Why Do Beats Occur When Tuning String Instruments?

[mhmil]

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.

 

[rl.bhat]

You are hearing 1 beat per 2.04 s.So the beat frequency f(beat) = 1/2.04 Hz.

 

[tomcue]

Your solution is correct, but you need to round your answer to the nearest whole number. So the final answer would be 18 Hz. Additionally, you should specify that the first string is playing at 453 Hz and the second string is playing at 471 Hz (since that is the frequency difference between the two strings). This will help clarify your answer and show the steps you took to arrive at it.