Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes gi

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Q. Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2022Subjective· 3mImportance★★★★★

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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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Add two sinusoids of close frequency; the amplitude envelope's periodicity is exactly the frequency difference.

Let two sound waves of equal amplitude $A$ and frequencies $f_1$ , $f_2$ ( $f_1 > f_2$ ) be:

$y_1 = A\sin(2\pi f_1 t), \quad y_2 = A\sin(2\pi f_2 t)$

By superposition, using $\sin C + \sin D = 2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$ :

$y = y_1+y_2 = \left[2A\cos\!\left(\pi(f_1-f_2)t\right)\right]\sin\!\left(2\pi\,\frac{f_1+f_2}{2}t\right)$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set ANN1 mark
$Q. What are beats?$
›Reveal solutionSolution

Beats are the periodic waxing and waning of sound loudness produced by the superposition of two nearly-equal-frequency sound waves.

When two sound waves of slightly different frequencies (f₁ and f₂, close to each other) but nearly equal amplitude travel in the same direction and superpose, the resultant amplitude at any point does not stay constant — it periodically rises to a maximum and falls to a minimum. This periodic variation in the intensity (loudness) of the resultant sound, heard as a regular waxing and waning, is called beats.

The number of beats produced per second (the beat frequency) equals the difference between the two individual frequencies:

fbeat = |f₁ − f₂|

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q.Which of the following phenomena in sound waves is responsible for the production of beats? (A) Reflection (B) Refraction (C) Interference (D) Resonance$
›Reveal solutionSolution

Beats are a periodic interference effect between two close-frequency waves.

When two sound waves of slightly different frequencies $f_1$ and $f_2$ travel through the same medium simultaneously, they superpose (interfere). At some instants their crests align (constructive interference, loud sound) and at others crest meets trough (destru …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Which property of sound waves is responsible for the formation of beats?$
(a) Amplitude
(b) Wavelength
(c) Frequency
(d) Speed
›Reveal solutionSolution

Beats are produced when two waves of nearly equal but slightly different frequencies superpose, causing periodic constructive/destructive interference — so it's the difference in FREQUENCY between the two waves that creates beats.

When two sound waves of frequencies f1 and f2 (close in value) overlap, their superposition produces a resultant wave whose amplitude rises and falls periodically. This is heard as a throbbing/pulsing loudness variation called 'beats,' occurring at a beat frequency = |f1 - f2|.

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Beats are heard clearly when the frequencies of the two waves are —$
(a) Equal
(b) Nearly equal
(c) Very different
(d) None of these
›Reveal solutionSolution

Beats are clearly heard when the two frequencies are nearly equal.

Beats are the periodic rise and fall of loudness produced when two waves of slightly different frequencies superpose; the beat frequency equals the difference of the two frequencies.

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. Beats are produced due to ?$
(a) Reflection
(b) Refraction
(c) Interference
(d) Resonance of sound waves
›Reveal solutionSolution

Beats are produced by interference of two sound waves of slightly different frequency.

When two waves of nearly equal frequencies travel in the same direction and superpose, they interfere constructively and destructively alternately, producing a periodic variation in intensity called beats …

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CBSE 2020Set ANNUAL1 markMCQ
$\frac{1}{4}$
›Reveal solutionSolution

Beat frequency = |f1 − f2| = 4 Hz, so the time between successive intensity maxima is 1/4 s.

Beat frequency $f_{beat} = |f_1 - f_2| = |454 - 450| = 4$ Hz. The time interval between two successive maxima (loudness peaks) equals the bea …

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CBSE 2019Set ANNUAL1 mark
$Q. When two sound sources are sounded together, then 2 beats are produced in .20 second. Find the frequency of the beats.$
›Reveal solutionSolution

Beat frequency = number of beats ÷ time taken = 2 beats / 0.20 s = 10 Hz.

When two sound sources of slightly different frequency are sounded together, the resultant intensity waxes and wanes periodically — this phenomenon is called beats. The beat frequency equals the difference between the two source frequencies, and is measured as the number of beats produced per second: …

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CBSE 2019Set ANNUAL1 mark
$Q. Write the formula of beat frequency.$
›Reveal solutionSolution

The beat frequency is simply the magnitude of the difference between the two individual frequencies: f_beat = |f1 - f2|.

When two sound waves of nearly equal but slightly different frequencies f1 and f2 travel together, their superposition produces a periodic waxing and waning of loudness called beats. The number of such loud-soft cycles heard per second (the beat frequency) …

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CBSE 2017Set ANNUAL1 mark
$Q. Define beats.$
›Reveal solutionSolution

Beats are the periodic loudness variation produced when two waves of nearly equal frequency interfere; beat frequency = |f1 - f2|.

When two sound waves of slightly different frequencies f1 and f2 (with f1 close to f2) travel through the same medium and superpose, at some instants they arrive in phase and reinforce each other (loud sound), and shortly after they arrive out of phase and cancel each other (faint/no sound) — because the phase difference between them keeps changing steadily with time. This periodic waxing and waning of the resultant sound intensity is called beats. The number of times the loudness becomes maximum per second is the beat frequency, and it equals the difference of the two individual frequencies: f_beat = |f1 - f2|. Beats are commonly used by musicians to tune instruments to the same pitch — as the two freque …

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Q.Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats.

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