# Question regarding beats

If two waves of different frequencies, say $$x_1=A\sin(\omega_1t)$$ and $$x_2=A\sin(\omega_2t)$$, are moving along in the same direction and these waves satisfy the condition $$|\omega_1-\omega_2|\ll \omega_1+\omega_2, \tag{1}$$ then we can observe a beat pattern.

Now the superposition of the wave can be given by $$x=A\sin\left(\dfrac{(ω_1-ω_2)t}{2}\right)\cos\left(\dfrac{(ω_1+ω_2)t}{2}\right).\tag{2}$$

Now I presume that the $$A\sin((ω_1-ω_2)t/2)$$ term of equation 2 doesn't change as fast as $$\cos((ω_1+ω_2)t/2)$$ because of condition $$(1)$$, so we can assume that the amplitude part of the equation is contributed by the term with $$ω_1-ω_2$$ and the frequency of the resultant wave is contributed by $$ω_1+ω_2$$. I understand up to this.

My question is, what is the beat frequency here? Why is it given by $$ω_1-ω_2$$ rather than $$\dfrac{ω_1-ω_2}{2}$$? As we discussed earlier, isn't the frequency of resultant wave given by $$\dfrac{ω_1+ω_2}{2}$$? Is beat frequency different from the frequency of resultant wave?

The equation $$x=A\sin\left(\dfrac{(ω_1-ω_2)t}{2}\right)\times \cos\left(\dfrac{(ω_1+ω_2)t}{2}\right)$$ can be thought of as consisting of two parts which are multiplied together.

The $$\cos\left(\dfrac{(ω_1+ω_2)t}{2}\right)$$ term which is an oscillation at a frequency which is the average of $$\omega_1$$ and $$\omega_2$$.

The $$A\sin\left(\dfrac{(ω_1-ω_2)t}{2}\right)$$ term which oscillates at a lower frequency, $$\dfrac{(ω_1-ω_2)}{2}$$, than the other term and has an amplitude $$A$$.

Adding some numbers, let $$A=1,\,\omega_1 = 2\,\pi\,100.5$$ and $$\omega_2 = 2\,\pi\,99.5$$, which is a difference of $$1\,\rm Hz$$.

Thus $$\dfrac{(ω_1+ω_2)}{2} = 2\,\pi\, 100$$ and $$\dfrac{(ω_1-ω_2)}{2} = 2\,\pi\, \frac 12$$ ie oscillations at $$100\,\rm Hz$$ are changing in amplitude at a frequency of $$\frac 12\,\rm Hz$$ as shown in the graph below.

The variation in amplitude of the wave of frequency $$100\,\rm Hz$$ is shown in the right hand diagram.
The $$100\,\rm Hz$$ wave cannot be properly drawn because between $$t=0\,\rm s$$ and $$t=1\,\rm s$$ one hundred oscillations would have to be displayed.

What you will note is that there is zero amplitude of the $$100\,\rm Hz$$ wave at $$t=0\,\rm s$$, maximum amplitude at $$t=\frac 12\,\rm s$$, zero amplitude at at $$t=1\,\rm s$$, maximum amplitude at $$t=\frac 32\,\rm s$$, zero amplitude at at $$t=2\,\rm s$$.
Thus in $$2$$ seconds (the period of the $$\sin \left(2\,\pi\,\frac12\right)$$ term, the amplitude of the $$100\rm \, Hz$$ goes through two oscillations ie the beat frequency is $$\dfrac{(ω_1-ω_2)}{2} \times 2 = ω_1-ω_2$$ which in the case is $$100.5-99.5=1\,\rm Hz$$.

Perhaps what you actually hear will become clearer after you use this rather nice simulation, Interference in Time and Beats Simulation?

When first started the simulation sets $$f_{\rm A}=300\,\rm Hz$$ and $$f_{\rm B}=305\,\rm Hz$$.

On running the simulation you will hear a sound wave of frequency $$302.5\rm\,Hz$$ $$pulsating$$ at a frequency of $$5\rm\,Hz$$ which is the beat frequency.

Keeping frequency $$f_{\rm A}$$ the same change $$f_{\rm B}$$ to:

$$\bf 1$$ $$301\,\rm Hz$$ to hear a $$1\rm\,Hz$$ beat,

$$\bf 2$$ $$300.1\,\rm Hz$$ to hear a $$0.1\rm\,Hz$$, period $$10\,\rm s$$, beat,

$$\bf 3$$ $$300.05\,\rm Hz$$ to hear a $$0.05\rm\,Hz$$, period $$20\,\rm s$$, beat. [Do not worry that the hundredth of a hertz are not displayed as the simulation will still work.]

$$\bf 4$$ Have fun noting that the lowest available frequency is $$40\,\rm Hz$$ and the highest $$1000\,\rm Hz$$ so it is not possible with this simulation to show two ultrasonic frequencies producing an audible beat frequency eg $$40,000\,\rm Hz$$ and $$40,100\,\rm Hz$$ producing an audible $$100\,\rm Hz$$ tone

• I am sorry but I don't get why we are considering the envelope function here ? I don't get what it is supposed to represent . Commented Feb 26, 2023 at 14:09
• You mean the $Acos(\dfrac{ω_1-ω_2}{2})$ part from $2Acos(\dfrac{ω_1-ω_2}{2})cos(\dfrac{ω_1+ω_2}{2})$ is the envelope part here ? So if my envelope part is of the form $Acos(πt)$(lets consider the resultant equation to be $Acos(πt)cos(6πt)$ ) then my beat frequency (ie) the number of times I hear the loudest amplitude is just 1 ,am I right to assume that ?But why can't I hear the amplitude that's quite a bit less than the max amplitude ,I am saying this because I just hear distinct sound rather than some continous form of sound when it comes to beats . Commented Feb 26, 2023 at 18:38
• @HarryCase I have rewritten my answer to try and address some of your queries. Commented Feb 27, 2023 at 14:36
• .Wow thank you so much .So we just hear the pulsating amplitude because it is the maximum value and it is hard for us to perceive a frequency like 100Hz Commented Feb 28, 2023 at 15:41

It's sort of a matter of definition, but the beat frequency is defined to be $$\omega_1-\omega_2$$ instead of half of that because that's the frequency at which the amplitude varies. That is, consider the following figure (from Wikipedia):

You can see that the period of the amplitude variation is actually half the period of the envelope function, and so the frequency is double what you would expect from the expression. This is the "correct" way to define it, because, for instance, if this wave was actually a sound, you would perceive the loudness of the sound vary with a frequency given by $$\omega_1-\omega_2$$, not half that.

• "You can see that the period of the amplitude variation is actually half the period of the envelope function"Sorry I am a bit dumb ,I don't get what you're trying to say here . Commented Feb 25, 2023 at 4:46
• I think I have some inkling as to what you're trying to say here (imgur.com/a/UQo8CUA) are you saying that the time period of the blue wave is twice of that of the wave indicated by two red dots ? But $Acos(\dfrac{(ω_1-ω_2)t}{2})$ isn't that indicated by a straight line because its value doesn't change much for a substansial amount of time ? Commented Feb 25, 2023 at 6:10
• @HarryCase Yes, that is what I meant: it's only the variation in the amplitude that matters, not the fact that the envelope function (your blue curve) completes one cycle (I think Farcher's answer is better for understanding this). As to your second question, I don't think I understand it. If the frequencies are very close to each other, the period of the envelope will be very large, and so you'll see many wiggles before the amplitude (envelope) varies appreciably. But eventually, the envelope function does go to zero. Commented Feb 25, 2023 at 21:35
• .I am quite confused why do we even need a $cos(πf_st)$ curve here(the one from the pic on your answer) ? What does it represent ? Why are we considering the oscillation of the envelope function ? Am I missing something important here ? Commented Feb 26, 2023 at 14:07