A guitar has 6 wires (called strings for some reason) all of which are 1 meter l

Question

A guitar has 6 wires (called strings for some reason) all of which are 1 meter long. Some are thicker than others, so each has a different linear mass density. The linear mass densities of the strings are 0.673, 1.15, 2.69, 4.88, 8.07 and 12.5; all in grams/meter.
A. Consecutive resonance frequencies have been found at 1650 Hz and 1980 Hz string #1 (the thinnest one). What is the frequency of this spring? Draw this wave.
B. Which resonances are 1650 Hz and 1980 Hz (i.e what are their values of 'n')? Draw these waves.
C. What is the tension in string #1?
D. Assuming the tension is the same in all 6 strings, what is the fundamental frequency for string #6 (the thickest one)? A guitar has 6 wires (called strings for some reason) all of which are 1 meter long. Some are thicker than others, so each has a different linear mass density. The linear mass densities of the strings are 0.673, 1.15, 2.69, 4.88, 8.07 and 12.5; all in grams/meter.
A. Consecutive resonance frequencies have been found at 1650 Hz and 1980 Hz string #1 (the thinnest one). What is the frequency of this spring? Draw this wave.
B. Which resonances are 1650 Hz and 1980 Hz (i.e what are their values of 'n')? Draw these waves.
C. What is the tension in string #1?
D. Assuming the tension is the same in all 6 strings, what is the fundamental frequency for string #6 (the thickest one)?
A. Consecutive resonance frequencies have been found at 1650 Hz and 1980 Hz string #1 (the thinnest one). What is the frequency of this spring? Draw this wave.
B. Which resonances are 1650 Hz and 1980 Hz (i.e what are their values of 'n')? Draw these waves.
C. What is the tension in string #1?
D. Assuming the tension is the same in all 6 strings, what is the fundamental frequency for string #6 (the thickest one)?

Explanation / Answer

A) We have to use this equation,

fn = nv/2L

Thus fn = nv/2L

fn+1 = (n+1)v/2L

fn+1 - fn = [(n+1)v/2L] – [nv/2L]

fn+1 - fn = = v/2L

1980 - 1650 = v/2L

v/2L = 330

Thus fundamental frequency = fn = nv/2L = fn = (1*v/2L ) = v/2L = 330 Hz

B)

fn = nv/2L = n*v/2L

fn = n*f1

1980 = n*330

n=1980/330 = 6

fn = n*f1

1650 = n*330

n=1650/330 = 5

C)

f1 = v/2L

f1 = sqrt[T/]/2L

330 = sqrt[T/0.00673]/(2*1)

Simplifying T = 2.93*10^3 N

D)

f1 = sqrt[(2.93*10^3)/(0.125)]/(2*1) = 76.6 Hz

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