Beats- time taken for waves to be in phase again

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The discussion centers on the relationship between beat frequency and the time taken for two waves to be in phase again. It explains that a low beat frequency results in a longer time for the waves to align, as indicated by the formula for the period of a beat, which is inversely proportional to the frequency difference. The phase difference between the waves is described mathematically, highlighting that constructive interference occurs at specific intervals. Participants emphasize that while the concepts are mathematically grounded, they lack a distinct physical interpretation beyond the mathematical descriptions. Ultimately, the conversation illustrates the interplay between physics and mathematics in understanding wave behavior.
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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' ?
 
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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 ?
 
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.
 
BvU said:
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.

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?
 
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 !
 
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