Why Do Beats Occur When Tuning String Instruments?

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SUMMARY

The discussion focuses on the phenomenon of beats in string instruments, specifically when tuning one string against another. When a player hears beats, it indicates a frequency difference between the two strings. In this case, with a fundamental frequency of 453 Hz for the first string and a beat frequency of 1 beat every 2.04 seconds, the calculated frequency difference for the second string is 18.12 Hz. However, the correct approach to determine the frequency of the second string involves using the formula for beat frequency, which is the inverse of the time period of the beats.

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Musicians, particularly string instrument players, music educators, and physics students interested in the acoustic properties of sound and tuning methods.

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Homework Statement



String instrument (e.g. guitar) players can tune one string off of another - once you are happy with the tension (and resulting pitch) in one (lower) string, you put your finger down partway up the first string at the appropriate spot to make a higher pitch, one which SHOULD match the fundamental pitch of the next string over. If the second string is almost, but not quite properly tuned, you will hear "beats". Suppose you are doing this, and hear one beat every 2.04 s when trying to adjust the second string. If the first string (with your finger on it) is playing at 453 Hz (this might not be right for a standard guitar, perhaps it's a more unusual string instrument!) how far off in frequency is the second string? (Figure out the answer in Hz, but do not enter units. If you think the second string is 5 Hz off, answer 5.0)

Homework Equations



I believe its f(beat)=f1-f2



The Attempt at a Solution



fb=(453*2.04) - (453*1)
fb= 471.12

The difference would be 471.12 - 453 = 18.12 Hz.

My homework is not accepting this as an answer and I'm not sure where to go from there.
 
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You are hearing 1 beat per 2.04 s.So the beat frequency f(beat) = 1/2.04 Hz.
 

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