An unfingered guitar string is 0.73 m long and is tuned to play E above middle C

Question

An unfingered guitar string is 0.73 m long and is tuned to play E above middle C (330 Hz). (a) How far from the end of this string must the finger be placed to play C above middle C (523 Hz)? m (b) What is the wavelength on the string of this 523 Hz wave? m (c) What are the frequency and wavelength of the sound wave produced in air at 20° C by this fingered string? Hz m An unfingered guitar string is 0.73 m long and is tuned to play E above middle C (330 Hz). (a) How far from the end of this string must the finger be placed to play C above middle C (523 Hz)? m (b) What is the wavelength on the string of this 523 Hz wave? m (c) What are the frequency and wavelength of the sound wave produced in air at 20° C by this fingered string? Hz m An unfingered guitar string is 0.73 m long and is tuned to play E above middle C (330 Hz). (a) How far from the end of this string must the finger be placed to play C above middle C (523 Hz)? m (b) What is the wavelength on the string of this 523 Hz wave? m (c) What are the frequency and wavelength of the sound wave produced in air at 20° C by this fingered string? Hz m

Explanation / Answer

At the fundamental, the string has two nodes at the end, and one anti-node in the middle, so therefore the vibrating length of the string (.73 m) is equal to 2/4 , so = 1.46 m

The wave speed on the string must be

v = f = (330 Hz)(1.46 m) = 481.8 m/s

Now, at 481.8 Hz, the wavelength must be

= v/f = (481.8 m/s)/(523 Hz) = 0.92 m, and since at the fundamental, only half the wavelength fits on the string, then we get that the string must now be (0.92 m)/2 = 0.4606 m long, which means the string must now be fingered 0.70 - 0.525 m = 0.26 m from the end.

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