Why Do Beats Occur When Tuning String Instruments?

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AI Thread Summary
Beats occur when two strings are slightly out of tune, creating a fluctuating sound perceived as beats. In this case, the player hears one beat every 2.04 seconds while tuning the second string to match the first string at 453 Hz. The beat frequency is calculated as 1 divided by the time interval, resulting in approximately 0.49 Hz. This indicates that the second string is off by about 0.49 Hz from the first string's frequency. The initial calculations provided were incorrect, and the player needs to adjust their approach to find the accurate frequency difference.
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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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