When two waves of almost equal frequencies $n_1$ and $n_2$ reach at a point simultaneously, what is the time interval between successive maxima?

When two waves of almost equal frequencies n₁ and n₂ reach at a point simultaneously, what is the time interva

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Concept understanding — Beats Frequency Analysis

Beats Frequency Analysis

The Intuition: When Two Tones "Wobble"

Imagine you're tuning a guitar. You pluck the string you're tuning, and at the same time, you play a 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 slightly different — say one is 440 Hz and the other is 442 Hz — you don't hear two separate notes. Instead, you hear a single tone that wobbles in loudness: it gets louder, then softer, then louder again, in a slow, rhythmic pulse.

That pulse is called a beat. The phenomenon is beats.

Why does this happen? Because the two sound waves are constantly going in and out of sync. When their crests align, they add up to a louder sound (constructive interference). When a crest meets a trough, they cancel partially (destructive interference). The result is a wave whose amplitude rises and falls at a rate equal to the difference between the two original frequencies.

Note

You don't hear the individual frequencies when they are very close. Your ear perceives the average frequency (around 441 Hz in the example), but the loudness fluctuates at the beat frequency.

The Precise Statement

Let two sound waves of slightly different frequencies $f_1$ and $f_2$ (with $f_1 > f_2$ ) travel through the same medium. Their displacements at a point can be written as:

y_1 = A \sin(2\pi f_1 t)

y_2 = A \sin(2\pi f_2 t)

By the principle of superposition, the resultant displacement is:

y = y_1 + y_2 = A \left[ \sin(2\pi f_1 t) + \sin(2\pi f_2 t) \right]

Using the trigonometric 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( 2\pi \frac{f_1 - f_2}{2} t \right) \sin\left( 2\pi \frac{f_1 + f_2}{2} t \right)

This is the key equation. It describes a wave with two parts:

The carrier wave: $\sin\left( 2\pi \frac{f_1 + f_2}{2} t \right)$ — this oscillates at the average frequency $f_{\text{avg}} = \frac{f_1 + f_2}{2}$ . This is the pitch you actually hear.
The envelope: $2A \cos\left( 2\pi \frac{f_1 - f_2}{2} t \right)$ — this modulates the amplitude of the carrier. The envelope oscillates at half the difference frequency.

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

Why $|f_1 - f_2|$ and not half of it? Because the cosine term goes through a full cycle (from maximum to minimum and back to maximum) when its argument changes by $2\pi$ . That happens when $\frac{f_1 - f_2}{2} t = 1$ , i.e., $t = \frac{2}{f_1 - f_2}$ . So the time period of the envelope is $T_{\text{envelope}} = \frac{2}{|f_1 - f_2|}$ . But the loudness (intensity) goes through two maxima per envelope cycle — one at each positive peak of the cosine and one at each negative peak (since squaring the amplitude gives intensity). So the period of the beat (the time between successive loudness maxima) is half of that: $T_{\text{beat}} = \frac{1}{|f_1 - f_2|}$ . Hence, the beat frequency is $f_{\text{beat}} = \frac{1}{T_{\text{beat}}} = |f_1 - f_2|$ .

Watch out

A common mistake is to think the beat frequency is $\frac{|f_1 - f_2|}{2}$ . That is the frequency of the envelope oscillation, not the beat. The ear detects two loudness peaks per envelope cycle, so the beat frequency is double the envelope frequency.

Key Conditions for Beats

Small difference: The two frequencies must be close (typically less than about 10–15 Hz apart). If the difference is too large, the ear perceives two separate tones instead of beats.
Comparable amplitudes: The amplitudes should be roughly equal for maximum contrast in loudness. If one is much louder, beats are still present but less noticeable.
Same medium: The waves must overlap in the same region of space.

Why This Matters for Exams …

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When two waves of nearly equal frequency interfere, they produce beats—periodic variations in amplitude. The time between successive maxima (loud sounds) is the reciprocal of the beat frequency: $\frac{1}{|n_1 - n_2|}$ .

Why beats occur

When two waves of slightly different frequencies superpose, they drift in and out of phase with each other. Sometimes their crests align and reinforce (constructive interference, maximum amplitude); half a beat cycle later, a crest meets a trough and they cancel (destructive interference, minimum amplitude). This periodic rise and fall in intensity is what we hear as beats.

The mathematics reveals that the combined wave oscillates at the average frequency $\frac{n_1 + n_2}{2}$ , but its amplitude itself oscillates slowly at the beat frequency $|n_1 - n_2|$ . Each complete cycle of the amplitude envelope—from one maximum through a minimum and back to the next maximum—takes one beat period.

Step-by-step derivation

Write the two waves. Assume both have the same amplitude $A$ and meet at a point. Their displacements are:

y_1 = A \sin(2\pi n_1 t), \quad y_2 = A \sin(2\pi n_2 t).

Superpose them. The resultant displacement is:

y = y_1 + y_2 = A[\sin(2\pi n_1 t) + \sin(2\pi n_2 t)].

Apply the sum-to-product identity. Recall that $\sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$ . Here:

y = 2A \cos\left(2\pi \frac{n_1 - n_2}{2} t\right) \sin\left(2\pi \frac{n_1 + n_2}{2} t\right).

Identify the modulation. The sine term oscillates rapidly at the average frequency $\frac{n_1 + n_2}{2}$ (the carrier). The cosine term oscillates slowly at frequency $\frac{|n_1 - n_2|}{2}$ and acts as a time-varying amplitude envelope.
Recognize that intensity depends on amplitude squared. The amplitude envelope is:

A_{\text{env}}(t) = 2A \left|\cos\left(\pi (n_1 - n_2) t\right)\right|.

The intensity (proportional to $A_{\text{env}}^2$ ) reaches a maximum whenever $\cos^2(\pi (n_1 - n_2) t) = 1$ , which happens twice per cycle of the cosine—once at each peak of $|\cos|$ . …

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

Showing the 12 most recent of 14 on this concept.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$Q. Two identical strings each of length 0.750 m are each tuned exactly to 440 Hz. The tension in one of the strings is then increased by 1.0 %. If they are now struck, then the beat frequency between the fundamentals of the two strings is nearly (A) 1 (B) 2 (C) 3 (D) No beats$
›Reveal solutionSolution
A small 1% increase in tension raises the string's frequency by about half that percentage (since $f\propto\sqrt{T}$ ), producing a beat frequency of roughly 2 Hz between the two strings.

Concept and Intuition

The fundamental frequency of a vibrating string is $f = \dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ , so for fixed length and linear mass density, $f \propto \sqrt{T}$ . A small fractional change in tension $T$ produces roughly half that fractional change in $f$ (by the binomial approximation $\sqrt{1+x}\approx 1+x/2$ for small $x$ ). When two originally identical strings' tensions differ slightly, they produce two slightly different frequencies, and superposing their sounds produces beats at the difference frequency.

Step-by-Step Solution

Both strings start identical: $f_1 = f_2 = 440\ \text{Hz}$ (same $L$ , $\mu$ , $T$ ).

Tension in one string is increased by $1\% = 0.01$ : $T' = 1.01\,T$ .

New frequency: $f' = f\sqrt{T'/T} = f\sqrt{1.01}$ .

Using the small-x approximation, $\sqrt{1.01} \approx 1 + \dfrac{0.01}{2} = 1.005$ .

$f' \approx 440 \times 1.005 = 442.2\ \text{Hz}$ . …
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AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
$y_1 = 2\sin(500\pi t - ax)$
›Reveal solutionSolution
Read off each wave's frequency from its angular frequency ( $\omega=2\pi f$ ), subtract to get the beat frequency (3 Hz), then convert to beats per minute — 180.

Concept and Intuition

When two waves of slightly different frequency superpose, the resulting amplitude modulation ("beats") occurs at a rate equal to the difference of the two frequencies. Reading $\omega$ straight from the argument of each sine function gives each wave's frequency directly.

Step-by-Step Solution

$y_1 = 2\sin(500\pi t - ax) \implies \omega_1 = 500\pi\text{ rad/s} \implies f_1 = \dfrac{\omega_1}{2\pi} = 250$ Hz.

$y_2 = 2\sin(506\pi t - bx) \implies \omega_2 = 506\pi\text{ rad/s} \implies f_2 = \dfrac{\omega_2}{2\pi} = 253$ Hz.

Beat frequency $= |f_2-f_1| = 253-250 = 3$ Hz $=3$ beats per second.

Beats per minute $= 3 \times 60 = 180$ . …
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AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
$Q. Ten tuning forks are arranged in increasing order of frequency. Any two nearest tuning forks produce 4 beats per second. The highest frequency of tuning fork is twice that of the lowest possible, then the highest and lowest frequencies, in Hz, are respectively (A) 80 and 40 (B) 100 and 60 (C) 72 and 32 (D) 72 and 36$
›Reveal solutionSolution
This tests setting up an arithmetic progression of frequencies from the "beats between adjacent forks" condition, then using the given ratio between highest and lowest to solve for the actual values.

Concept and Intuition

When ten tuning forks are arranged in increasing order of frequency and any two adjacent forks produce a fixed beat frequency (here 4 Hz), their frequencies form an arithmetic progression with common difference equal to that beat frequency. With $n=10$ forks there are $n-1 = 9$ gaps between the lowest and the highest fork.

Step-by-Step Solution

Let the lowest frequency be $f$ (Hz). Successive fork frequencies: $f,\ f+4,\ f+8,\ \dots,\ f+9\times4$ (10 terms total, so 9 steps of 4 Hz).

Highest frequency $= f + 36$ .

Given: highest $= 2\times$ lowest, i.e. $f + 36 = 2f$ .

Solve: $36 = 2f - f = f \Rightarrow f = 36\ \text{Hz}$ .

Highest frequency $= f + 36 = 36+36 = 72\ \text{Hz}$ . …
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AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
$\mathrm{Hz}$
›Reveal solutionSolution
Beat frequency fixes $f_2$ to 250 Hz or 262 Hz; the effect of wax (frequency decreases) picks 262 Hz as the only consistent choice.

Concept and Intuition

Beat frequency is $|f_1 - f_2|$ , so it never tells you which fork is higher — that ambiguity is resolved by a physical perturbation like adding wax, filing, or loading with tape, all of which only ever reduce a tuning fork's frequency (added mass/wax lowers the natural frequency).

Step-by-Step Solution

Given $f_1 = 256$ Hz and beat frequency $= 6$ Hz: $f_2 = 256 \pm 6$ , i.e. $f_2 = 250$ Hz or $f_2 = 262$ Hz.

Wax is attached to fork 2, and its frequency can only decrease.

Case $f_2 = 262$ Hz: after wax, $f_2' < 262$ . New beats $= 2 \Rightarrow f_2' = 254$ or $258$ Hz. Since $f_2'$ decreased from 262, and moving down towards 256 Hz reduces the beat count, $f_2' = 258$ Hz (still above 256, beats $=2$ ) is consistent with a small decrease. …
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AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
$306\,\mathrm{ms^{-1}}$
›Reveal solutionSolution
Beat frequency is the difference of the two individual frequencies; solving $\Delta f=v(1/\lambda_1-1/\lambda_2)=12$ gives $v\approx306\,\mathrm{ms^{-1}}$ .

Concept and Intuition

When two waves of slightly different frequency (or wavelength) superpose, the resulting beat frequency is the magnitude of the difference between the two individual frequencies: $f_{beat}=|f_1-f_2|$ . Using $f=v/\lambda$ lets us relate the beat frequency directly to the speed of sound.

Step-by-Step Solution

$f_1=\dfrac{v}{\lambda_1}=\dfrac{v}{0.50}$ , $f_2=\dfrac{v}{\lambda_2}=\dfrac{v}{0.51}$ (in metres).

Beat frequency: $f_{beat}=f_1-f_2=v\left(\dfrac1{0.50}-\dfrac1{0.51}\right)$ .

$\dfrac1{0.50}=2$ , $\dfrac1{0.51}\approx1.960784$ .

$f_{beat}=v(2-1.960784)=v(0.039216)$ . …
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AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
$330\ \mathrm{m\,s^{-1}}$
›Reveal solutionSolution
Two close wavelengths produce a beat frequency computed from their corresponding frequencies; 10 beats occur in 3 seconds.

Concept and Intuition

Beats arise when two sound waves of slightly different frequency interfere, producing a periodic rise and fall in loudness at the difference frequency $|f_1-f_2|$ . Frequency is related to wavelength via $f=v/\lambda$ , so a small difference in wavelength (99 cm vs 100 cm) translates into a small but nonzero beat frequency.

Step-by-Step Solution

$f_1 = v/\lambda_1 = 330/0.99\ \mathrm{Hz}$ , $f_2 = v/\lambda_2 = 330/1.00 = 330\ \mathrm{Hz}$ .

Beat frequency: $\Delta f = f_1-f_2 = v\left(\dfrac{1}{\lambda_1}-\dfrac{1}{\lambda_2}\right) = 330\left(\dfrac{1}{0.99}-1\right) = 330\times\dfrac{1-0.99}{0.99} = 330\times\dfrac{0.01}{0.99} = \dfrac{3.3}{0.99} = \dfrac{10}{3}\ \mathrm{Hz}$ . …
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AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
$Q. When a stretched string is vibrated simultaneously with a 440 Hz tuning fork, a beat frequency of 5 Hz is produced. If the experiment is repeated with a tuning fork of 437 Hz, the beat frequency produced is 8 Hz. The frequency of the string is (A) 445 Hz (B) 435 Hz (C) 429 Hz (D) 448 Hz$
›Reveal solutionSolution
Beats only tell you the magnitude of the frequency difference, giving two candidate string frequencies per tuning fork; the value common to both trials, 445 Hz, is the answer.

Concept and Intuition

When two frequencies close to each other are sounded together, they produce beats at a rate equal to the magnitude of their difference: $f_{beat}=|f_1-f_2|$ . This is ambiguous on its own (the unknown could be above or below the known frequency), so a second trial with a different reference frequency is used to break the tie — only one candidate value will be consistent with both observations.

Step-by-Step Solution

With the 440 Hz fork, beat frequency 5 Hz $\Rightarrow$ string frequency $f = 440\pm5 = 445$ or $435$ Hz.

With the 437 Hz fork, beat frequency 8 Hz $\Rightarrow$ string frequency $f = 437\pm8 = 445$ or $429$ Hz. …
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AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$y_1 = 3\sin 250\pi t$
›Reveal solutionSolution
This tests the beats phenomenon: two close frequencies superpose to give a periodic intensity maximum, recurring at the reciprocal of the beat frequency — here $0.2\ \text{s}$ .

Concept and Intuition

When two waves of slightly different frequencies $f_1$ and $f_2$ superimpose, their resultant amplitude waxes and wanes at the beat frequency $f_{beat}=|f_1-f_2|$ . Since intensity depends on amplitude squared, a maximum of loudness (maximum intensity) occurs once every beat period $T_{beat}=1/f_{beat}$ .

Step-by-Step Solution

Identify angular frequencies from $y_1=3\sin(250\pi t)$ and $y_2=2\sin(260\pi t)$ : $\omega_1=250\pi$ , $\omega_2=260\pi$ .

Convert to ordinary frequency using $f=\omega/2\pi$ : $f_1 = 125\ \text{Hz}$ , $f_2=130\ \text{Hz}$ .

Beat frequency: $f_{beat}=|f_2-f_1| = 5\ \text{Hz}$ . …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$20\ \mathrm{Hz}$
›Reveal solutionSolution
Using $f\propto \dfrac{1}{Ld}\sqrt{T/\rho}$ for each wire and the given ratios, the two frequencies are 340 Hz and 350 Hz, giving 10 Hz beats.

Concept and Intuition

A stretched wire's fundamental frequency is $f=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$ where $\mu=\rho A=\rho\dfrac{\pi d^2}{4}$ is mass per unit length. So $f\propto\dfrac{1}{Ld}\sqrt{\dfrac{T}{\rho}}$ (constants cancel in a ratio). Comparing the two wires via their given ratios of L, d, ρ, T lets us find the frequency ratio directly.

Step-by-Step Solution

Let wire-1: $L_1=35,\ d_1=4,\ \rho_1=1,\ T_1=8$ ; wire-2: $L_2=34,\ d_2=1,\ \rho_2=2,\ T_2=1$ .

$\dfrac{f_1}{f_2}=\dfrac{L_2 d_2}{L_1 d_1}\sqrt{\dfrac{T_1\rho_2}{T_2\rho_1}}=\dfrac{34\times1}{35\times4}\sqrt{\dfrac{8\times2}{1\times1}}=\dfrac{34}{140}\sqrt{16}=\dfrac{34}{140}\times4=\dfrac{136}{140}=\dfrac{34}{35}$ . …
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AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
$Q. Two stretched strings A and B when vibrated together produce 4 beats per second. If the tension applied to the string A increased, the number of beats produced per second is increased to 7. If the frequency of string B is 480 Hz initially, the frequency of string A is (A) 473 Hz (B) 476 Hz (C) 484 Hz (D) 487 Hz$
›Reveal solutionSolution
The tension-increase test resolves the two possible frequencies of string A; since beats grow when tension (and hence $f_A$ ) increases, A's frequency must already exceed B's.

Concept and Intuition

Beat frequency only tells you $|f_A-f_B|$ , not which is larger — that ambiguity is exactly what the "increase tension and watch the beats" test resolves, since raising tension always raises a string's fundamental frequency.

Step-by-Step Solution

$|f_A-480|=4 \Rightarrow f_A=476$ Hz or $f_A=484$ Hz.

Increasing the tension in A increases $f_A$ .

If $f_A=476$ Hz (below 480), increasing $f_A$ moves it toward 480, which would first decrease the beat frequency — contradicting the given increase to 7 Hz. …
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AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
$\upsilon_1$
›Reveal solutionSolution
The time between two successive beat-minima equals one full beat period, $\dfrac{1}{\upsilon_1-\upsilon_2}$ .

Concept and Intuition

When two waves of close frequencies superpose, the resultant amplitude is modulated by an envelope $|\cos(\pi\Delta\upsilon\, t)|$ , where $\Delta\upsilon=\upsilon_1-\upsilon_2$ . This envelope reaches zero (silence/minimum) once per beat period — so the time between two consecutive minima is exactly one beat period, not half of it.

Step-by-Step Solution

Beat frequency $= \upsilon_1-\upsilon_2$ (given $\upsilon_1>\upsilon_2$ ).

Resultant intensity envelope $\propto \cos^2(\pi\Delta\upsilon\,t)$ (or equivalently $|\cos(\pi\Delta\upsilon t)|$ for amplitude), which is zero when $\pi\Delta\upsilon\, t = \pi/2, 3\pi/2, 5\pi/2,\dots$

Consecutive zero times: $t_n = \dfrac{(2n-1)}{2\Delta\upsilon}$ . The spacing between consecutive $t_n$ is $\dfrac{1}{\Delta\upsilon}$ . …
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AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
$\mathrm{ms^{-1}}$
›Reveal solutionSolution
Setting the difference of the two open-pipe fundamental frequencies equal to the observed 4 Hz beat rate gives the speed of sound in the medium as 204 m/s.

Concept and Intuition

An open organ pipe of length $L$ has fundamental frequency $f=\dfrac{v}{2L}$ (both ends being antinodes, so the pipe holds half a wavelength). Two pipes of slightly different lengths therefore have slightly different fundamentals, and their difference is the beat frequency heard when both sound together.

Step-by-Step Solution

$f_1 = \dfrac{v}{2(0.50)} = \dfrac{v}{1.00}$ (for $L_1=50\,\mathrm{cm}=0.50\,\mathrm m$ ).

$f_2 = \dfrac{v}{2(0.51)} = \dfrac{v}{1.02}$ (for $L_2=51\,\mathrm{cm}=0.51\,\mathrm m$ ).

Beats per second $=\dfrac{40\text{ beats}}{10\text{ s}}=4\,\mathrm{Hz}$ .

$f_1 - f_2 = v\left(\dfrac1{1.00}-\dfrac1{1.02}\right) = v\cdot\dfrac{1.02-1.00}{1.00\times1.02} = v\cdot\dfrac{0.02}{1.02}$ . …
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Q.When two waves of almost equal frequencies n1n_1n1​ and n2n_2n2​ reach at a point simultaneously, what is the time interval between successive maxima?

Why beats occur

Step-by-step derivation

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Q.When two waves of almost equal frequencies $n_1$ and $n_2$ reach at a point simultaneously, what is the time interval between successive maxima?

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