = 2.25 cm