# Beat problem

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1. Jul 4, 2017

### Granger

1. The problem statement, all variables and given/known data
A guitar string with 0.60 m and 0.012 of mass vibrates with frequencies that are multiples of 109 Hz. Approaching to the string a tuning fork of 440 Hz we verify beats between the sound signals of the string and the tuning fork. Calculate the time interval between consecutive maximum of sound intensity. Identify the harmonic responsible for these beats.

2. Relevant equations
$$f=\abs{f_2-f_1}$$

3. The attempt at a solution
I don't have any idea on how to determine the harmonic responsible for these beats. But without that there is no way I can determine the frequency and therefore I can not determine the period (time interval between to consecutive maximum).
There's something here that is escaping me, can someone clarify me please...

2. Jul 4, 2017

### BvU

HEllo Granger,

Beats have to do with small differences in frequencies that cause a low-frequency amplitude envelope. So in your case you would expect something to happen with the third harmonic (436 Hz) of the string and the ground tone of the fork. See if you can write the sum of these two tones as a product as in the second link. And think carefully what the beat frequency value means for the time between consecutive maxima

3. Jul 4, 2017

### Granger

Thanks! Oh ok I didn't understand the concept. It makes sense it's small frequencies. After calculation I concluded it's the 4th harmonic (436 Hz).
And I think I understand now. Because we have a product of cosines (so (-1)(-1)=1) we will have 2 maxima in a period (inverse of the beat frequency). Therefor the time between consecutive maxima is half the inverse of the beat frequency, right?

Last edited: Jul 4, 2017
4. Jul 4, 2017

### BvU

sorry, my mistake (counted: ground tone - 1st - 2nd - 3rd instead of ground - 2nd - 3rd - 4th)

Nevertheless, I think you understand it quite well !

5. Jul 10, 2017

### rude man

Actually, the time between consecutive maxima is 1/|f1 - f2|. This is because a maximum occurs twice in each amlitude cycle which has a frequency of |f1 - f2|/2.