Beats

Q: What is the formula for beat frequency?

Beat frequency equals the absolute difference of the two frequencies, and the beat period is its reciprocal. For forks of 256 and 260 hertz there are 4 beats per second, one every quarter of a second, and the tone heard has the average frequency, 258 hertz.

PhysicsWavesFor JEE aspirants

Beats are the regular rise and fall in loudness heard when two sounds of nearly equal frequency are played together. The number of beats per second equals the difference of the two frequencies, $f_{b} = ∣ f_{1} - f_{2} ∣$ , while the pitch heard is their average. Beats come from superposition in time: this page derives them, explains why only slow beats are heard, and shows how musicians and examiners use beats, with waxing and filing, to find an unknown frequency. A favourite of JEE Main and NEET.

On this page1What beats are2Derivation3Beat period4Hearing limit5Unknown frequency6Beats vs interference

Key Formulas - Quick Reference

★ Must learnBeat frequency: $f_{b} = ∣ f_{1} - f_{2} ∣$ beats per second; beat period $T_{b} = \frac{1}{∣ f _{1} - f _{2} ∣}$
Resultant: $y = 2 A cos [2 π \frac{f _{1} - f _{2}}{2} t] sin [2 π \frac{f _{1} + f _{2}}{2} t]$
★ Must learnFrequency (pitch) heard: $\frac{f _{1} + f _{2}}{2}$ ; amplitude varies as $R = 2 A cos π (f_{1} - f_{2}) t$
Loudest at $t = 0, \frac{1}{Δ f}, \frac{2}{Δ f}, \dots$ ; softest at $t = \frac{1}{2Δ f}, \frac{3}{2Δ f}, \dots$
Intensity: $I = 4 I_{0} cos^{2} π (f_{1} - f_{2}) t$ (equal tones); unequal: $A$ swings between $A_{1} + A_{2}$ and $∣ A_{1} - A_{2} ∣$
★ Must learnUnknown fork: $f_{X} = f_{0} \pm f_{b}$ ; wax lowers $f_{X}$ , filing raises it
★ Must learnDistinct beats are heard only for $f_{b} ≲ 10 Hz$

1. What Are Beats?

Sound two sources of almost the same frequency together, such as two tuning forks of $256$ and $260 Hz$ . You hear one tone whose pitch is the average of the two, but its loudness repeatedly grows and dies away instead of staying steady. These periodic variations in loudness are called beats.

If the frequency of one source is changed, the rate of the loudness variation changes too. This rate is the beat frequency. As the two frequencies come closer the beats slow down, and when they are equal the beats stop. A musician tunes a guitar string against a reference note exactly this way: adjusting the tension while listening until the beats become so slow that none are heard, when the two are in tune.

2. Mathematical Treatment

Let the two waves reaching a point have frequencies $f_{1}$ and $f_{2}$ and equal amplitudes $A$ :

$y_{1} = A sin 2 π f_{1} t$ and $y_{2} = A sin 2 π f_{2} t$ .
Superpose, using $sin C + sin D = 2 cos \frac{C - D}{2} sin \frac{C + D}{2}$ :
$y = 2 A cos [2 π \frac{f _{1} - f _{2}}{2} t] sin [2 π \frac{f _{1} + f _{2}}{2} t]$
Write $y = R sin [2 π \frac{f _{1} + f _{2}}{2} t]$ : the particle vibrates at the average frequency $\frac{f _{1} + f _{2}}{2}$ with a slowly varying amplitude
$R = 2 A cos [2 π \frac{f _{1} - f _{2}}{2} t]$

Figure 1: Where the two tones are in step the sum is loud (max); half a beat period later they are in opposite phase and the sum is nearly silent (min). The envelope repeats every

T_{b} = \frac{1}{∣ f _{1} - f _{2} ∣}

2.1 Times of maximum and minimum loudness

Maximum when $∣ R ∣ = 2 A$ , i.e. $cos [π (f_{1} - f_{2}) t] = \pm 1$ : $π (f_{1} - f_{2}) t = N π$ , so $t = 0, \frac{1}{f _{1} - f _{2}}, \frac{2}{f _{1} - f _{2}}, \dots$
Minimum when $R = 0$ : $π (f_{1} - f_{2}) t = (2 N + 1) \frac{π}{2}$ , so $t = \frac{1}{2 ( f _{1} - f _{2} )}, \frac{3}{2 ( f _{1} - f _{2} )}, \dots$ , exactly midway between the maxima.

★ Must learnBeat period (time between successive maxima, or minima):

T_{b} = \frac{1}{∣ f _{1} - f _{2} ∣}

. Beat frequency (beats heard per second):

f_{b} = \frac{1}{T _{b}} = ∣ f_{1} - f_{2} ∣

Figure 2:

I = 4 I_{0} cos^{2} π (f_{1} - f_{2}) t

. One loud-soft cycle is one beat, so the number of beats per second is

∣ f_{1} - f_{2} ∣

, not half of it.

Exam Trick

One beat = one loud and one soft. The amplitude $cos π Δ f t$ repeats every $2/Δ f$ , but loudness depends on $∣ R ∣$ (or $R^{2}$ ), which repeats every $1/Δ f$ . So the beat frequency is $∣ f_{1} - f_{2} ∣$ , not $∣ f_{1} - f_{2} ∣/2$ . In $t$ seconds you hear $∣ f_{1} - f_{2} ∣ t$ beats.

2.2 Unequal amplitudes

If the amplitudes are $A_{1} \neq = A_{2}$ , the resultant amplitude swings between $A_{1} + A_{2}$ (loud) and $∣ A_{1} - A_{2} ∣$ (soft) at the same beat frequency. The sound never goes completely silent, so the beats are less distinct: $\frac{I _{m a x}}{I _{m i n}} = (\frac{A _{1} + A _{2}}{A _{1} - A _{2}})^{2}$ .

Figure 3: Beats with phasors. Seen from a frame turning with the

f_{1}

phasor, the

f_{2}

phasor turns slowly, once every

T_{b} = \frac{1}{∣ f _{1} - f _{2} ∣}

. The resultant goes

2 A \to 2 A \to 0 \to 2 A

: one loud-soft cycle per turn. With unequal amplitudes it swings between

A_{1} + A_{2}

and

∣ A_{1} - A_{2} ∣

Key idea

You hear one tone at the average frequency

\frac{f _{1} + f _{2}}{2}

, whose loudness pulses

∣ f_{1} - f_{2} ∣

times a second.

3. Why Only Slow Beats Are Heard

The sensation of a sound persists in the ear for about $\frac{1}{10} s$ (the same fact that sets the minimum distance for an echo in the Sound Waves concept). If the loudness rises and falls more than about $10$ times a second, successive maxima merge and the ear cannot count them. So distinct beats are heard only when $∣ f_{1} - f_{2} ∣$ is less than about $10 Hz$ . For a larger difference the ear hears a rough or harsh tone, and for a large difference it hears the two notes separately.

Interference

Two waves of the same frequency. Loud and soft points are fixed in space (path difference decides). Pattern is steady in time.

Beats

Two waves of slightly different frequencies. Loud and soft moments alternate in time at one point. Also called interference in time.

Quick Recall: tap to check

Forks of

384 Hz

and

388 Hz

sound together. What is heard?

386 Hz

tone whose loudness peaks 4 times a second.

Can beats be heard between

300 Hz

and

350 Hz

No:

50 Hz

is far above about

10 Hz

Two equal tones of intensity

I_{0}

beat. Maximum and minimum intensity?

4 I_{0}

and

0

4. Finding an Unknown Frequency with Beats

An unknown fork $X$ sounded with a known fork $f_{0}$ gives $b$ beats per second. Then $∣ f_{X} - f_{0} ∣ = b$ , so $f_{X} = f_{0} + b$ or $f_{0} - b$ . Beats alone cannot tell which. A small, known change to $X$ decides it.

Figure 4: The standard beats set-up. The known fork (

f_{0} = 256 Hz

) and fork

X

are struck together and the beats are counted:

f_{X} = f_{0} \pm b

. A blob of wax on a prong of

X

lowers

f_{X}

; filing a prong raises it. Whether the beats then rise or fall settles the sign.

Figure 5:

f_{b} = ∣ f - 256∣

. Four beats per second fits

f = 252

260 Hz

. Lowering

f

(waxing) moves left on the V: beats rise from 252 but fall from 260, which settles the answer.

Change made	Effect on frequency	Reason
Load a fork's prong with wax	Decreases	Larger vibrating mass
File a fork's prong	Increases	Smaller mass (filing the prong tips)
Increase the tension of a string	Increases	$f \propto T$
Increase the vibrating length of a string or pipe	Decreases	$f \propto 1/ L$
Warm the air in a pipe	Increases	$v \propto T$ , $f \propto v$

Figure 6: Sonometer and fork. The wire's frequency

f = \frac{1}{2 l} \frac{T}{μ}

is compared with a

256 Hz

fork: 4 beats per second means

f = 252

260 Hz

. Tightening the wire raises

f

; the beats fall only if it started at

252 Hz

(Solved Example 6).

Figure 7: Wax lowers the frequency, filing raises it. Ask: does the change move

f_{X}

away from

f_{0}

(beats rise) or towards it (beats fall)? (Special case: if the beats stay the same after waxing,

f_{X}

has jumped from

f_{0} + b

f_{0} - b

Key idea

Move

f_{X}

a little and watch the beats: towards

f_{0}

means fewer beats, away from

f_{0}

means more.

JEE Advanced

When the beats do not change. If $f_{X} = f_{0} + b$ and waxing lowers it by exactly $2 b$ , it lands on $f_{0} - b$ and the beat count stays $b$ . So "the beats remain the same after waxing" means $f_{X}$ was $f_{0} + b$ . With filing, the same result means $f_{X}$ was $f_{0} - b$ . Beats also arise when a moving source's Doppler-shifted note meets the original note (Doppler Effect concept).

Exam Trick

List both, then nudge. Write $f_{X} = f_{0} + b$ and $f_{0} - b$ side by side. Move each a little in the stated direction (wax, a longer string or lower tension: down; filing, a shorter string, higher tension or a warmer pipe: up) and keep the one whose beat count changes the way the question says.

Figure 8: Beats from a Doppler-shifted echo. Running at

2 m s^{- 1}

towards a wall with a

512 Hz

fork (

v = 340 m s^{- 1}

), he hears the fork at

512 Hz

and the echo at

f \frac{v + u}{v - u} \approx 518.1 Hz

: about

6

beats per second (Doppler Effect concept).

Quick Recall: tap to check

A fork gives 5 beats per second with a

400 Hz

fork; after filing it gives 7. Find its frequency.

405 Hz

: filing raised it further from 400.

Why does wax lower a fork's frequency?

It adds mass to the prong, so the prong vibrates more slowly.

Two tones beat 3 times a second. How long is one beat?

T_{b} = \frac{1}{3} s

Figure 9: Revision map: cause,

f_{b} = ∣ f_{1} - f_{2} ∣

, intensity, hearing limit, unknown frequency and common set-ups.

5. Solved Examples

Solved Example 1

Two tuning forks of frequencies

256 Hz

and

260 Hz

are sounded together. Find (a) the beat frequency (b) the beat period (c) the frequency heard (d) the times of maximum loudness if they are in phase at

t = 0

Solution:

(a) $f_{b} = 260 - 256 = 4 Hz$ . (b) $T_{b} = \frac{1}{4} = 0.25 s$ . (c) $\frac{256 + 260}{2} = 258 Hz$ . (d) $t = 0, 0.25, 0.50, 0.75 s, \dots$ (minima at $0.125, 0.375 s, \dots$ ).

Solved Example 2

The displacement at a point due to two sounds is

y = 10 sin 400 π t + 10 sin 408 π t

(in

μ m

t

in s). Find the beat frequency, the frequency heard and the maximum amplitude.

Solution:

$f_{1} = \frac{400 π}{2 π} = 200 Hz$ , $f_{2} = 204 Hz$ . Beat frequency $= 4 Hz$ ; frequency heard $= 202 Hz$ ; maximum amplitude $= 10 + 10 = 20 μ m$ .

Solved Example 3

A fork of unknown frequency gives

4

beats per second with a

256 Hz

fork. When its prong is loaded with a little wax, it gives

6

beats per second. Find its original frequency.

Solution:

$f_{X} = 256 \pm 4 = 252$ or $260 Hz$ . Wax lowers $f_{X}$ . From $260$ it would move towards $256$ and the beats would fall; from $252$ it moves away from $256$ and the beats rise, as observed.

Answer: $252 Hz$ .

Solved Example 4

A fork gives

5

beats per second with a

512 Hz

fork. After its prongs are filed slightly, it gives

3

beats per second. Find its original frequency.

Solution:

$f_{X} = 507$ or $517 Hz$ . Filing raises $f_{X}$ . From $507$ it moves towards $512$ (beats fall to 3, as observed); from $517$ it would move away (beats rise).

Answer: $507 Hz$ .

Solved Example 5

A fork gives

4

beats per second with a

256 Hz

fork. After waxing it still gives

4

beats per second. Find its original frequency.

Solution:

Before: $252$ or $260 Hz$ . Waxing lowers the frequency. From $252$ the beats would only rise. From $260$ , a drop of $8 Hz$ takes it to $252 Hz$ , again $4$ beats below $256$ .

Answer: $260 Hz$ (it has moved from $4$ above to $4$ below).

Solved Example 6

A sonometer wire gives

4

beats per second with a

256 Hz

fork. When the tension in the wire is increased slightly, the beats decrease. Find the original frequency of the wire.

Solution:

$f = 252$ or $260 Hz$ . More tension raises $f$ . The beats fall only if $f$ moves towards $256$ : from $252$ .

Answer: $252 Hz$ .

Solved Example 7

Two open organ pipes of lengths

50 cm

and

51 cm

sound their fundamentals together. How many beats per second are heard? (

v = 340 m s^{- 1}

)

Solution:

$f_{1} = \frac{340}{2 \times 0.50} = 340 Hz$ ; $f_{2} = \frac{340}{2 \times 0.51} \approx 333.3 Hz$ .

Answer: $f_{b} \approx 6.7$ beats per second.

Solved Example 8

Twenty-six tuning forks are arranged in order of increasing frequency. Each gives

3

beats per second with the next, and the last fork has twice the frequency of the first. Find the frequencies of the first and the last forks.

Solution:

The last fork is $25$ steps above the first: $f_{26} = f_{1} + 25 \times 3 = f_{1} + 75$ . Also $f_{26} = 2 f_{1}$ .

So $2 f_{1} = f_{1} + 75$ : $f_{1} = 75 Hz$ and $f_{26} = 150 Hz$ .

Solved Example 9

Two sounds of frequencies

200 Hz

and

250 Hz

are played together. The number of distinct beats heard per second is
(A)

50

(B)

25

(C)

450

(D) none

Solution:

The difference is $50 Hz$ , far above the roughly $10 Hz$ the ear can follow, so no distinct beats are heard.

Answer: (D).

Solved Example 10

Two sounds of slightly different frequencies have amplitudes in the ratio

3 : 1

. Find the ratio of maximum to minimum intensity during the beats.

Solution:

$\frac{I _{m a x}}{I _{m i n}} = (\frac{3 + 1}{3 - 1})^{2} = 4$ .

Answer: $4 : 1$ (the sound never becomes silent).

Solved Example 11

Forks of

512 Hz

and

516 Hz

are sounded together for

10 s

. How many beats are heard?

Solution:

$f_{b} = 4 Hz$ ; in $10 s$ : $4 \times 10 = 40$ beats.

Answer: 40.

Solved Example 12

A string gives

4

beats per second with a

512 Hz

fork. When its tension is slightly decreased, it gives

6

beats per second. The original frequency of the string is
(A)

508 Hz

(B)

516 Hz

(C)

504 Hz

(D)

520 Hz

Solution:

$f = 508$ or $516 Hz$ . Lower tension lowers $f$ . From $508$ it moves away from $512$ and the beats rise, as observed; from $516$ it would move towards $512$ and the beats would fall.

Answer: (A).

Practice Questions

Forks of $440 Hz$ and $437 Hz$ : find the beat period and the frequency heard.Answer: $T_{b} = 1/3 s$ ; $438.5 Hz$ .
A fork gives $6$ beats per second with a $400 Hz$ fork; after waxing, $4$ beats per second. Find its frequency.Answer: $406 Hz$ (waxing moved it towards $400$ ).
A string gives $5$ beats per second with a $300 Hz$ fork; decreasing the tension increases the beats. Frequency of the string?Answer: $295 Hz$ .
Two strings of frequencies $f$ and $f + 2$ beat. After $5 s$ how many loud moments have been heard (starting loud at $t = 0$ )?Answer: $11$ (at $t = 0, 0.5, \dots, 5 s$ ).
Two equal tones beating have $I_{m a x} = 8 units$ . Find the intensity of each tone and the average intensity.Answer: $2$ units each; average $4$ units.
Twenty forks in increasing order, each $4$ beats per second above the previous; the last is an octave of the first. Find the first.Answer: $f_{1} = 19 \times 4 = 76 Hz$ ; last $152 Hz$ .

Common Mistakes to Avoid

Watch out

Taking the beat frequency as $∣ f_{1} - f_{2} ∣/2$ . One beat is one loud-soft cycle: $f_{b} = ∣ f_{1} - f_{2} ∣$ .
Thinking the pitch heard is $f_{1}$ or $f_{2}$ . It is the average, $(f_{1} + f_{2}) /2$ .
Forgetting the two possibilities $f_{0} \pm b$ for an unknown fork.
Mixing up waxing and filing: wax (added mass) lowers the frequency, filing raises it.
Expecting beats for any two frequencies. Distinct beats need $∣ f_{1} - f_{2} ∣ ≲ 10 Hz$ .
Assuming the sound goes silent in every beat. It does only for equal amplitudes.
Confusing beats with interference: beats vary in time at one point; interference varies in space.

Frequently Asked Questions

What are beats in sound?

Beats are the periodic rise and fall in loudness heard when two sounds of slightly different frequencies are played together. The two waves drift in and out of step, so the resultant amplitude grows and dies away. The number of beats per second equals the difference of the two frequencies.

What is the formula for beat frequency?

Beat frequency is $f_{b} = ∣ f_{1} - f_{2} ∣$ , and the beat period is $1/∣ f_{1} - f_{2} ∣$ . For forks of 256 and 260 hertz there are 4 beats per second, one every quarter of a second, and the tone heard has the average frequency, 258 hertz.

Why can't we hear beats if the frequency difference is large?

The sensation of sound lasts about a tenth of a second in the ear. If loudness rises and falls more than about ten times a second, the maxima merge and cannot be counted. So distinct beats are heard only for frequency differences below about 10 hertz.

How do you find an unknown frequency using beats?

Sound the unknown fork with a known one and count b beats per second, so the unknown is the known frequency plus or minus b. Then wax or file the unknown fork. Waxing lowers its frequency and filing raises it; whether the beats increase or decrease decides the sign.

Does loading a tuning fork with wax increase or decrease its frequency?

Loading a prong with wax adds mass, so the fork vibrates more slowly and its frequency decreases. Filing the prongs removes mass and increases the frequency. These two small changes are used to decide the sign in beat problems.

What is the difference between beats and interference?

Interference comes from two waves of the same frequency and gives loud and soft places fixed in space. Beats come from two waves of slightly different frequencies and give loud and soft moments alternating in time at the same place, so beats are sometimes called interference in time.

How are beats asked in JEE Main?

JEE Main asks for beat frequency and period, the frequency heard, unknown frequency from waxing or filing, beats between strings or pipes of slightly different lengths or tensions, and series of forks. Always list both possibilities, known frequency plus or minus beats, before using the extra clue.

What should NEET students learn about beats?

For NEET learn that beat frequency is the difference of the two frequencies, that the ear hears beats only below about 10 per second, that waxing lowers and filing raises a fork's frequency, and how to decide an unknown frequency from the change in beats.

Previous year questions on Beats

5 questions from past papers, each with a step-by-step solution.

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