Describe the formation of beats.

III. Long Answer Questions · Q8

Q. Describe the formation of beats.

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Concept understanding — Beats

The Intuition: When Two Almost-Identical Notes Clash

Imagine you're tuning a guitar. You pluck the string you want to tune, and simultaneously strike the reference note from a tuning fork. If the two notes are exactly the same pitch, you hear a single, steady tone. But if they are almost the same — say one is 440 Hz and the other is 442 Hz — something strange happens. The sound doesn't stay steady. Instead, it swells in loudness, fades away, swells again, fades again, in a slow, throbbing rhythm. That rhythmic pulsing is what we call beats.

Why does this happen? Because the two waves, travelling together, sometimes add up constructively (making a loud sound) and sometimes cancel each other out (making a near-silence). They are constantly shifting in and out of phase.

The Precise Physics: Superposition of Two Slightly Different Frequencies

Let two waves of equal amplitude $A$ but slightly different angular frequencies $\omega_1$ and $\omega_2$ arrive at your ear. Their displacements at a point can be written as:

y_1 = A \sin(\omega_1 t)

y_2 = A \sin(\omega_2 t)

By the principle of superposition, the resultant displacement is:

y = y_1 + y_2 = A[\sin(\omega_1 t) + \sin(\omega_2 t)]

Using the sum-to-product identity:

\sin P + \sin Q = 2 \sin\left(\frac{P+Q}{2}\right) \cos\left(\frac{P-Q}{2}\right)

we get:

y = 2A \cos\left(\frac{\omega_1 - \omega_2}{2} t\right) \sin\left(\frac{\omega_1 + \omega_2}{2} t\right)

This is the key result. The resultant wave has two parts:

A fast oscillation at the average frequency $\frac{\omega_1 + \omega_2}{2}$ (which is nearly the same as the original frequencies). This is what your ear hears as the pitch.
A slowly varying amplitude given by $2A \cos\left(\frac{\omega_1 - \omega_2}{2} t\right)$ . This envelope modulates the loudness.

f_{\text{beat}} = |f_1 - f_2|

The beat frequency is the absolute difference of the two original frequencies. Your ear perceives one loud-soft cycle for every complete cycle of the cosine envelope. Since the cosine goes through a full cycle when its argument changes by $2\pi$ , the time period of one beat is:

T_{\text{beat}} = \frac{2\pi}{|\omega_1 - \omega_2|/2} = \frac{2\pi}{|\omega_1 - \omega_2|}

And since $f = \omega/2\pi$ , the beat frequency in hertz is simply:

f_{\text{beat}} = |f_1 - f_2|

What You Actually Hear

Your ear does not follow the rapid $(\omega_1 + \omega_2)/2$ oscillations individually — that's just the pitch you perceive. What you notice is the envelope: the amplitude rises and falls at the beat frequency. So with 440 Hz and 442 Hz, you hear a note of roughly 441 Hz that grows louder and softer 2 times every second. …

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Step 1. Take two sound waves of slightly different frequency but equal amplitude, at a fixed point: $y_1=A\sin(\omega_1t)$ and $y_2=A\sin(\omega_2t)$ .

Step 2. Their sum, using the sum-to-product identity, is $y=y_1+y_2=2A\cos\!\big(2\pi\tfrac{f_1-f_2}{2}t\big)\sin(2\pi f_{avg}t)$ , where $f_{avg}=(f_1+f_2)/2$ .

Step 3. Since $|f_1-f_2|$ is much smaller than $f_1+f_2$ , the amplitude envelope $y_P=2A\cos(\ldots)$ varies slowly in time compared with the fast carrier oscillation at $f_{avg}$ . …

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Q.Describe the formation of beats.

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