In the last article, we have studied numerical problems on beats involving the sounding of two tuning forks. In this article, we shall study numerical problems on beats involving the sounding of two sound notes.

Example – 01:

Two sound waves having wavelengths of 87 cm and 88.5 cm when superimposed produce 10 beats per second. Find the velocity of sound.

Given: Wavelength of first wave = λ₁ = 87 cm = 87 × 10^-2 m, Wavelength of second wave = λ₂ = 88.5 cm = 88.5 × 10^-2 m, No. of beats = 10 per second.

To Find: Velocity of sound = v =?

Solution:

We have v= n λ, Hence n = v/ λ

Now n ∝ 1/ λ   given λ₂ >   λ₁ Hence n₁ >   n₂

Ans:  Velocity of sound 513.3 m/s

Example – 02:

Wavelengths of two sound waves in a gas are 2.0 m and 2.1m respectively. They produce 8 beats per second when sounded together. Calculate the velocity of sound in the gas and the frequencies of the two waves.

Given: Wavelength of first wave = λ₁ = 2.0 m, Wavelength of second wave = λ₂ = 2.1m, No. of beats = 8 per second.

To Find: Velocity of sound = v =? Frequencies of notes =?

Solution:

We have v= n λ, Hence n = v/ λ

Now n ∝ 1/ λ   given λ₂ >   λ₁ Hence n₁ >   n₂

Now n₁ = v/λ₁ = 336 /2.0 = 168 Hz and

n₂ = v/λ₂ = 336 /2.1 = 160 Hz

Ans:  Velocity of sound 336 m/s, the frequencies of notes are 168 Hz and 160 Hz.

Example – 03:

Two sound waves of lengths 1m and 1.01m produce 6 beats in two seconds when sounded together in the air. Find the velocity of sound in air.

Given: Wavelength of first wave = λ₁ = 1 m, Wavelength of second wave = λ₂ = 1.01m, No. of beats = 6 per two second = 3 per second.

To Find: Velocity of sound = v =?

Solution:

We have v= n λ, Hence n = v/ λ

Now n ∝ 1/ λ   given λ₂ >   λ₁ Hence n₁ >   n₂

Ans:  Velocity of sound 303 m/s

Example – 04:

Two tuning forks of frequencies 320 Hz and 340 Hz produce sound waves of lengths differing by 6cm in a medium. Find the velocity of sound in the medium.

Given: Frequency of first wave = n₁ = 320 Hz, Frequency of second wave = n₂ = 340 Hz, Difference in wavelengths = 6 cm = 6 × 10^-2 m

To Find: Velocity of sound = v =?

Solution:

We have v= n λ, Hence λ = v/ n

Now λ ∝ 1/ n   given n₂ >   n₁ Hence λ₁ > λ₂

Ans:  Velocity of sound 326.4 m/s

Example – 05:

Wavelengths of two sound waves in air are 81/174m and 81/175 m. When these waves meet at a point simultaneously, they produce 4 beats per second. Calculate the velocity of sound in air.

Given: Wavelength of first wave = λ₁ = 81/174 m, Wavelength of second wave = λ₂ = 81/175 m, No. of beats = 4 per second.

To Find: Velocity of sound = v =?

Solution:

We have v= n λ, Hence n = v/ λ

Now n ∝ 1/ λ   given λ₁ >   λ₂ Hence n₂ >   n₁

Ans:  Velocity of sound 324 m/s

Example – 06:

Wavelengths of two sound waves in air are 81/173m and 81/170 m. When these waves meet at a point simultaneously, they produce 10 beats per second. Calculate the velocity of sound in air.

Given: Wavelength of first wave = λ₁ = 81/173 m, Wavelength of second wave = λ₂ = 81/170m, No. of beats = 10 per second.

To Find: Velocity of sound = v =?

Solution:

We have v= n λ, Hence n = v/ λ

Now n ∝ 1/ λ   given λ₂ >   λ₁ Hence n₁ >   n₂

Ans:  Velocity of sound 270 m/s

Example – 07:

Wavelengths of two notes in the air are 70/153 m and 70 /157 m. Each of these notes produces 8 beats per second with the third note of fixed frequency. What are the velocity of sound in air and the frequency of the third note?

Given: Wavelength of first wave = λ₁ = 70/153 m, Wavelength of the second wave = λ₂ = 70/157 m, No. of beats with the third note = 8 per second.

To Find: Velocity of sound = v =? The frequency of the third note =?

Solution:

We have v= n λ, Hence n = v/ λ

Now n ∝ 1/ λ   given λ₁ >   λ₂ Hence n₂ >   n₁

Let n be the frequency of the third note, such that n₂ > n >   n₁

Given n₂ – n   = 8 ………. (1)

and n  –  n₁= 8  ………. (2)

Adding equations (1) and (2) we get

Now n₁ = v/λ₁ = 280 /(70/153) = 4 × 153 = 612 Hz

and n  = n₁ + 8 = 512 + 8 = 520 Hz

Ans:  Velocity of sound 280 m/s and frequency of the third note is 620 Hz.

Example – 08:

Wavelengths of two notes in the air are 80/179 m and 80/177 m. Each note produces 4 beats per second with the third note of fixed frequency. Calculate the velocity of sound in air.

Given: Wavelength of the first wave = λ₁ =80/179 m, Wavelength of the second wave = λ₂ = 80/177 m, No. of beats with the third note = 4 per second.

To Find: Velocity of sound = v =?

Solution:

We have v= n λ, Hence n = v/ λ

Now n ∝ 1/ λ   given λ₁ <   λ₂ Hence n₁ >   n₂

Let n be the frequency of the third note, such that n₁ > n >   n₂

Given n₁ – n   = 4 ………. (1)

and n  –  n₂= 4  ………. (2)

Adding equations (1) and (2) we get

Ans:  Velocity of sound 320 m/s

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Science > Physics > Wave Motion > Numerical Problems on Beats

The post Numerical Problems on Beats – 02 appeared first on The Fact Factor.

In this article, we shall study solving problems on formation of beats by sounding two notes together

Example – 01:

A tuning fork C produces 8 beats per second with a fork D of frequency 340 Hz. When the prongs of C are filed a little, the beats per second decrease to 4. Find the frequency of C before filing.

Solution:

Consider case – 1, before filing the prongs

Frequency of Tuning fork D = n_D= 340 Hz

Number of beats heard = 8 per second

∴ Frequency of Tuning fork C before filing the prongs = n_C= (340 ± 8) Hz

∴ Frequency of Tuning fork C before filing the prongs = 332 Hz or 348 Hz

Consider case – 2, when prongs of C are filed

Number of beats heard = 4 per second

∴ Frequency of Tuning fork C after filing the prongs = n_C= (340 ± 4) Hz

∴ Frequency of Tuning fork C after filing the prongs = 336 Hz or 344 Hz

When prongs of the tuning fork are filed the frequency of tuning fork increases

If frequency of fork C is 348 Hz, on filing its frequency should be more than 348 Hz

It means it cannot take lower values of 344 Hz and 336 Hz and hence the frequency of C cannot be 348 Hz

Hence frequency of tuning fork C should be 332 Hz, which satisfies the two cases and it increases to 336 Hz after filing the prongs

Ans: Frequency of fork C is 332 Hz

Example – 02:

A tuning fork B produces 4 beats per second with another tuning fork C of frequency 512 Hz. When the prongs of B are filed a little, the beats occur at shorter intervals. What was the frequency of B before filing?

Solution:

Consider case – 1, before filing the prongs

Frequency of Tuning fork C = n_C= 512 Hz

Number of beats heard = 4 per second

∴ Frequency of Tuning fork B before filing the prongs = n_B= (512 ± 4) Hz

∴ Frequency of Tuning fork B before filing the prongs = 508 Hz or 516 Hz

Consider case – 2, when prongs of B are filed

Now beats are heard at shorter interval thus the number of beats increase

If frequency of fork B is 508 Hz, on filing its frequency should be more than 508 Hz

It means the frequency of beat should decrease, but in this case the frequency of beats is increasing.

Hence the frequency of B cannot be 508 Hz

Hence frequency of tuning fork B should be 516 Hz, which satisfies the two cases and it increases above 516 Hz after filing the prongs giving more number of beats

Ans: Frequency of fork B is 516 Hz

Example – 03:

Two tuning forks A and B when sounded together give 4 beats per second. The frequency of A is 480 Hz. When B is filed and the two forks sounded together, 4 beats per second are heard again. Find the frequency of B before it was filed?

Solution:

Consider case – 1, before filing the prongs

Frequency of Tuning fork A = n_A= 480 Hz

Number of beats heard = 4 per second

∴ Frequency of Tuning fork B before filing the prongs = n_B= (480 ± 4) Hz

∴ Frequency of Tuning fork B before filing the prongs = 476 Hz or 484 Hz

Consider case – 2, when prongs of B are filed

Number of beats heard = 4 per second

∴ Frequency of Tuning fork B after filing the prongs = n_B= (480 ± 4) Hz

∴ Frequency of Tuning fork B after filing the prongs = 476 Hz or 484 Hz

When prongs of tuning fork are filed the frequency of tuning fork increases

If frequency of fork B is 484 Hz, on filing its frequency should be more than 484 Hz

It means number of beats will be more than 4 and hence the frequency of C cannot be 484 Hz

Hence frequency of tuning fork B should be 476 Hz, which satisfies the two cases

Initially, the number of beats decrease and then increase and can give 4 beats per second when its frequency is 484 Hz.

Ans: Frequency of fork B is 476 Hz

Example – 04:

A tuning fork B produces 8 beats per second with another tuning fork C of frequency 512 Hz. When the prongs of B are loaded with a little wax, the number of beats increases to 6 per second. What was the frequency of B before it was loaded?

Solution:

Consider case – 1, before loading with wax

Frequency of Tuning fork C = n_C= 512 Hz

Number of beats heard = 8 per second

∴ Frequency of Tuning fork B before loading with wax = n_B= (512 ± 8) Hz

∴ Frequency of Tuning fork B before loading with wax = 504 Hz or 520 Hz

Consider case – 2, when prongs of B are loaded with wax

Number of beats heard = 6 per second

∴ Frequency of Tuning fork B after loading with wax = n_B= (512 ± 6) Hz

∴ Frequency of Tuning fork B after loading with wax = 506 Hz or 518 Hz

When prongs of the tuning fork are loaded with wax the frequency of tuning fork decreases

If frequency of fork B is 504 Hz, on loading with wax its frequency should be less than 504 Hz

It means it cannot take values of 506 Hz and 520 Hz and hence the frequency of B can not be 504 Hz

Hence the frequency of tuning fork B should be 520 Hz, which satisfies the two cases and it decreases to 518 Hz 0r 506 Hz after loading with wax

Ans: Frequency of fork B is 520 Hz

Example – 05:

A tuning fork B produces 4 beats per second with a tuning fork C of frequency 384 Hz. When the prongs of B are loaded with a little wax, the number of beats increases to 6 per second. What was the frequency of B before loading?

Solution:

Consider case – 1, before loading with wax

Frequency of Tuning fork C = n_C= 384 Hz

Number of beats heard = 4 per second

∴ Frequency of Tuning fork B before loading with wax = n_B= (384 ± 4) Hz

∴ Frequency of Tuning fork B before loading with wax = 380 Hz or 388 Hz

Consider case – 2, when prongs of B are loaded with wax

Number of beats heard = 6 per second

∴ Frequency of Tuning fork B after loading with wax = n_B= (384 ± 6) Hz

∴ Frequency of Tuning fork B after loading with wax = 378 Hz or 390 Hz

When prongs of the tuning fork are loaded with wax the frequency of tuning fork decreases

If frequency of fork B is 388 Hz, on loading with wax its frequency should be less than 388 Hz and the number of beats should decrease but beats are increasing to 6 per second.

Hence the frequency of B cannot be 388 Hz

Hence frequency of tuning fork B should be 380 Hz, which satisfies the two cases and it decreases to 376 Hz after loading with wax

Ans: Frequency of fork B is 380 Hz

Example – 06:

Two tuning forks A and B produce 5 beats per second when sounded together. The frequency of A is 256 Hz. If one of the prongs of B is loaded with a little wax, the number of beats is found to increase. Calculate the frequency of B.

Solution:

Consider case – 1, before loading with wax

Frequency of Tuning fork A = n_A= 256 Hz

Number of beats heard = 5 per second

∴ Frequency of Tuning fork B before loading with wax = n_B= (256 ± 5) Hz

∴ Frequency of Tuning fork B before loading with wax = 251 Hz or 261 Hz

Consider case – 2, when prongs of B are loaded with wax

The number of beats heard is increasing

If the frequency of fork B is 261 Hz, on loading with wax its frequency should be less than 261 Hz and the number of beats should decrease but beats are increasing.

Hence the frequency of B cannot be 261 Hz

Hence frequency of tuning fork B should be 251 Hz, which satisfies the two cases and its frequency decreases below 251 Hz after loading with wax and number of beats increases

Ans: Frequency of fork B is 251 Hz

Example – 07:

A note produces 4 beats with a tuning fork of frequency 512 Hz and 6 beats with a fork of frequency 514 Hz. Find the frequency of the note.

Solution:

Consider case – 1, sounding with tuning fork of frequency 512 Hz

Number of beats heard = 4 per second

∴ Frequency of the note = n = (512 ± 4) Hz

∴ Frequency of tuning fork B before loading with wax = 508 Hz or 516 Hz

Consider case – 2 sounding with tuning fork of frequency 514 Hz

Number of beats heard = 6 per second

∴ Frequency of the note = n = (514 ± 6) Hz

∴ Frequency of Tuning fork B after loading with wax = 508 Hz or 520 Hz

If we study the two cases 508 Hz is a common answer

Ans: Frequency of the note is 508 Hz

Example – 08:

A set of 8 tuning forks is arranged in a series of increasing frequencies. If each fork gives 4 beats per second with the preceding one and the frequency of the last fork is an octave of the first, find the frequencies of the first and last fork.

Solution:

Let n₁ and n₈ be the frequencies of the first and the last fork respectively

Given: arrangement is of tuning forks in increasing frequencies and n₈ = 2n₁

Number of Beats heard with consecutive tuning forks = X = 4 per second

Frequency of n^th fork in ascending series is given by

n_n = n₁ +  (n – 1) × X

∴ n₈ = n₁ + (8 -1) × 4

∴ n₈ = n₁ + 28

But given, n₈ = 2n₁

∴ 2n₁ = n₁ + 28

∴ n₁ = 28 Hz

n₈ = 2 × 28 = 56 Hz

Ans: The frequency of first tuning fork is 28 Hz and that of the last fork is 56 Hz.

Example – 09:

A set of 11 tuning forks is kept in ascending order of frequencies. Each tuning fork gives 5 beats/s with the previous one and the frequency of the last fork is 1.5 times that of the first.  Find the frequency of the first and the last fork.

Solution:

Let n₁ and n₁₁ be the frequencies of the first and the last fork respectively

Given: arrangement is of tuning forks in increasing frequencies and n₁₁ = 1.5n₁

Number of Beats heard with consecutive tuning forks = X = 5 per second

Frequency of n^th fork in ascending series is given by

n_n = n₁ +  (n – 1) × X

∴ n₁₁ = n₁ + (11 – 1) × 5

∴ n₁₁ = n₁ + 50

But given, n₁₁ = 1.5n₁

∴ 1.5n₁ = n₁ + 50

∴ 0.5 n₁ = 50

∴   n₁ = 50/0.5 = 100 Hz

n₁₁ = 1.5 × 100 = 150 Hz

Ans: The frequency of first tuning fork is 50 Hz and that of the last fork is 150 Hz.

Example – 10:

A set of 28 tuning forks is arranged in a series of decreasing frequencies. Each fork gives 4 beats per second with the preceding one and the frequency of the first fork is an octave of the last. Calculate the frequency of the first and 22^nd fork.

Solution:

Let n₁ and n₂₈ be the frequencies of the first and the last fork respectively

Given: arrangement is of tuning forks in decreasing frequencies and n₁ = 2 n₂₈

Number of Beats heard with consecutive tuning forks = X = 4 per second

Frequency of n^th fork in descending series is given by

n_n = n₁ –  (n – 1) × X

∴ n₂₈ = n₁ – (28 – 1) × 4

∴ n₂₈ = n₁ – 108

But given, n₁ = 2 n₂₈

∴ n₂₈ = 2 n₂₈ – 108

∴   n₂₈ = 108 Hz

∴   n₁ = 2 n₂₈ = 2 × 108 = 216 Hz

Frequency of 22^nd fork

n₂₂ = 216 – (22-1) × 4

n₂₂ = 216 – 84 = 132 Hz

Ans: The frequency of first tuning fork is 216 Hz and that of the 22^nd fork is 132 Hz.

Example – 11:

A set of 31 tuning forks is kept in descending order of frequencies. Each tuning fork gives 5 beats/s with the previous one and the frequency of the first fork is twice that of the last.  Find the frequency of the first and the last fork.

Solution:

Let n₁ and n₃₁ be the frequencies of the first and the last fork respectively

Given: arrangement is of tuning forks in decreasing frequencies and n₁ = 2 n₃₁

Number of Beats heard with consecutive tuning forks = X = 5 per second

Frequency of n^th fork in descending series is given by

n_n = n₁ –  (n – 1) × X

∴ n₃₁ = n₁ – (31 – 1) × 5

∴ n₃₁ = n₁ – 150

But given, n₁ = 2 n₃₁

∴ n₂₈ = 2 n₃₁ – 150

∴   n₂₈ = 150 Hz

∴   n₁ = 2 n₃₁ = 2 × 150 = 300 Hz

Ans: The frequency of first tuning fork is 300 Hz and that of the last fork is 150 Hz

Previous Topic: Theory of Formation of Beats

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Science > Physics > Wave Motion > Numerical Problems on Beats

The post Numerical Problems on Beats 01 appeared first on The Fact Factor.

In this article, we shall study the formation of beats, conditions required for its formation, and expression for the period and frequency of beats.

Principle of Superposition of Waves:

When two waves arrive at a point simultaneously, each wave produces its own effect at that point as if that wave alone was passing through the point. Hence, the resultant displacement of the particle at that point is given by the vector sum of the displacements due to individual waves meeting at that point.

Explanation:

If y₁ and y₂ are the displacements produced by two waves arriving at a point simultaneously, then the resultant displacement of the particle is the vector sum of the displacement due to two waves.

Case – I:

When two waves arrive at a point simultaneously in phase such that the crest of the first wave matches with the crest of the second and trough of the first wave matches with the trough of the second then the resultant displacement of the particle is the vector sum of the displacement produced by the particle by each wave.

In this case, the two waves meet each other in phase and hence the two displacements adds up and the resultant wave has a higher amplitude than any of the two waves. If the amplitudes of the two waves are equal then the resultant amplitude is double the amplitude of each wave.

Case – II:

When two waves arrive at a point simultaneously in opposite phase such that the crest of the first wave matches with the trough of the second wave and the trough of the first wave matches with the crest of the second then the resultant displacement of the particle is vector sum of the displacement produced by the particle by each wave.

In this case, the two waves are in opposite phase and hence the two displacement tries to cancel each other and the resultant displacement is the difference between the displacements produced by the two waves. If the amplitudes of the two waves are equal, then the resultant amplitude is zero.

Phenomenon of Beats:

The physical effect of the superposition of two or more waves is called interference.  The phenomenon of beats is an example of interference in time because in this case, we consider the variation in the intensity of sound with time at a given, place.

When two sound waves having equal amplitudes but of slightly different frequencies travel in a medium in the same direction and arrive at a point simultaneously, they interfere and we hear alternate maxima and minima in the resultant intensity of sound. The maximum of sound is called ‘Waxing’ and the minimum of sound is called ‘’Waning’.

The formation of periodic waxing and waning of sound due to interference (superposition) of two sound waves of the same amplitude but of slightly different frequencies, is called beats. One waxing and one waning which are consecutive form one beat.

The time interval between two successive maxima or minima (waxing or waning) is called the period of beats. The number of beats heard per second is called the frequency of beats.

Conditions of Formation of Beats:

• The amplitude of the two interfering waves should be the same.
• The difference between the frequencies of interfering waves should be small. The beats can be heard only if the frequency difference is less than 10.

Terminology of Beats:

• Waxing: When two sound waves having equal amplitudes but of slightly different frequencies travel in a medium in the same direction and arrive at a point simultaneously, they interfere and produce alternate maxima and minima in the resultant intensity of sound. The maximum of sound is called ‘Waxing’
• Waning: When two sound waves having equal amplitudes but of slightly different frequencies travel in a medium in the same direction and arrive at a point simultaneously, they interfere and produce alternate maxima and minima in the resultant intensity of sound. The minimum of sound is called ‘Waning.
• Beats: The formation of periodic waxing and waning of sound due to interference (superposition) of two sound waves of the same amplitude but of slightly different frequencies, is called beats.  One waxing and one waning form one beat.
• Periods of beats: The time interval between two successive maxima or minima (waxing or waning) is called the period of beats.
• Frequency of Beats: The number of beats heard per second is called the frequency of beats.

Quincke’s Experiment to Study Beats Phenomenon:

The phenomenon of interference between two longitudinal waves in the air can be demonstrated by Quincke’s tube shown in the figure. Quincke’s tube consists of U shaped glass tubes A and B. The tube SAR has two openings at S (source)  and R (receiver or listener). The other tube B can slide over the tube A. A sound wave from S (source) travels along both the paths SAR  and  SBR  in the opposite directions and meet at R (receiver end).

If the path difference between the two waves i.e. SAR ~ SBR is an integral multiple of the wavelength, the intensity of the sound will be maximum due to constructive interference. In this case, the path difference between the waves is even multiples of λ/2. The corresponding phase difference between the two waves is even multiples of π. i.e.   2pπ where p = 0, 1, 2, 3 ….

If the tube B is gradually slid over A, a stage is reached when the intensity of sound is zero at R due to destructive interference. Then no sound will be heard at R, In this case, the path difference between the waves is odd multiples of λ/2, the intensity of the sound will be minimum. The corresponding phase difference f between the two waves is odd multiples of π. i.e. (2p + 1)π where p = 0, 1, 2, 3 …..

Expression for Frequency of Beats / Period of Beats:

Consider two sources of sound-emitting sound waves of equal amplitudes ‘a’ and traveling in the same direction at the same speed.  They have slightly different frequencies say n₁ and n₂ where n₁ > n₂. The displacements produced by the two waves is given by

y₁ = a Sin ω₁t   = a Sin 2πn₁t

y₂ = a Sin ω₂t   = a Sin 2πn₂t

By the principle of superposition of waves, the resultant displacement is given by

y = y₁ + y₂

∴   y = a Sin 2πn₁t + a Sin 2πn₂t

∴ y =   a (Sin 2πn₁t + Sin 2πn₂t )

By using the following formula of trigonometry

The form of the equation shows that the resultant motion is also a Simple Harmonic Wave of mean frequency but its amplitude R changes with time. As the intensity is proportional to the square of the amplitude, the resultant intensity will also vary and it will be maximum when R is maximum.

Waxing:

The maximum value of R is ± 2 a and it occurs when,

We find that at instants t₀ , t₁, t₂₁ , t₃ the intensity is maximum or there is waxing of sound. From the above relations, the time interval between two successive maxima (waxing) is given by

Frequency of beats   = Difference between the frequencies of the two waves.

Waning:

The minimum value of R is 0 and it occurs when,

We find that at instants t₀, t₁, t₂₁ , t₃ the intensity is minimum or there is waning of sound. From the above relations, the time interval between two successive minima (waning) is given by

Frequency of beats   = Difference between the frequencies of the two waves.

From the above expressions, we also find that between two successive maxima (waxing), there is a minimum (waning) i.e. Waxing and waning are alternate and equally spaced. Thus the number of beats produced per second is the difference between the frequencies of the two waves.

Applications of Beats:

• Beats are used in tuning musical instruments like sitar, violin, etc. The musical instrument is sounded with another instrument of known frequency.  Generally, the beats are heard.  Then it is slightly adjusted so that there are no beats.  This is called tuning.
• In the Sonometer experiment, beats can be used to adjust the vibrating length between the two bridges.
• To find the frequency (N) of the given tuning fork beats can be used.
• Detection of harmful gases in mines. Two identical organ pipes, one filled with pure air and the other filled with air from the mine are blown together.  If there are no beats then the mine air is pure but if beats are heard the mine-air is impure.

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The post Formation of Beats appeared first on The Fact Factor.