# Beats- time taken for waves to be in phase again

1. Sep 16, 2014

### somecelxis

1. The problem statement, all variables and given/known data
i read some online notes. from there , i know that if the beat frequnecy ( difference in frequncies between 2 waves ) is low , then the time taken for two waves to be in phase again is very long.... which means more number of waves has to be produced for two waves to be in phase again. Period of beat = 1/(f1-f2) ....Why is it so ? can someone explain in a 'physics way' ?

2. Sep 16, 2014

### BvU

Has to do with simple math rules for addition of sines:
$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$
$\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$
$\sin(\alpha + \beta)+\sin(\alpha - \beta) = 2\sin\alpha\cos\beta$

Now if $\alpha/(2\pi\,t)$ is the average frequency $f_1+f_2\over 2$ and $\beta/(2\pi\,t)$ is the frequency difference $f_1-f_2\over 2$, what do you see ?

3. Sep 16, 2014

### BvU

Another way: constructive interference if phase difference is $2k\pi$ ($k$ integer).

Phase difference of $\sin( 2\pi f_1 t) + \sin (2\pi f_2 t)$ is $2\pi (f_1-f_2)\,t$, so it is $2k\pi$ at intervals $\Delta t = 1/(f_1-f_2)$ as you found.

4. Sep 16, 2014

### somecelxis

as the (f1-f2) increases, Period of beat = 1/(f1-f2) decreases. i knew this ... but this only involve maths ... can you explain in a 'physics' way?

5. Sep 16, 2014

### BvU

There's nothing physical. The sines are just a description in math language.
See the ball n+1 catching up with ball n in the video or this one . A whole bunch of (co)sines going in and out of phase. The stunning group effect has "nothing" to do with the movement of the individual pendula.

It's yet another approach, but still math: if f2= f1 * (1+ε), the second wave catches up a fraction ε of a period for each full period of the first. After 1/ε of these periods the two are in phase again. Physics = mathematics here, sorry about that !