Chapter 14 ยท Waves

14.4 Beats

Two notes of nearly equal frequency, heard together, swell and fade in loudness: beats. The loudness peaks $|\nu_1-\nu_2|$ times a second, which lets musicians tune instruments and lets you find an unknown frequency to a fraction of a hertz.

Strike two tuning forks that are very nearly, but not exactly, alike and hold them near your ear. You do not hear two notes. You hear a single note whose loudness rises and falls โ€” wahโ€ฆ wahโ€ฆ wahโ€ฆ โ€” a few times a second. Musicians use exactly this effect to tune: a guitarist plays the same note on two strings and turns a peg until the throbbing slows down and stops; a piano tuner counts these throbs to set each string of the piano to a precise fraction of a hertz. The steady drone of a twin-engine propeller aircraft pulses in the same way when its two engines run at slightly different speeds.

The rise and fall is called beats. It is the principle of superposition of Section 14.3 once more, but now applied at one place to two waves of slightly different frequency: the two sounds drift slowly into step and out of step, reinforcing each other and then cancelling. This short section finds the beat frequency, $|\nu_1-\nu_2|$, and puts it to work โ€” finding unknown frequencies, deciding whether a fork is above or below a reference, and tuning strings and pipes with the formulas of the last two sections.

Two notes of nearly equal frequency

Two sound waves of the same amplitude $a$ and frequencies $\nu_1$ and $\nu_2$ (with $\nu_1\gt\nu_2$, and the two close together) arrive at your ear. At the ear, take their displacements as $s_1=a\cos\omega_1t$ and $s_2=a\cos\omega_2t$, with $\omega=2\pi\nu$. By the principle of superposition and the identity $\cos A+\cos B=2\cos\dfrac{A-B}2\cos\dfrac{A+B}2$:

$$s=s_1+s_2=\left[2a\cos\left(\frac{\omega_1-\omega_2}{2}\,t\right)\right]\cos\left(\frac{\omega_1+\omega_2}{2}\,t\right).$$

Read this as an oscillation at the average angular frequency $\dfrac{\omega_1+\omega_2}2$ โ€” the pitch you hear โ€” whose amplitude, the factor in square brackets, changes slowly with time. The loudness depends on the size of that amplitude, $2a\left|\cos\pi(\nu_1-\nu_2)t\right|$, not on its sign. It is greatest ($2a$) whenever $\cos\pi(\nu_1-\nu_2)t=\pm1$, i.e. at

$$t=0,\ \frac1{\nu_1-\nu_2},\ \frac2{\nu_1-\nu_2},\dots$$

and zero half-way between. So the loudness reaches a maximum $\nu_1-\nu_2$ times every second. The same result in words: the phase difference between the two waves, $(\omega_1-\omega_2)t$, grows steadily; each time it passes a whole multiple of $2\pi$ the waves are in step (loud), and each time it passes an odd multiple of $\pi$ they cancel (soft).

๐Ÿ“˜ Definition 14.9 โ€” Beats

Beats are the periodic rise and fall in the loudness of the sound heard when two waves of slightly different frequencies superpose. The number of loudness maxima per second is the beat frequency; the time between successive maxima is the beat period.

๐Ÿ“ Result 14.14 โ€” Beat frequency

Two sound waves of frequencies $\nu_1$ and $\nu_2$ that differ only slightly, heard together, produce a tone of frequency $\dfrac{\nu_1+\nu_2}2$ whose loudness rises and falls with

$$\nu_\text{beat}=|\nu_1-\nu_2| ,$$

i.e. a beat period $T_\text{beat}=\dfrac1{|\nu_1-\nu_2|}$.

Notice the factor of two that catches people out: the amplitude factor $\cos\pi(\nu_1-\nu_2)t$ completes one cycle in $\dfrac{2}{\nu_1-\nu_2}$, but it peaks twice in that time (once positive, once negative), and the ear hears both as loud. The ear can follow beats up to roughly $10$โ€“$15$ per second. With a larger difference the sound becomes rough, and beyond a few tens of hertz you simply hear two separate notes.

โœ๏ธ Example 1 โ€” Two tuning forks
Two tuning forks of frequencies $256\ \text{Hz}$ and $260\ \text{Hz}$ are sounded together. Find (a) the beat frequency; (b) the frequency of the note heard; (c) the time between two successive maxima of loudness; (d) the time from a maximum to the next silence.
(a) $\nu_\text{beat}=260-256=4\ \text{Hz}$: four beats per second.
(b) $\dfrac{256+260}{2}=258\ \text{Hz}$.
(c) $T_\text{beat}=\dfrac14=0.25\ \text{s}$.
(d) Half a beat period, $0.125\ \text{s}$: in that time the phase difference grows by $2\pi\times4\times0.125=\pi$, from in step to antiphase.
Check In $0.25\ \text{s}$ the forks make $64$ and $65$ vibrations: exactly one extra vibration per beat โœ“.
โœ๏ธ Example 2 โ€” Reading a beat equation
At a point, the sound from two loudspeakers is $s=0.020\cos(1000\pi t)+0.020\cos(1010\pi t)$, in pascals of excess pressure ($t$ in seconds). Find the two frequencies, the beat frequency, the frequency of the tone heard and the largest excess pressure.
Frequencies $\nu=\dfrac{\omega}{2\pi}$: $500\ \text{Hz}$ and $505\ \text{Hz}$.
Beats $\nu_\text{beat}=5\ \text{Hz}$; the tone heard is $502.5\ \text{Hz}$.
Peak When the waves are in step, $0.020+0.020=0.040\ \text{Pa}$.
Form $s=0.040\cos(5\pi t)\cos(1005\pi t)$: the slowly varying factor $\cos5\pi t$ has frequency $2.5\ \text{Hz}$, but $|\cos5\pi t|$ peaks $5$ times a second โœ“.
Interactive 1 โ€” The Beats Generator

Choose two frequencies. The top panel zooms in on $20\ \text{ms}$ around the moving cursor and shows the two waves (thin) and their sum (thick) โ€” watch them slide into and out of step. The bottom panel shows $2\ \text{s}$ of the sum: its amplitude $2a|\cos\pi(\nu_1-\nu_2)t|$ swells and fades. Count the swells in one second. Press ๐Ÿ”Š to hear the beats (use headphones or a speaker; it plays only while switched on).

440 Hz444 Hz
beat frequency $|\nu_1-\nu_2|$โ€”
beat periodโ€”
tone heard $\tfrac12(\nu_1+\nu_2)$โ€”
loudness now (amplitude)โ€”

Using beats: unknown frequencies and tuning

Beats measure a frequency difference very precisely โ€” counting $3$ beats in a second against a $512\ \text{Hz}$ fork pins down a difference of $3\ \text{Hz}$ out of $512$, better than $1\%$, with nothing but your ears. But a beat count alone cannot say which note is higher: $512\pm3\ \text{Hz}$ give the same $3$ beats per second. To decide, change one frequency slightly in a known direction and see whether the beats speed up or slow down.

๐Ÿงญ Problem-solving strategy โ€” deciding which note is higher
  1. From the beat frequency $b$ against a known frequency $\nu_0$, write both possibilities: $\nu=\nu_0+b$ or $\nu=\nu_0-b$.
  2. Decide whether the change (wax, file, tension, length, temperature) raises or lowers $\nu$.
  3. Test each possibility: does the change move $\nu$ towards $\nu_0$ (beats slow down) or away from it (beats speed up)? Keep the one that matches what was heard.
  4. Check the numbers: if the beats change from $b$ to $b'$, the frequency must have changed by $|b-b'|$ (or by $b+b'$ if it crossed $\nu_0$).
โœ๏ธ Example 3 โ€” Wax on a fork
An unknown tuning fork $X$ gives $4$ beats per second with a standard fork of $512\ \text{Hz}$. When a prong of $X$ is loaded with a little wax, it gives $6$ beats per second with the standard. Find the frequency of $X$ before it was loaded. What would have happened if the prong had been filed instead?
Options $\nu_X=516\ \text{Hz}$ or $508\ \text{Hz}$.
Wax Wax lowers $\nu_X$. From $516$ it would move towards $512$ and the beats would slow; from $508$ it moves away and the beats speed up โ€” as observed.
Answer $\nu_X=508\ \text{Hz}$ (after waxing, $506\ \text{Hz}$).
Filing Filing would raise $\nu_X$ towards $512\ \text{Hz}$: the beats would slow, stop at $512\ \text{Hz}$, and then speed up again if filing continued past it.
โœ๏ธ Example 4 โ€” Tuning a violin string
A violinist's A string ($L=0.33\ \text{m}$, $\mu=0.60\ \text{g/m}$) is sounded with a $440\ \text{Hz}$ tuning fork and $3$ beats per second are heard. Tightening the string slightly makes the beats faster. (a) What is the string's frequency? (b) What is its tension, and what tension would put it exactly in tune?
(a) $437$ or $443\ \text{Hz}$. Tightening raises $\nu$; the beats speed up, so $\nu$ moved away from $440$: $\nu=443\ \text{Hz}$.
(b) $T=\mu(2L\nu)^2=6.0\times10^{-4}\times(2\times0.33\times443)^2=6.0\times10^{-4}\times292.4^2=51.3\ \text{N}$. In tune: $6.0\times10^{-4}\times(0.66\times440)^2=50.6\ \text{N}$.
Fraction $\dfrac{T'}{T}=\left(\dfrac{440}{443}\right)^2=0.9865$: let the tension off by about $1.4\%$ ($0.7\ \text{N}$).
Check A $1\%$ change in tension changes $\nu$ by about $\tfrac12\%$ (here $2.2\ \text{Hz}$) โ€” small adjustments, which is why tuning pegs turn so little. Interactive 2 starts with this string.
โœ๏ธ Example 5 โ€” Beats between two organ pipes
Two organ pipes, open at both ends, are $0.500\ \text{m}$ and $0.510\ \text{m}$ long and sound their fundamentals together. Take $v=340\ \text{m/s}$. Find the beat frequency.
Frequencies $\nu=\dfrac v{2L}$: $\dfrac{340}{1.000}=340\ \text{Hz}$ and $\dfrac{340}{1.020}=333.3\ \text{Hz}$.
Beats $340-333.3=6.7\ \text{Hz}$ โ€” about $7$ beats per second, clearly audible.
Check A $2\%$ difference in length gives a $2\%$ difference in frequency: $0.02\times340=6.8\ \text{Hz}$, close to the exact $6.7$ โœ“ (the shorter pipe is the higher).
โœ๏ธ Example 6 โ€” Beats from a warm pipe
Two identical organ pipes each sound $300\ \text{Hz}$ at $20\ ^\circ\text{C}$. One of them is warmed by sunlight through a window to $30\ ^\circ\text{C}$. How many beats per second are heard when they are sounded together? (Ignore the expansion of the pipe.)
Idea Same length, so $\nu\propto v\propto\sqrt T$ (Section 14.2).
Numbers $\nu'=300\sqrt{\dfrac{303}{293}}=300\times1.0169=305.1\ \text{Hz}$.
Answer About $5$ beats per second ($5.1$).
Note This is why an organist tunes the organ at the temperature at which it will be played โ€” and why wind instruments go sharp as they warm up in a concert.
Interactive 2 โ€” Tune the String by Ear

A violin A string ($L=0.33\ \text{m}$, $\mu=0.60\ \text{g/m}$) is played against a $440\ \text{Hz}$ tuning fork. The graph shows $2\ \text{s}$ of the combined sound and the counter counts the loudness maxima as they pass. Adjust the tension until the beats slow to nothing. Does tightening first speed the beats up or slow them down โ€” and what does that tell you? (๐Ÿ”Š plays the string and the fork together.)

51.3 N
string frequency $\tfrac1{2L}\sqrt{T/\mu}$โ€”
beat frequencyโ€”
beats countedโ€”
tension for zero beatsโ€”
โš ๏ธ Common mistakes

1. Beat frequency $=\dfrac{\nu_1-\nu_2}2$. That is the frequency of the amplitude factor $\cos\pi(\nu_1-\nu_2)t$; the loudness peaks twice per cycle of it, so the beat frequency is $|\nu_1-\nu_2|$.

2. "Beats tell you which note is higher." They give only the difference; you must change one frequency in a known direction to decide.

3. Wax raises the frequency. Wax adds mass and lowers it; filing the prongs raises it.

4. Beats between very different frequencies. $256\ \text{Hz}$ and $320\ \text{Hz}$ do not give $64$ audible beats a second; they are heard as two notes. Beats are heard for differences up to roughly $10$โ€“$15\ \text{Hz}$.

5. Confusing beats with standing waves. Standing waves come from equal frequencies travelling in opposite directions and vary in space (nodes); beats come from slightly different frequencies at one place and vary in time.

๐Ÿš€ Beyond the board syllabus (JEE Advanced) โ€” the Doppler effect

The Doppler effect was removed from the NCERT syllabus and is not listed in the recent JEE Main or NEET UG syllabi, but JEE Advanced still includes the Doppler effect in sound (always check the official syllabus for your exam year). An ambulance siren sounds higher as it approaches and lower as it moves away. With the air at rest, sound speed $v$, a source moving at $v_s$ and an observer moving at $v_o$ along the line joining them, the observed frequency is

$$\nu'=\nu\,\frac{v+v_o}{v-v_s}$$

with $v_o$ positive when the observer moves towards the source and $v_s$ positive when the source moves towards the observer (change the signs for motion away). A moving source squashes the wavelength ahead of it to $\dfrac{v-v_s}{\nu}$; a moving observer meets the wave crests more often. For a $680\ \text{Hz}$ siren at $20\ \text{m/s}$ and $v=340\ \text{m/s}$, a bystander hears $680\times\dfrac{340}{320}=722.5\ \text{Hz}$ as it approaches and $680\times\dfrac{340}{360}=642\ \text{Hz}$ as it recedes.

Doppler beats. Sound reflected from a moving object comes back shifted, and the echo beats with the original: for an object approaching at a small speed $u$, the beat frequency is about $\dfrac{2u}{v}\nu$. Police radar guns and medical Doppler ultrasound (measuring the speed of blood in an artery) work this way, with the frequency difference measured electronically.

Exercises

Take the speed of sound in air as $340\ \text{m/s}$ unless told otherwise. Give answers to 2โ€“3 significant figures.

I
Beat frequency.
  • 1 Two tuning forks of $384\ \text{Hz}$ and $380\ \text{Hz}$ are sounded together. Find the beat frequency and the beat period.
  • 2 A fork sounded with a $440\ \text{Hz}$ standard gives a loudness maximum every $0.20\ \text{s}$. What are the possible frequencies of the fork?
  • 3 Sources of $498\ \text{Hz}$ and $502\ \text{Hz}$ sound together. What frequency is heard, and how many beats per second?
  • 4 Why are no beats heard when notes of $256\ \text{Hz}$ and $320\ \text{Hz}$ are played together?
  • 5 At a point, $s=0.010\cos(600\pi t)+0.010\cos(606\pi t)$ (SI). Find the beat frequency and the greatest value of $s$.
  • 6 How many beats are heard in $10\ \text{s}$ when forks of $256\ \text{Hz}$ and $259.5\ \text{Hz}$ sound together?
  • 7 For a beat frequency of $5\ \text{Hz}$, how long is it from a loudness maximum to the next silence?
  • 8 In $s=2a\cos\pi(\nu_1-\nu_2)t\,\cos\pi(\nu_1+\nu_2)t$, the slowly varying factor has frequency $\dfrac{\nu_1-\nu_2}{2}$. Explain why the beat frequency is nevertheless $\nu_1-\nu_2$.
1 $4\ \text{Hz}$; $T_\text{beat}=0.25\ \text{s}$.
2 $\nu_\text{beat}=\dfrac1{0.20}=5\ \text{Hz}$: $435\ \text{Hz}$ or $445\ \text{Hz}$.
3 $500\ \text{Hz}$; $4$ beats per second.
4 The difference, $64\ \text{Hz}$, is far too fast for the ear to follow as a rise and fall of loudness; the two notes are heard separately (as a musical interval).
5 $300\ \text{Hz}$ and $303\ \text{Hz}$: $3$ beats per second; $s_\text{max}=0.020$ (in step).
6 $3.5\ \text{Hz}\times10\ \text{s}=35$ beats.
7 Half a beat period: $\dfrac{1}{2\times5}=0.10\ \text{s}$.
8 Loudness depends on the size of the amplitude, $|2a\cos\pi(\nu_1-\nu_2)t|$. During one cycle of the cosine, its size reaches $2a$ twice (at $+2a$ and at $-2a$), so the loudness peaks twice per cycle: $2\times\dfrac{\nu_1-\nu_2}2=\nu_1-\nu_2$ times a second.
II
Finding unknown frequencies.
  • 1 Fork $B$ gives $6$ beats per second with fork $A$ of $256\ \text{Hz}$. Loading $B$ with wax reduces the beats to $4$ per second. Find the frequency of $B$.
  • 2 In part 1, suppose instead that waxing $B$ increased the beats to $8$ per second. Find the frequency of $B$.
  • 3 A fork $X$ gives $5$ beats per second with a $512\ \text{Hz}$ fork. Filing the prongs of $X$ makes the beats faster. Find the frequency of $X$.
  • 4 A sonometer wire sounding $200\ \text{Hz}$ gives $5$ beats per second with a fork. When the tension is increased slightly the beats fall to $3$ per second. Find the fork's frequency.
  • 5 Twenty-five tuning forks are arranged in order of increasing frequency; each gives $3$ beats per second with the next, and the last has twice the frequency of the first. Find the frequencies of the first and the last.
  • 6 Two strings of a guitar meant to sound $220\ \text{Hz}$ give $2$ beats per second; one is exactly in tune and the other is known to be sharp. By what percentage should the tension of the sharp string be reduced?
  • 7 A fork gives $4$ beats per second with the fundamental of an open pipe $0.50\ \text{m}$ long. When the pipe is made slightly longer, the beats become faster. Find the fork's frequency.
  • 8 Explain why loading a fork with wax lowers its frequency, and filing its prongs raises it.
1 $250$ or $262\ \text{Hz}$. Wax lowers $\nu_B$; the beats slow, so $\nu_B$ moved towards $256$: $\nu_B=262\ \text{Hz}$.
2 The beats speed up, so $\nu_B$ moved away from $256$ downwards: $\nu_B=250\ \text{Hz}$.
3 $507$ or $517\ \text{Hz}$. Filing raises $\nu_X$; the beats speed up, so it moved away from $512$ upwards: $\nu_X=517\ \text{Hz}$.
4 $195$ or $205\ \text{Hz}$. Raising the tension raises the wire's frequency; the beats slow, so the wire moved towards the fork: the fork is above, $205\ \text{Hz}$.
5 $\nu_{25}=\nu_1+24\times3=2\nu_1$: $\nu_1=72\ \text{Hz}$, $\nu_{25}=144\ \text{Hz}$.
6 Sharp string: $222\ \text{Hz}$. $\dfrac{T'}{T}=\left(\dfrac{220}{222}\right)^2=0.982$: reduce the tension by $1.8\%$.
7 Pipe: $\dfrac{340}{1.0}=340\ \text{Hz}$; fork $336$ or $344\ \text{Hz}$. Lengthening lowers the pipe's frequency and the beats speed up, so the pipe moved away from the fork: the fork is above, $344\ \text{Hz}$.
8 A prong vibrates like a springโ€“block system with $\omega=\sqrt{k/m}$: wax adds mass and lowers $\omega$; filing the tips removes mass from where the motion is largest and raises it.
III
Applications.
  • 1 (Modelling) A piano tuner hears $1.5$ beats per second between two of the three strings that sound A$_4$; one of them is exactly $440\ \text{Hz}$. What are the possible frequencies of the other?
  • 2 (Modelling) The propellers of a twin-engine aircraft produce tones of $100\ \text{Hz}$ and $103\ \text{Hz}$. Describe what the passengers hear, and how the pilot can stop it.
  • 3 (Modelling) Two identical open pipes $0.80\ \text{m}$ long sound their fundamentals; one is at $20\ ^\circ\text{C}$ ($v=343\ \text{m/s}$), the other at $30\ ^\circ\text{C}$. Find the beat frequency.
  • 4 Two identical wires carry tensions of $100\ \text{N}$ and $104\ \text{N}$. The first has a fundamental frequency of $200\ \text{Hz}$. Find the beat frequency when both sound their fundamentals.
  • 5 A pipe closed at one end, $0.170\ \text{m}$ long, gives $4$ beats per second with a $504\ \text{Hz}$ fork. By how much must the pipe be shortened to remove the beats?
  • 6 Sources of $300\ \text{Hz}$ and $304\ \text{Hz}$ are in phase at $t=0$. When are they first in antiphase, and when next in phase?
1 $438.5\ \text{Hz}$ or $441.5\ \text{Hz}$.
2 A single drone near $101.5\ \text{Hz}$ whose loudness throbs $3$ times a second. The pilot adjusts one engine's speed until the throbbing stops (the propellers are "synchronised").
3 $\nu_{20}=\dfrac{343}{1.60}=214.4\ \text{Hz}$; $v_{30}=343\sqrt{\dfrac{303}{293}}=348.8\ \text{m/s}$, $\nu_{30}=218.0\ \text{Hz}$: $3.6$ beats per second.
4 $\nu\propto\sqrt T$: $\nu_2=200\sqrt{1.04}=204.0\ \text{Hz}$: $4.0$ beats per second ($3.96$).
5 Pipe: $\dfrac{340}{4\times0.170}=500\ \text{Hz}$. For $504\ \text{Hz}$: $L=\dfrac{340}{4\times504}=0.1687\ \text{m}$: shorten by $1.3\ \text{mm}$.
6 The phase difference grows at $2\pi\times4\ \text{rad/s}$: antiphase ($\pi$) at $t=0.125\ \text{s}$, in phase again ($2\pi$) at $t=0.25\ \text{s}$.
IV
Harder problems.
  • 1 A sonometer wire gives $4$ beats per second with a tuning fork when its vibrating length is $0.48\ \text{m}$, and also $4$ beats per second when its length is $0.50\ \text{m}$ (same tension). Find the frequency of the fork.
  • 2 Two open pipes $0.500\ \text{m}$ and $0.510\ \text{m}$ long give $6.7$ beats per second when sounded together in their fundamental modes. Find the speed of sound.
  • 3 Using $s=2a\cos\pi(\nu_1-\nu_2)t\,\cos\pi(\nu_1+\nu_2)t$, find all the times at which the sound is silent, and show that successive silences are $\dfrac1{|\nu_1-\nu_2|}$ apart.
1 $\nu\propto\dfrac1L$: the shorter wire is higher, so $\nu_f+4$ at $0.48\ \text{m}$ and $\nu_f-4$ at $0.50\ \text{m}$. $(\nu_f+4)\times0.48=(\nu_f-4)\times0.50$ gives $0.02\nu_f=3.92$, $\nu_f=196\ \text{Hz}$.
2 $\dfrac v2\left(\dfrac1{0.500}-\dfrac1{0.510}\right)=6.7$: $\dfrac v2\times0.0392=6.7$, $v=3.4\times10^2\ \text{m/s}$ ($342$).
3 Silent when $\cos\pi(\nu_1-\nu_2)t=0$: $\pi(\nu_1-\nu_2)t=\dfrac\pi2,\dfrac{3\pi}2,\dots$, i.e. $t=\dfrac{2m+1}{2|\nu_1-\nu_2|}$, $m=0,1,2,\dots$; successive values differ by $\dfrac1{|\nu_1-\nu_2|}$.
๐ŸŽฏ Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board ยท 3 marks What are beats? Two waves $s_1=a\cos2\pi\nu_1t$ and $s_2=a\cos2\pi\nu_2t$, with $\nu_1$ slightly greater than $\nu_2$, reach the ear together. Show that the beat frequency is $\nu_1-\nu_2$.

Definition The periodic rise and fall of loudness when two sounds of slightly different frequency are heard together. [1]
Sum $s=2a\cos\pi(\nu_1-\nu_2)t\,\cos\pi(\nu_1+\nu_2)t$: a tone of frequency $\dfrac{\nu_1+\nu_2}2$ with amplitude $A(t)=2a\cos\pi(\nu_1-\nu_2)t$. [1]
Beats $|A|=2a$ when $\pi(\nu_1-\nu_2)t=m\pi$, i.e. $t=\dfrac{m}{\nu_1-\nu_2}$: maxima are $\dfrac1{\nu_1-\nu_2}$ apart, so $\nu_\text{beat}=\nu_1-\nu_2$. [1]
Exam tip Say explicitly that the loudness depends on $|A|$, which peaks twice per cycle of the cosine โ€” that is the step that turns $\dfrac{\nu_1-\nu_2}2$ into $\nu_1-\nu_2$.

E2. Board ยท 2 marks A tuning fork $A$ of $320\ \text{Hz}$ gives $5$ beats per second with fork $B$. When $B$ is filed slightly, the beats fall to $3$ per second. Find the original frequency of $B$, giving your reasoning.

Options $\nu_B=315\ \text{Hz}$ or $325\ \text{Hz}$. [ยฝ]
Reason Filing raises $\nu_B$. The beats slowed, so $\nu_B$ moved towards $320\ \text{Hz}$ from below. [1]
Answer $\nu_B=315\ \text{Hz}$ (after filing, $317\ \text{Hz}$). [ยฝ]
Exam tip Write both candidates first, then eliminate one โ€” a bare answer without the reasoning loses most of the marks.

E3. NEET A tuning fork of $480\ \text{Hz}$ produces $10$ beats per second with a sonometer wire. When the tension of the wire is increased slightly, the beats fall to $8$ per second. The original frequency of the wire was (a) $470\ \text{Hz}$ (b) $490\ \text{Hz}$ (c) $480\ \text{Hz}$ (d) $460\ \text{Hz}$

Idea $470$ or $490\ \text{Hz}$. More tension raises the wire's frequency; fewer beats means it moved towards $480$, so it was below.
Answer (a) $470\ \text{Hz}$ (now $472\ \text{Hz}$).
Exam tip (b) is the answer if the tension had been decreased; (d) doubles the beat count.

E4. JEE Main (Numerical value) Twenty-six tuning forks are arranged in increasing order of frequency. Each produces $4$ beats per second with the next, and the frequency of the last fork is twice that of the first. Find the frequency of the first fork in hertz.

Idea $26$ forks have $25$ gaps of $4\ \text{Hz}$: $\nu_{26}=\nu_1+100$.
Answer $\nu_1+100=2\nu_1$: $\nu_1=100$ (and $\nu_{26}=200\ \text{Hz}$).
Exam tip The common slip is $26$ gaps instead of $25$ โ€” count fence posts, not fence panels.

E5. JEE Advanced (One or more correct) Two sound waves of equal amplitude and frequencies $\nu_1$ and $\nu_2$ ($\nu_1-\nu_2=\Delta\nu$, small) are heard together. Which statements are correct? (A) The loudness reaches a maximum $\Delta\nu$ times per second. (B) The amplitude of the resultant varies as a cosine of frequency $\Delta\nu$. (C) The pitch heard has frequency $\dfrac{\nu_1+\nu_2}2$. (D) At an instant of silence the two waves at the ear are in antiphase.

(A) $\nu_\text{beat}=\Delta\nu$ โœ“.
(B) The amplitude factor is $2a\cos\pi\Delta\nu\,t$, of frequency $\dfrac{\Delta\nu}2$ โœ—.
(C) The rapidly varying factor is $\cos\pi(\nu_1+\nu_2)t$ โœ“.
(D) Silence means the displacements cancel at every instant near then: phase difference $\pi$ โœ“.
Answer (A), (C), (D).
Exam tip (B) is the classic trap: "amplitude frequency" $\dfrac{\Delta\nu}{2}$ versus "beat frequency" $\Delta\nu$.

E6. JEE Advanced (Numerical value; beyond NCERT โ€” Doppler effect) A student walks at $1.7\ \text{m/s}$ towards a large wall, holding a vibrating $340\ \text{Hz}$ tuning fork. The speed of sound is $340\ \text{m/s}$. How many beats per second does she hear between the sound coming directly from the fork and its echo from the wall? (Give one decimal place.)

Note The Doppler effect is beyond the NCERT syllabus (see the ๐Ÿš€ box); this is JEE Advanced practice.
At the wall Source approaching: the wall receives $\nu'=340\times\dfrac{340}{340-1.7}=341.71\ \text{Hz}$ and reflects it.
Echo heard She moves towards this "source": $\nu''=341.71\times\dfrac{340+1.7}{340}=343.42\ \text{Hz}$. The direct sound reaches her at $340\ \text{Hz}$ (no relative motion).
Answer $343.42-340=3.4$ beats per second.
Exam tip For small speeds, $\nu_\text{beat}\approx\dfrac{2u}{v}\nu=\dfrac{2\times1.7}{340}\times340=3.4$ โ€” a quick check worth remembering.
๐Ÿ”ฌ Researchโ€“Teaching NexusResearch-tutored ยท optional, about 45 minutes

Research question. How small a difference in frequency can your ears detect using beats โ€” and at what difference do beats turn into "roughness" and then into two separate notes?

Why it matters. Beats are how pianos, guitars and orchestras are tuned the world over, and the same mixing of two close frequencies (the "heterodyne" principle) is at the heart of every radio and mobile-phone receiver. Psychoacoustics โ€” how the ear and brain hear such mixtures โ€” guides the design of hearing aids, audio compression (MP3 and its successors) and the tuning systems of music from Indian ragas to Western harmony.

What to do. Use two free tone-generator apps (or two phones), each able to play a steady sine tone at a set frequency.

  1. Play $440\ \text{Hz}$ on one and $441\ \text{Hz}$ on the other, held close together. Count the beats in $30\ \text{s}$ and compare with the prediction.
  2. Increase the second frequency in steps: $442$, $444$, $448$, $455$, $470$, $500\ \text{Hz}$. For each, note whether you hear clear beats, a rough buzzing sound, or two separate tones.
  3. Record the difference at which beats can no longer be counted, and the difference at which you first hear two separate pitches.
  4. Repeat around $220\ \text{Hz}$ and $880\ \text{Hz}$. Do the thresholds stay the same in hertz, or the same as a fraction of the frequency?
  5. Pool results from your class and plot each threshold against the base frequency.

Tools: Two phones or tone-generator apps, a stopwatch, a spreadsheet.

Hand in: A table of your observations; a graph of the thresholds against base frequency; a paragraph explaining why beats are useful for tuning but useless for telling apart two very different notes.

Going further. Find out what the "critical band" of the ear is and how its width changes with frequency. Why does a piano tuner listen for beats between harmonics (e.g. the 3rd harmonic of one note and the 2nd of another) when tuning a fifth?

Instructor note: most students count beats reliably up to about $10$โ€“$15\ \text{Hz}$; between about $15$ and $50$โ€“$100\ \text{Hz}$ (depending on base frequency) the sound is "rough"; two pitches are heard once the difference exceeds roughly the critical bandwidth, which is about $100\ \text{Hz}$ below $500\ \text{Hz}$ and about $15$โ€“$20\%$ of the frequency above it. Phone speakers distort at high volume, so keep the level moderate.

๐Ÿงช Self-check quiz

0/5 correct

Q1. Tuning forks of $256\ \text{Hz}$ and $262\ \text{Hz}$ are sounded together. The number of beats per second is

Q2. When notes of $498\ \text{Hz}$ and $502\ \text{Hz}$ are heard together, the pitch heard is

Q3. A prong of a tuning fork is loaded with a little wax. Its frequency

Q4. A fork of unknown frequency gives $4$ beats per second with a $300\ \text{Hz}$ fork. Loading the unknown fork with wax makes the beats faster. The unknown frequency was

Q5. At the moments of silence between beats, the two sound waves at the ear are