0
$\begingroup$

. . .it is the combined vibration or disturbance basically having the average of the combining frequencies, but with an amplitude that varies periodically with time-one cycle of this variation including many cycles of the basic vibration.

Now, what does the author want to meany by the above bolded statement?

Also another:

. . .description as a beat phenomenon is physically meaningful only if $|\omega_1 - \omega_2| \ll \omega_1 + \omega_2$; ie. if, over some substantial number of cycles, the vibration approximates to sinusoidal vibration with constant amplitude and with angular frequency $\dfrac{{\omega_1} + {\omega_2}}{2}$.

At first part, it was said that amplitude does vary with time; however here it is mentioning about constant amplitude! What does this mean? And also difference of two numbers is always less than their sum; what is so special about that? Why does the author emphasize on $|\omega_1 - \omega_2| \ll \omega_1 + \omega_2$? Plz help.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ the $\ll$ symbol means much less than - the criterion then is that $\omega_1$ and $\omega_2$ are approximately equal. If instead $\omega_2 \approx 0$ you'd have $\omega_1 - \omega_2 \approx \omega_1 + \omega_2$. $\endgroup$
    – Sten
    Commented Jan 27, 2015 at 9:41
  • $\begingroup$ @Sten: Sir, can you please explain to me the amplitude issue? It'll great help:) $\endgroup$
    – user36790
    Commented Jan 27, 2015 at 10:56
  • $\begingroup$ You might find the diagrams and explanation at en.wikipedia.org/wiki/Beat_(acoustics) helpful. $\endgroup$
    – N. Virgo
    Commented Jan 28, 2015 at 2:05

2 Answers 2

Reset to default
1
$\begingroup$

but with an amplitude that varies periodically with time - one cycle of this variation including many cycles of the basic vibration.

There the author is talking about this kind of situation:

Amplitude of vibration goes down slowly, then amplitude of vibration goes up slowly, then amplitude of vibration goes down slowly, and so on. Slowly means slowly compared to how quickly the basic vibration goes up and down.

over some substantial number of cycles, the vibration approximates to sinusoidal vibration with constant amplitude

There the author repeats that amplitude must go up and down slowly, by saying that amplitude must stay approximately constant over a few cycles of basic vibration.

Why does the author emphasize on $|\omega_1 − \omega_2| \ll \omega_1 + \omega_2$?

The idea there is that difference of frequencies should be small compared to frequencies.

A more clear way to say that: $|\omega_1 − \omega_2| < \omega_1 \& |\omega_1 − \omega_2| <\omega_2$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Sir,can you tell why the author mentioned about constant amplitude? Beat has no constant amplitude,right? Please help what the author wanted to tell. $\endgroup$
    – user36790
    Commented Jan 28, 2015 at 12:37
  • $\begingroup$ And sir, what does basic vibration mean? Is it the vibration that contributes in the beat's formation?? $\endgroup$
    – user36790
    Commented Jan 28, 2015 at 12:50
  • 1
    $\begingroup$ With "constant amplitude" author is referring to slowly changing amplitude. Seriously. I am not misunderstanding the author. Slowly changing amplitude approximates constant amplitude. What is "basic vibration"? Beats are amplitude changes, basic vibration is the vibration whose amplitude changes. Author says that putting two waves together results one wave whose frequency is the average of the frequencies of the original waves, and that wave the author calls "basic vibration". $\endgroup$
    – stuffu
    Commented Jan 28, 2015 at 14:48
  • $\begingroup$ +1. Basic vibration then means the wave whose amplitude changes in the beat. Then there are many basic vibrations in the beat,right,sir??? $\endgroup$
    – user36790
    Commented Jan 28, 2015 at 16:12
  • $\begingroup$ It is possible to find more than one frequency in the basic vibration by doing a Fourier transform of the basic vibration. But if we believe the Fourier transform then there are no amplitude changes. Fourier transform does not acknowledge existence of any amplitude changes. So if we want that there are beats we say that there is just one frequency whose amplitude is changing, even if it would be possible to say that there are two frequencies whose amplitudes are not changing. $\endgroup$
    – stuffu
    Commented Jan 28, 2015 at 19:11
0
$\begingroup$

Suppose you have two clocks, one looses a minute over an hour. The phase difference between the two clocks would be over a period of sixty hours. If something was driven by the difference of the two hands as a vector, its intensity would wax and wane over 60 hours.

If you have a pair of waves that are at say, 612000 and 612440 Hz, it's like our example of the clock loosing a minute an hour. The wave distribution appears to follow the net cycle of 612000 Hz, just like the clocks still go around an hour a time, but they add intensities, and they drift in and out of phase at 440 Hz.

This is what happens with the radio. You can't hear a signal of 612000 Hz, and the signal would look like a filled in wave of whatever the difference of signals are. But you can hear 440 Hz, and the course design of the radio is pushed around by the signal strength, which is set by the phase difference of the two signals. (the difference of minute hands in the clocks).

Improve this answer
$\endgroup$
1
  • $\begingroup$ What does the author want to tell by the constant amplitude? Beat has no constant amplitude. Sir,can you please help? $\endgroup$
    – user36790
    Commented Jan 28, 2015 at 12:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.