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 C (65.4 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. Number inches Unfortunately, none of the strings in your collection have such a large diameter. In fact, the largest diameter you possess is 0.05650 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 C note? Number after before

Explanation / Answer

f=(1/2L)(T/)frequency is proportional to (1/)

is proportional to the square of the radius

82.4Hz --> 65.4Hz ,( 0.05inch -->??)

65.4/82.4=(1/)

1/= 0.6299

= 1/0.6299= r^2/0.025^2

r^2= 0.025^2/0.6299

r= 0.0315

diameter= 2r= 2*0.0315

= 0.063 inches

The frequency is given by f = (1/(2L))(T/)

For 2 different strings we can write:

f = (1/(2L))(T/)

f = (1/(2L))(T/)

Squaring gives:

f² = (1/(4L²))(T/)

f² = (1/(4L²))(T/)

Since L = L, taking the ratios of the equations gives:

(f/f)² = (T/T)(/) (equation 1)

f/f = 82.4/65.4

is proportional to cross-sectional area squared, hence is proportional to diameter squared:

/ = (0.05650/0.0500)² (work out the value)

Plug in these values into equation 1. Then you can find T/T.

T/T. = (f/f)²/(/)

= (82.4/65.4)2/(0.05650/0.0500)²

= 1.2432

Thus,

T1/T2 = 1/1.2432

= 0.8044

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