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Excerpts from A.P.French's Vibrations & Waves:

. . . It may be seen that the combined displacement can be fitted within an envelope defined by the pair of equations$$ x = 2\mathit{A} \cos\dfrac{(\omega_1 - \omega_2).t}{2} \cdots (2-6)$$ because the rapidly oscillating factor i.e. $\mathit{A} = \cos \dfrac{(\omega_1 + \omega_2)t}{2}$ always lies between $[-1,+1]$ , & the Eq. $(2-6)$ describes a relatively slow amplitude-modulation of the oscillation. One will see that the time between successive zeros of the modulating disturbance is one-half the period of the modulating factor.

Now, that word modulation has maddened me! Though I know its meaning in english language, I evidently can't getting what the author wanted to tell by the bolded phrases. Plz help me explaining these intuitively.

[N.B. I am not an expert; just have started wave-mechanics. So, . . . :)]

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    $\begingroup$ Have I asked a wrong question here? Do downvote generously but please post a comment before or after doing so . . . $\endgroup$
    – user36790
    Commented Jan 30, 2015 at 16:35
  • $\begingroup$ . . .writting a comment doesn't bother the time. I am a student; don't know everything . So, I'm here. Downvoting without any plea is a great slap on my face. $\endgroup$
    – user36790
    Commented Jan 30, 2015 at 16:46
  • $\begingroup$ I suspect the downvoter thinks you should have Googled amplitude modulation. I suppose I basically agree, though of course Googling is always easier when you already know what you're Googling for. Anyhow have an upvote from me to cancel the downvote :-) $\endgroup$
    – John Rennie
    Commented Jan 30, 2015 at 16:47
  • $\begingroup$ @John Rennie: Thanks, sir; actually I did google it but it was a horrendous experience after I read the wiki article ; so intricate. $\endgroup$
    – user36790
    Commented Jan 30, 2015 at 17:01
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    $\begingroup$ possible duplicate of In order for the occurence of beat, why is it compulsory that $|{\omega_2 - \omega_1}| \ll \omega_1 + \omega_2$? $\endgroup$
    – Carl Witthoft
    Commented Jan 30, 2015 at 19:48

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The term modulation as used here means amplitude modulation.

If you have some wave with a high frequency, call this $f(t)$, then you can multiply it by another signal with a much lower frequency, call this $g(t)$. The effect of $g$ is basically to modulate the intensity of $f$ up and down as a function of time.

In this case when you sum your two waves with frequencies $\omega_1$ and $\omega_2$ you get a wave $f(t)$ with a frequency $(\omega_1 + \omega_2)/2$, multiplied by a wave $g(t)$ with frequency $(\omega_1 - \omega_2)/2$. So the signal you observe is the product $g(t)f(t)$. If $\omega_1 \approx \omega_2$, which it is in this case, then $f$ is high frequency and $g$ is low frequency. So $g$ modulates $f$ exactly as in amplitude modulation.

It's the variation in the intensity of $f$ caused by $g$ that we hear as the beat.

Response to comment:

Have a look at this image from this article about amplitude modulation:

Amplitude modulation

When the modulation is described as slow this means slow compared to the carrier frequency. That is, it means the frequency of the modulating wave $g$ must be a lot less than the frequency of the carrier wave $f$.

In your system the carrier signal is the function I've called $f(t)$ with a frequency $(\omega_1 + \omega_2)/2$. The modulating signal is the function I've called $g(t)$ with a frequency $(\omega_1 - \omega_2)/2$. Hopefully it's obvious that the frequency of $f$ has to be much higher than the frequency of $g$ because that gives a nice clear modulation. If the frequencies are too similar then when you multiply $f$ and $g$ you just get an unholy mess with no clear beat pattern.

If $\omega_1 \approx \omega_2$ the the average of the two frequencies is a high frequency and the difference is a small frequency.

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  • $\begingroup$ +1. Sir, can you tell why Mr.French mentioned about slow modulation? And sir, one thing more, why is it important to have $\omega_1 \approx \omega_2$? $\endgroup$
    – user36790
    Commented Jan 30, 2015 at 18:00
  • $\begingroup$ @user36790: I've edited my answer to respond to your cmment. $\endgroup$
    – John Rennie
    Commented Jan 30, 2015 at 18:33

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