An unfingered guitar string is 0.72 mlong and is tuned to play E above middle C
ID: 1751570 • Letter: A
Question
An unfingered guitar string is 0.72 mlong and is tuned to play E above middle C (330 Hz). (a) How far from the end of this string mustthe finger be placed to play G# abovemiddle C (415 Hz)? m(b) What is the wavelength on the string of this 415 Hz wave?
m
(c) What are the frequency and wavelength of the sound waveproduced in air at 20° C by this fingered string?
Hz
m
(a) How far from the end of this string mustthe finger be placed to play G# abovemiddle C (415 Hz)? m
(b) What is the wavelength on the string of this 415 Hz wave?
m
(c) What are the frequency and wavelength of the sound waveproduced in air at 20° C by this fingered string?
Hz
m
Explanation / Answer
given length of guitar string L =0.72m the wavelength of the fundamental frequency for a string is1 = 2L1 frequency of the string is f = 415 Hz as the speed of the wave on the string does not change v =f11 =f22 f12L1 = f22L2 f1L1 = f2L2 thus we get 330*0.72 = 415*L2 orL2 = 0.572m therefore the fingure must be placed at (0.72-0.572) m fromthe nut of the guitar that is L = 0.148m form the end. b) The string is fixed at both ends and is vibrating in itsfundamental mode. thus the wavlength is = 2L2 = 2(0.572m) = 1.144m c) the frequency of sound = frequency of the string fs= 415Hz the wavlength is = v/f = 343/415 = 0.826m orL2 = 0.572m therefore the fingure must be placed at (0.72-0.572) m fromthe nut of the guitar that is L = 0.148m form the end. b) The string is fixed at both ends and is vibrating in itsfundamental mode. thus the wavlength is = 2L2 = 2(0.572m) = 1.144m c) the frequency of sound = frequency of the string fs= 415Hz the wavlength is = v/f = 343/415 = 0.826mRelated Questions
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