What is the relationship between beat period and cosine cycles in a graph?

[somecelxis]

Homework Statement

please refer to the notes , since we know that a cycle of cosine is as in the photo 1 , why the beat period has 8 complete cycle of cosine?

Homework Equations

The Attempt at a Solution

 

[CWatters]

This is talking about beating (mixing) two slightly different frequencies/wavelengths together.

If both start off in sync at t=0 they will become out of sync and then back into sync. The time it takes to "come back into sync" depends on the difference between the two. The closer they are to being identical the longer and more cycles it takes. In this case it takes 8 cycles.

As an exercise try plotting two sine waves on graph paper with slightly different frequencies. Then add a third waveform representing the sum of the amplitudes of the other two. You might learn more doing it by hand but you can also do it in excel.

 

[somecelxis]

why The closer they are to being identical the longer and more cycles it takes. In this case it takes 8 cycles.
? can you please explain further?

 

[haruspex]

why The closer they are to being identical the longer and more cycles it takes. In this case it takes 8 cycles.
? can you please explain further?

Pick a point where the sum has maximum magnitude. Here the two sources are 'in phase'. Go along to the next cycle of one of them. The two are now slightly out of phase. Each additional cycle you move along, the phase difference increases. The closer the two sources are in frequency, the smaller the phase shift each cycle, so the more cycles you have to go through for the phase difference to reach 2π (or thereabouts). When the phase shift reaches that value, the two are in phase again. That completes one beat.

 

[somecelxis]

Pick a point where the sum has maximum magnitude. Here the two sources are 'in phase'. Go along to the next cycle of one of them. The two are now slightly out of phase. Each additional cycle you move along, the phase difference increases. The closer the two sources are in frequency, the smaller the phase shift each cycle, so the more cycles you have to go through for the phase difference to reach 2π (or thereabouts). When the phase shift reaches that value, the two are in phase again. That completes one beat.

what do you mean by The two are now slightly out of phase. Each additional cycle you move along, the phase difference increases. The closer the two sources are in frequency, the smaller the phase shift each cycle,

do you mean pick 2 points on the graph of resultant displacement? how can two of them are sligtly out of phase?

 

[NascentOxygen]

There are 3 waveforms under consideration here. There is one sinewave of frequency f1, and another almost identical but having a slightly different frequency, f2. Then there is the sum of these.

You have been advised to sketch the f1 and f2 sinusoids to illustrate how they can go from being in sync at zero phase to some time later being in sync at zero phase again. It seems you have not yet tried this?

 

[haruspex]

what do you mean by The two are now slightly out of phase. Each additional cycle you move along, the phase difference increases. The closer the two sources are in frequency, the smaller the phase shift each cycle,

do you mean pick 2 points on the graph of resultant displacement? how can two of them are sligtly out of phase?

No, I wrote

Go along to the next cycle of one of [the two sources]

In your diagram, the two sources are not shown.

 

[somecelxis]

No, I wrote

In your diagram, the two sources are not shown.

i think i got what do you mean . based on the diagram , can i say that at 0.5T , the both grpah with f1 and f2 in phase again? whereas at T , both the graph is out of phase again? (180 degree difference)

 

[haruspex]

i think i got what do you mean . based on the diagram , can i say that at 0.5T , the both grpah with f1 and f2 in phase again? whereas at T , both the graph is out of phase again? (180 degree difference)

Yes.