On a guitar, the lowest toned string is usually strung to the E note, which prod

Question

On a guitar, the lowest toned string is usually strung to the E note, which produces sound at 82.4 Hz. The diameter of E guitar strings is typically 0.0500 inches and the scale length between the bridge and nut (the effective length of the string) is 25.5 inches. Various musical acts tune their E strings down to produce a "heavier" sound or to better fit the vocal range of the singer. As a guitarist you want to detune the E on your guitar to D (73.4 Hz). If you were to maintain the same tension in the string as with the E string, what diameter of string would you need to purchase to produce the desired note? Assume all strings available to you are made of the same material.

Unfortunately, none of the strings in your collection have such a large diameter. In fact, the largest diameter you possess is 0.05307 inches. If the tension on your existing string is denoted Tbefore, by what fraction will you need to detune (that is, lower the tension) of this string to achieve the desired D note?

Explanation / Answer

Part 1:

Use equation related frequency, tension, length, and mass density to get the constants (T and L) on one side of the equation

F = (1/2L)*sqrt(T/u)

F2 = 1/4L2*(T/u)

FE2uE = T/4L2 = FD2uD

The mass per unit length is equal to the product of the density (d) and the cross-sectional area of the string. The density remains constant in both strings. The cross sectional area is equal to pi*D2/4, where capital D represents diameter. Therefore

FE2*DE2 = FD2*DD2

(82.4)2(0.05)2 = (73.4)2DD2

DD = 0.05613 inches

Part 2:

F2u = T/4L2

State 1 represents the D-string with 0.05613 inch diameter and state 2 represents the D-string with 0.05307 inch diameter

T1/u1 = F2/4L2 = T2/u2

T1/D12 = T2/D22

Solving for T2 gives

0.89T1

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