An unfingered guitar string is 0.70 mlong and is tuned to play E above middle C

Question

An unfingered guitar string is 0.70 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)?
1 m
(b) What is the wavelength on the string of this 415 Hz wave?
2 m
(c) What are the frequency and wavelength of the sound waveproduced in air at 20° C by this fingered string?
3 Hz
4 m (a) How far from the end of this string mustthe finger be placed to play G# abovemiddle C (415 Hz)?
1 m
(b) What is the wavelength on the string of this 415 Hz wave?
2 m
(c) What are the frequency and wavelength of the sound waveproduced in air at 20° C by this fingered string?
3 Hz
4 m

Explanation / Answer

Since f= v/2L where v - velocity of the wave propagating in the string L- length of the string and v=(T/) and is constant so long as tension T isconstant - linear density of the string Also v= 2Lf0 Now a) How far from the end of this string must the finger beplaced to play G# above middle C(415 Hz)?
fa= v/2L= 2Lf0 / 2La La= L f0 / fa b) What is the wavelength on the string of this 415 Hz wave a = v/fa c)What are the frequency and wavelength of the sound waveproduced in air at 20° C by this fingered string? The frequency will remain the same however the speed of soundis medium dependent and therefore the wavelength has to be adjustedaccordingly. vair = 331 + 0.606 C (C= 20°C) = vair /f ( f=415Hz) Just plug and play Please let me know if you have any questions. c)What are the frequency and wavelength of the sound waveproduced in air at 20° C by this fingered string? The frequency will remain the same however the speed of soundis medium dependent and therefore the wavelength has to be adjustedaccordingly. vair = 331 + 0.606 C (C= 20°C) = vair /f ( f=415Hz) Just plug and play Please let me know if you have any questions.

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