Question

A guitar string can be considered a string fixed at both ends when tuned and played properly. The fundamental frequency of the E-string on a guitar is 328 Hz. The person playing the guitar shortens the string length to four-fifths of its original length by placing their finger on the fret at that location and securing the string there. What is the fundamental frequency of that shortened string (in Hz) assuming wave speed in the string remains constant?

Answer 1

Answer:

$f' = 410 Hz$

Explanation:

Initially when fundamental frequency of the string is

f = 328 Hz

let the length of the string is Lo at this time

so we will have

$f = \frac{2v}{L_o}$

now the length is shorten so that the frequency will change

final length is four fifth of the original length

so here we have

$L = \frac{4L_o}{5}$

now we have

$f' = \frac{2v}{4L_o/5}$

$f' = \frac{5v}{2L_o}$

now we have

$\frac{f'}{f} = \frac{5v/2L_o}{2v/L_o}$

so we have

$\frac{f'}{f} = \frac{5}{4}$

so the new fundamental frequency will be

$f' = \frac{5}{4}(328)$

$f' = 410 Hz$