Question

The fist (thin) string in a guitar should be two octaves higher than the sixth (thick) string, that is, its fundamental frequency is four times higher. If both strings are equal in length and under the same tension, what would the ratio of their linear mass densities be?

Answer 1

the fundamental frequency of the first string is 4 times higher than that of the sixth string,

so we can say that f1 = 4f6

now, we know that f = 1/2L (T/)

where L is the length of the string, T is the tension and is the linear mass density

so,

f1 =1/2L (T/1)

f6 = 1/2L (T/6)

therefore

f1/f6 = (6/1)

4 = (6/1)

1/6 = 1/16