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)