Beats

Superposition of Waves · Maharashtra HSC

Theory

What Beats Are

When two sound waves of the same amplitude but slightly different frequencies, travelling in the same direction, superpose, the intensity of the sound varies periodically with time. This phenomenon is the production of beats.

The occurrences of maximum intensity are called waxing; those of minimum intensity are called waning.
One waxing and the successive waning together constitute one beat.
The number of beats heard per second is the beat frequency.

Beats can be heard only if the frequency difference between the two superimposed waves is very small — in practice less than about $6$ to $7$ Hz for the normal human ear. At larger differences the individual beats cannot be distinguished.

Analytical Method to Determine Beat Frequency

Consider two sound waves of the same amplitude $a$ and slightly different frequencies $n_1$ and $n_2$ , arriving in phase at some point $x$ of the medium. The displacement due to each wave at that point at any instant is

$y_1 = a\sin\left[2\pi\left(n_1t - \dfrac{x}{\lambda_1}\right)\right], \qquad y_2 = a\sin\left[2\pi\left(n_2t - \dfrac{x}{\lambda_2}\right)\right]$

Take the listener at $x = 0$ for simplicity:

$y_1 = a\sin(2\pi n_1t), \qquad y_2 = a\sin(2\pi n_2t)$

By the principle of superposition,

$y = y_1 + y_2 = a\sin(2\pi n_1t) + a\sin(2\pi n_2t)$

Using $\sin C + \sin D = 2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$ ,

$y = 2a\sin\left[2\pi\left(\dfrac{n_1 + n_2}{2}\right)t\right]\cos\left[2\pi\left(\dfrac{n_1 - n_2}{2}\right)t\right]$

Rearranging,

$y = 2a\cos\left[\dfrac{2\pi(n_1 - n_2)}{2}t\right]\sin\left[\dfrac{2\pi(n_1 + n_2)}{2}t\right]$

Substituting

$A = 2a\cos\left[\dfrac{2\pi(n_1 - n_2)}{2}t\right] \qquad \text{and} \qquad n = \dfrac{n_1 + n_2}{2}$

we get

$y = A\sin(2\pi nt)$

This is the equation of a progressive wave of frequency $n$ and amplitude $A$ . The frequency $n$ is the mean of the frequencies $n_1$ and $n_2$ of the arriving waves, while the amplitude $A$ varies periodically with time.

Waxing

The intensity of sound is proportional to the square of the amplitude, so the resultant intensity is maximum when the amplitude is maximum. For maximum amplitude, $A = \pm 2a$ :

$2a\cos\left[2\pi\left(\dfrac{n_1 - n_2}{2}\right)t\right] = \pm 2a \qquad \therefore\ \cos\left[2\pi\left(\dfrac{n_1 - n_2}{2}\right)t\right] = \pm 1$

$\therefore\ 2\pi\left(\dfrac{n_1 - n_2}{2}\right)t = 0,\ \pi,\ 2\pi,\ 3\pi,\ \ldots$

$\therefore\ t = 0,\ \dfrac{1}{n_1 - n_2},\ \dfrac{2}{n_1 - n_2},\ \dfrac{3}{n_1 - n_2},\ \ldots$

The time interval between two successive maxima of sound is always $\dfrac{1}{n_1 - n_2}$ , so the period of beats is

$T = \dfrac{1}{n_1 - n_2}$

The number of waxings heard per second is the reciprocal of the period of waxing, so the beat frequency is

$N = n_1 - n_2$

Waning

The intensity is minimum when the amplitude is zero. For $A = 0$ :

$2a\cos\left[2\pi\left(\dfrac{n_1 - n_2}{2}\right)t\right] = 0 \qquad \therefore\ \cos\left[2\pi\left(\dfrac{n_1 - n_2}{2}\right)t\right] = 0$

$\therefore\ 2\pi\left(\dfrac{n_1 - n_2}{2}\right)t = \dfrac{\pi}{2},\ \dfrac{3\pi}{2},\ \dfrac{5\pi}{2},\ \ldots$

$\therefore\ t = \dfrac{1}{2(n_1 - n_2)},\ \dfrac{3}{2(n_1 - n_2)},\ \dfrac{5}{2(n_1 - n_2)},\ \ldots$

The time interval between two successive minima is also $\dfrac{1}{n_1 - n_2}$ , as expected, so the beat frequency in waning is again $N = n_1 - n_2$ .

Comparing the instants of successive waxing and waning shows that waxing and waning occur alternately with equal frequency. Hence the beat frequency is equal to the difference between the frequencies of the two interfering waves.

Diagram in Words

Draw two sine curves one below the other against a common time axis: the upper of frequency $n_1$ , the lower of slightly greater frequency $n_2$ , so that they start in step, drift out of step, and come back into step. Below them draw the resultant: a rapid oscillation at the mean frequency whose peaks are enclosed between two slowly varying dashed envelope curves that bulge out at the waxings and pinch to zero at the wanings. Mark the interval between two successive bulges as the period of beats, $T = \dfrac{1}{n_1 - n_2}$ .

Applications of Beats

Matching the frequencies of different musical instruments. Artists tune until no beats are heard; when the beat frequency is zero the instruments are in unison, their frequencies identical, and playing them together gives pleasant music.
Determining the speed of an aeroplane using Doppler RADAR. A microwave pulse of known frequency is sent towards the moving aeroplane; the Doppler principle applies twice, once for the signal received by the aeroplane and once for the signal reflected back and received at the source. The beats arising from the difference between the transmitted and the returned frequency give the speed. Traffic police use the same principle for the speed of a vehicle, and SONAR uses it for the speed of submarines.
Determining an unknown frequency. The unknown note is sounded together with a source of adjustable known frequency, which is adjusted until the beat frequency falls to zero; the two frequencies are then equal, so the unknown frequency is known.

Doppler Effect

When a source of sound and the listener are in relative motion, the listener detects a sound whose frequency is different from the actual or original frequency of the source. This is the Doppler effect. Four applications:

Traffic police use it to determine the speed of a vehicle and check whether the speed limit is exceeded.
SONAR works on it, determining the speed of submarines using a sound source and sensitive microphones.
Doppler ultrasonography and the echo cardiogram work on it; doctors use it to assess the direction and speed of blood flow and to identify circulation problems.
The speed of an aeroplane can be determined using Doppler RADAR.

Set of Tuning Forks

For a set of $P$ tuning forks in which each fork gives $Y$ beats per second with the next:

$n_L = n_F + (P-1)Y \qquad \text{(arranged in increasing order of frequency)}$

$n_L = n_F - (P-1)Y \qquad \text{(arranged in decreasing order of frequency)}$

$n_F$ and $n_L$ are the frequencies of the first and the last fork; for the $p$ -th fork replace $P$ by $p$ . "The last fork is an octave of the first" means $n_L = 2n_F$ .

Standard Results

One waxing and the successive waning together make one beat.
$y = 2a\cos\left[2\pi\left(\dfrac{n_1 - n_2}{2}\right)t\right]\sin\left[2\pi\left(\dfrac{n_1 + n_2}{2}\right)t\right]$ ; resultant frequency $= \dfrac{n_1 + n_2}{2}$ .
Beat frequency $N = n_1 - n_2$ ; period of beats $T = \dfrac{1}{n_1 - n_2}$ .
Beats are audible only for a frequency difference less than about $6$ to $7$ Hz.

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
What are beats?
Maharashtra HSC board question — asked in October 2008Answer 1 ↓
2
Explain what is Doppler effect in sound and state its any 'four' applications.
Maharashtra HSC board question — asked in March 2015Answer 2 ↓
3
Show that beats frequency is equal to frequency difference between two interfering waves. OR Prove that the frequency of beats is equal to the difference between the frequencies of the two sound notes giving rise to beats. OR Show that the beat frequency of two interfering sound waves is the difference between the individual frequencies of the two sound waves.
Maharashtra HSC board question — asked in July 2018; July 2023; March 2022; October 2008Answer 3 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

4
Wavelength of two sound notes in air are 85/170 m and 85/172 m. Each of these notes produce 4 beats per second with a third note of fixed frequency. Find the frequency of third note and velocity of sound in air.
Maharashtra HSC board question — asked in March 2008Answer 4 ↓
5
Wavelengths of two notes in the air are $\left(\dfrac{70}{153}\right)$ m and $\left(\dfrac{70}{157}\right)$ m. Each of these notes produces 8 beats per second with a tuning fork of fixed frequency. Find the velocity of sound in the air and frequency of the tuning fork.
Maharashtra HSC board question — asked in October 2014Answer 5 ↓
6
In a set, 21 tuning forks are arranged in a series of decreasing frequencies. Each tuning fork produces 4 beats per second with the preceding fork. If the first fork is an octave of the last fork, find the frequencies of the first and tenth fork.
Maharashtra HSC board question — asked in October 2015Answer 6 ↓
7
A set of 48 tuning forks is arranged in a series of descending frequencies such that each fork gives 4 beats per second with preceding one. The frequency of first fork is 1.5 times the frequency of the last fork, find the frequency of the first and 42 $^{nd}$ tuning fork.
Maharashtra HSC board question — asked in March 2016Answer 7 ↓
8
Two sound notes have wavelengths $\dfrac{83}{170}$ m and $\dfrac{83}{172}$ m in the air. These notes when sounded together produce 8 beats per second. Calculate the velocity of sound in the air and frequencies of the two notes.
Maharashtra HSC board question — asked in March 2009; March 2017Answer 8 ↓
9
A set of 12 tuning forks is arranged in order of increasing frequencies. Each fork produces 'Y' beats per second with the previous one. The last is an octave of the first. The fifth fork has a frequency of 90 Hz. Find 'Y' and frequency of the first and the last tuning forks.
Maharashtra HSC board question — asked in July 2017Answer 9 ↓
10
The wavelengths of two sound waves in air are $\dfrac{81}{173}$ m and $\dfrac{81}{170}$ m. They produce 10 beats per second. Calculate the velocity of sound in air.
Maharashtra HSC board question — asked in March 2018Answer 10 ↓
11
Two tuning forks of frequencies 320 Hz and 340 Hz are sounded together to produce sound wave. The velocity of sound in air is 326.4 m/s. Calculate the difference in wavelengths of these waves.
Maharashtra HSC board question — asked in February 2023Answer 11 ↓
12
Two tuning forks having frequencies 320 Hz and 340 Hz are sounded together to produce sound waves. The velocity of sound in air is 340 m/s. Find the difference in wavelength of these waves.
Maharashtra HSC board question — asked in February 2024Answer 12 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
One waxing and successive waning together constitute one beat.
2.↑ back to the question
i. When a source of sound and the listener are in relative motion, the listener detects a sound whose frequency is different from the actual or original frequency of the sound source. This is Doppler effect. ii. Application of Doppler effect: a. The principle of Doppler effect is used by traffic police to determine the speed of a vehicle to check whether speed limit is exceeded. b. SONAR works on the principle of Doppler effect for determining the speed of submarines using a sound source and sensitive microphones. c. Doppler ultrasonography and echo cardiogram work on Doppler effect d. Speed of an airplane can be determined using doppler RADAR.
3.↑ back to the question
i. Consider two sound waves, having same amplitude and slightly different frequencies $n_1$ and $n_2$ . ii. Let x be a point of the medium where they arrive in phase. iii. The displacement due to each wave at any instant of time at that point is given as $y_1=a\sin\left[2\pi\left(n_1t-\dfrac{x}{\lambda_1}\right)\right]$ $y_2=a\sin\left[2\pi\left(n_2t-\dfrac{x}{\lambda_2}\right)\right]$ iv. Let us assume for simplicity that the listener is at x = 0. $\therefore y_1=a\sin(2\pi n_1t)$ and $y_2=a\sin(2\pi n_2t)$ v. According to the principle of superposition of waves, y = $y_1+y_2$ $\therefore y=a\sin(2\pi n_1t)+a\sin(2\pi n_2t)$ $\therefore y=2a\sin\left[2\pi\left(\dfrac{n_1+n_2}{2}\right)t\right]\cos\left[2\pi\left(\dfrac{n_1-n_2}{2}\right)t\right]$ [By using formula, $\sin C+\sin D=2\sin\left(\dfrac{C+D}{2}\right)\cos\left(\dfrac{C-D}{2}\right)$ ] Rearranging the above equation, we get $y=2a\cos\left[\dfrac{2\pi(n_1-n_2)}{2}t\right]\sin\left[\dfrac{2\pi(n_1+n_2)}{2}t\right]$ vi. Let $2a\cos\left[\dfrac{2\pi(n_1-n_2)}{2}t\right]=A$ $\dfrac{n_1+n_2}{2}=n$ $\therefore y=A\sin(2\pi nt)$ This is the equation of a progressive wave having frequency 'n' and amplitude 'A'. The frequency 'n' is the mean of the frequencies $n_1$ and $n_2$ of arriving waves while the amplitude A varies periodically with time. vii. The intensity of sound is proportional to the square of the amplitude. Hence the resultant intensity will be maximum when the amplitude is maximum. viii. For maximum amplitude (waxing), A = ± 2a $\therefore 2a\cos\left[\dfrac{2\pi(n_1-n_2)}{2}\right]t=\pm2a$ $\therefore \cos\left[\dfrac{2\pi(n_1-n_2)}{2}\right]t=\pm a$ i.e., $\left[2\pi\left(\dfrac{n_1-n_2}{2}\right)t\right]=0,\pi,2\pi,3\pi,….$ $\therefore t=0,\dfrac{1}{n_1-n_2},\dfrac{2}{n_1-n_2},\dfrac{3}{n_1-n_2},….$ Thus, the time interval between two successive maxima of sound is always $\dfrac{1}{n_1-n_2}$ . Hence, the period of beats is $T=\dfrac{1}{n_1-n_2}$ . The number of waxing heard per second is the reciprocal of period of waxing. $\therefore$ Beat frequency in waxing, $N=n_1-n_2$ ix. The intensity of sound will be minimum when amplitude is zero (waning): For minimum amplitude, A = 0, $\therefore 2a\cos\left[2\pi\left(\dfrac{n_1-n_2}{2}\right)t\right]=0$ $\therefore \cos\left[2\pi\left(\dfrac{n_1-n_2}{2}\right)t\right]=0$ $\therefore \left[2\pi\left(\dfrac{n_1-n_2}{2}\right)t\right]=\dfrac{\pi}{2},\dfrac{3\pi}{2},\dfrac{5\pi}{2},….$ $\therefore t=\dfrac{1}{2(n_1-n_2)},\dfrac{3}{2(n_1-n_2)},\dfrac{5}{2(n_1-n_2)},….$ Therefore, time interval between two successive minima is also $\dfrac{1}{(n_1-n_2)}$ . The number of waning heard per second is the reciprocal of period of waning. Period of beat, $T=\dfrac{1}{n_1-n_2}$ Beat frequency in waning, $N=n_1-n_2$ x. Beats frequency is equal to frequency difference between two interfering waves.
4.↑ back to the question
Given: $\lambda_1=\dfrac{85}{172}$ m, $\lambda_2=\dfrac{85}{170}$ m To find: Frequency of third note (n), Velocity of sound (v) Formula: v = nλ Calculation: For two notes in air, $n_1\lambda_1=n_2\lambda_2$ $n_1\times\dfrac{85}{172}=n_2\times\dfrac{85}{170}$ $\therefore \lambda_2>\lambda_1$ $\therefore n_1>n_2$ From the given condition of the problem, $n_1=n+4$ and $n_2=n-4$ $\therefore (n+4)\dfrac{85}{172}=(n-4)\dfrac{85}{170}$ $\therefore 170(n+4)=172(n-4)$ $\therefore n=$ 684 Hz From formula, $v=(n+4)\dfrac{85}{172}=(684+4)\dfrac{85}{172}$ $\therefore v=$ 340 m/s Ans: i. The frequency of the third note is 684 Hz. ii. Velocity of sound in air is 340 m/s.
5.↑ back to the question
Given: $\lambda_1=\dfrac{70}{153}$ m, $\lambda_2=\dfrac{70}{157}$ m To find: Frequency (n), Velocity of sound (v) Formula: v = nλ Calculation: Using formula, For two notes in air, $n_1\lambda_1=n_2\lambda_2$ $n_1\times\dfrac{70}{153}=n_2\times\dfrac{70}{157}$ $\therefore \lambda_1>\lambda_2\quad\therefore n_2>n_1$ Considering the given condition in problem, $n_2=n+8$ and $n_1=n-8$ $\therefore (n+8)\dfrac{70}{157}=(n-8)\dfrac{70}{153}$ $\therefore 153(n+8)=157(n-8)$ $\therefore n=$ 620 Hz Using formula, $v=(n+8)\dfrac{70}{157}=(620+8)\dfrac{70}{157}$ $\therefore v=628\times\dfrac{70}{157}=$ 280 m/s Ans: i. The frequency of tuning fork is 620 Hz ii. The velocity of sound in air is 280 m/s.
6.↑ back to the question
Given: P = 21, Y = 4 beat/s, $n_F=2n_L$ To find: i. Frequency of first fork $(n_F)$ ii. Frequency of tenth fork $(n_{10})$ Formula: $n_L=n_F-(P-1)Y$ Calculation: i. When tuning forks are arranged in the decreasing order of frequencies, the frequency of the $p^{th}$ tuning fork is, $n_L=n_F-(P-1)Y=n_1-(21-1)4$ $\therefore n_L=n_F-80\qquad....(1)$ As frequency of first fork is an octave of last, $\therefore n_F=2n_L$ $\therefore n_L=\dfrac{n_F}{2}$ From equation (1), $\dfrac{n_F}{2}=n_F-80$ $\therefore n_F-\dfrac{n_F}{2}=80$ $\therefore \dfrac{n_F}{2}=80$ $\therefore n_F=$ 160 Hz ii. For 10 $^{th}$ fork, $n_{10}=n_1-(10-1)Y=160-9\times4=160-36$ $\therefore n_{10}=$ 124 Hz Ans: i. The frequency of the first fork is 160 Hz. ii. The frequency of the tenth fork is 124 Hz.
7.↑ back to the question
Given: P = 48, Y = 4, $n_F=1.5n_L$ To find: i. Frequency of first tuning fork $(n_F)$ ii. Frequency of 42 $^{nd}$ tuning fork $(n_{42})$ Formula: $n_L=n_F-(P-1)Y$ Calculation: i. When tuning forks are arranged in the decreasing order of frequencies, the frequency of the $p^{th}$ tuning fork is given by, $n_L=n_F-(P-1)Y=n_1-(48-1)4$ $\therefore n_L=n_F-188\qquad....(1)$ As frequency of first fork is 1.5 times of last, $\therefore n_F=1.5n_L$ $\therefore n_L=\dfrac{n_F}{1.5}$ From equation (1), $\dfrac{n_F}{1.5}=n_F-188$ $\therefore n_F-\dfrac{n_F}{1.5}=188$ $\therefore \dfrac{n_F}{3}=188\;\Rightarrow\;n_F=$ 564 Hz ii. For 42 $^{nd}$ fork, $n_{42}=n_1-(42-1)Y=564-41\times4=564-164$ $\therefore n_{42}=$ 400 Hz Ans: i. Frequency of first fork is 564 Hz. ii. Frequency of 42 $^{nd}$ fork is 400 Hz.
8.↑ back to the question
Given: $\lambda_1=\dfrac{83}{170}$ m, $\lambda_2=\dfrac{83}{172}$ m, Beat frequency, N = 8 beats per second To find: i. Velocity (v), ii. Frequency $(n_1, n_2)$ Formula: v = nλ Calculation: From formula, $n_1=\dfrac{v}{\lambda_1}$ and $n_2=\dfrac{v}{\lambda_2}$ . But $\lambda_1>\lambda_2$ $\therefore n_2>n_1$ $\therefore n_2-n_1=8$ $\therefore v\left(\dfrac{1}{\lambda_2}-\dfrac{1}{\lambda_1}\right)=8$ or $v\left(\dfrac{172}{83}-\dfrac{170}{83}\right)=8$ $\therefore v=\dfrac{8\times83}{2}=$ 332 m/s $\therefore n_1=332\times\dfrac{170}{83}=$ 680 Hz and $n_2=\dfrac{332\times172}{83}=$ 688 Hz Ans: i. The velocity of sound in air is 332 m/s. ii. The frequencies of the two notes are 680 Hz and 688 Hz.
9.↑ back to the question
Given: P = 12, $n_L=2n_F$ , $n_5=90$ Hz To find: Number of beats (Y), Frequency of last fork $(n_L)$ Formula: $n_L=n_F+(P-1)Y$ Calculation: From formula, $n_5=n_F+(5-1)Y$ $\therefore n_F+4Y=90\qquad....(1)$ $n_L=n_F+(12-1)Y$ $\therefore n_L=n_F+11Y\qquad....(2)$ $n_L=2n_F$ $\therefore n_F=11Y\qquad....[\text{From (2)}]$ Substituting in equation (1), 15Y = 90 or Y = 6 beat/s $\therefore n_F=11\times6=$ 66 Hz $\therefore n_L=2\times66=$ 132 Hz Ans: i. The value of Y is 6 beat/s ii. The frequency of the first and last tuning fork is 66 Hz and 132 Hz respectively.
10.↑ back to the question
Given: $\lambda_1=\dfrac{81}{170}$ m, $\lambda_2=\dfrac{81}{173}$ m, Beat frequency, N = 10 beats per second To find: Velocity (v) Formula: v = nλ Calculation: From formula, $n_1=\dfrac{v}{\lambda_1}$ and $n_2=\dfrac{v}{\lambda_2}$ . But $\lambda_1>\lambda_2$ $\therefore n_2>n_1$ $\therefore n_2-n_1=10$ $v\left(\dfrac{1}{\lambda_2}-\dfrac{1}{\lambda_1}\right)=10$ $v\left(\dfrac{173}{81}-\dfrac{170}{81}\right)=10$ $\therefore v=\dfrac{10\times81}{3}=$ 270 m/s Ans: The velocity of sound in air is 270 m/s.
11.↑ back to the question
Given: $n_1=320$ Hz, $n_2=340$ Hz, v = 326.4 ms $^{-1}$ To find: Difference of wavelength of the waves Formula: $\lambda=\dfrac{v}{n}$ Calculation: From formula, $\lambda_1=\dfrac{v}{n_1}$ $\lambda_2=\dfrac{v}{n_2}$ $\because n_1<n_2$ $\therefore \lambda_1>\lambda_2$ Difference in wavelength, $\lambda_1-\lambda_2=\dfrac{v}{n_1}-\dfrac{v}{n_2}$ $\therefore \lambda_1-\lambda_2=v\left[\dfrac{1}{n_1}-\dfrac{1}{n_2}\right]$ $\therefore \lambda_1-\lambda_2=326.4\left[\dfrac{1}{320}-\dfrac{1}{340}\right]$ $\therefore \lambda_1-\lambda_2=326.4\left[\dfrac{340-320}{320\times340}\right]$ $=\dfrac{6528}{320\times340}$ $=\dfrac{192}{320\times10}$ $=\dfrac{6}{100}=$ 0.06 m Ans: The difference in wavelength is 0.06 m.
12.↑ back to the question
Given: $n_1=320$ Hz, $n_2=340$ Hz, v = 340 ms $^{-1}$ To find: Difference of wavelength of the waves Formula: $\lambda=\dfrac{v}{n}$ Calculation: From formula, $\lambda_1=\dfrac{v}{n_1}$ and $\lambda_2=\dfrac{v}{n_2}$ $\because n_1<n_2$ $\therefore \lambda_1>\lambda_2$ Difference in wavelength, $\lambda_1-\lambda_2=\dfrac{v}{n_1}-\dfrac{v}{n_2}$ $=v\left[\dfrac{1}{n_1}-\dfrac{1}{n_2}\right]$ $=340\left[\dfrac{1}{320}-\dfrac{1}{340}\right]$ $=340\left[\dfrac{340-320}{320\times340}\right]$ $=\dfrac{340\times20}{320\times340}$ $=\dfrac{1}{16}=$ 0.0625 m Ans: The difference in wavelength is 0.0625 m.