The 6 strings on a guitar all have about the same length and are stretched with about the same tension. The highest string vibrates with a frequency that is 4 times that of the lowest string.

If the strings are made of the same material, how would you expect the diameters of the lowest and highest strings to compare?

A. d(low)=2d(high)
B. d(low)=4d(high)
C. d(low)=16d(high)

B. d(low)=4d(high)

Explanation:

Frequency of a string can be written as;

f = v/2L

Where;

v = sound velocity

L = string length

Frequency can be further expanded to;

f = v/2L = (1/2L)√(T/u) 1

Where;

m= mass,

u = linear density of string,

T = tension

p = density of string material

A = cross sectional area of string

d = string diameter

u = m/L 2

m = pAL = p(πd^2)L/4 (since Area = (πd^2)/4)

f = (1/2L)√(T/u) = (1/2L)√(T/(m/L))

f = (1/2L)√(T/((p(πd^2)L/4)/L))

f = (1/2L)√(4T/pπd^2)

f = (1/L)(1/d)√(4T/pπ)

Since the length of the strings are the same, the frequency is inversely proportional to the string diameter.

f ~ 1/d

So, if

4f(low) = f(high)

Then,

d(low) = 4d(high)

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