In this article, we shall study to calculate the apparent frequency of a source using the Doppler effect.

Example – 01:

A whistle of frequency 420 Hz is sounded on the roadside. Find the apparent frequency of the whistle for a motorist driving at 60 km/h when approaching and receding. The velocity of sound in air =320 m/s.

Given: Actual Frequency of source = n = 420 Hz, Source stationary v_S = 0, Speed of listener = v_L = 60 km/h = 60 ×5/18 = 16.67 m/s, Velocity of sound in air = v = 320 m/s.

To Find: Apparent frequency = n_a =? when approaching and receding.

Solution:

Part – I: When the listener is approaching the stationary source

Part – II: When the listener is receding the stationary source

Ans: Apparent frequency when approaching = 441.9 Hz, Apparent frequency when receding = 398.1 Hz

Example – 02:

An engine sounding a whistle of frequency 2000Hz is receding from the stationary observer at 72 km/h. what is the apparent frequency for the observer? The velocity of sound in air =340 m/s

Given: Actual Frequency of source = n = 2000 Hz, Observer stationary V_L = 0, Speed of source = v_S = 72 km/h = 72 ×5/18 = 20 m/s, Velocity of sound in air = v = 340 m/s.

To Find: Apparent frequency = n_a = ? when receding.

Solution:

Ans: Apparent frequency when receding = 1889 Hz

Example – 03:

A source of sound having a frequency of 90 Hz is moving towards a stationary observer with a speed of 1/10ththat of sound. What is the apparent frequency of sound as heard by the observer?

Given: Actual Frequency of source = n = 90 Hz, Observer stationary V_L = 0, Speed of source = v_S = 1/10 v,

To Find: Apparent frequency = n_a = ? when approaching.

Solution:

Ans: Apparent frequency when approaching = 100 Hz

Example – 04:

A source of sound is approaching a stationary observer at 26.4 m/s if the frequency of the source is 512 Hz and the velocity of sound is 340 m/s, find the note as heard by the observer.

Given: Actual Frequency of source = n = 512 Hz, Observer stationary V_L = 0, Speed of source = v_S = 26.4 m/s, Velocity of sound in air = v = 340 m/s.

To Find: Apparent frequency = n_a = ? when approaching.

Solution:

Ans: Apparent frequency when approaching  = 555.1 Hz

Example – 05:

A locomotive passing a stationary observer at 30 m/s is sounding a whistle. Determine the ratio of the frequencies of the notes heard by the observer as the engine approaches and recedes. The velocity of sound in air =320m/s.

Given: Observer stationary V_L = 0, Speed of source = v_S = 30 m/s, Velocity of sound in air = v = 320 m/s.

To Find: Ratio of frequencies n_a approaching /n_a receding   =?

Solution:

Ans: Required ratio is 1.21:1

Example – 06:

The frequency of a sounding source is 1000Hz. An observer is moving at 20 m/s directly towards the stationary source. Find the apparent frequency for the observer. What would the apparent frequency be if the observer is at rest and the source is directly moving towards him at 20 m/s? Velocity of sound =340 m/s.

Given: Actual Frequency of source = n = 1000 Hz, Velocity of sound = v = 340 m/s. 

Part – I: Source stationary v_S = 0, Speed of listener = v_L = 20 m/s, Condition: Approaching.

Part – II: Listener stationary v_L = 0 , Speed of source = v_S = 20 m/s, Condition: Approaching.

To Find: Apparent frequency = n_a = ? when approaching and receding.

Solution:

Part – I: When the listener is approaching the stationary source

Part – II: When the source is approaching the stationary observer

Ans: Apparent frequency when source stationery and listener approaching = 1059 Hz and the apparent frequency when listener stationery and source approaching = 1062.5 Hz

Example – 07:

Two engines pass each other in opposite direction. One of the engines is blowing a whistle of frequency 540 Hz. Calculate the apparent frequency of the whistle for an observer on the other train as they approach and recede. The velocity of both engines is 40 m/s. The velocity of sound in air = 340 m/s.

Given: Speed of listener V_L = 40 m/s, Speed of source = v_S = 40 m/s, Velocity of sound in air = v = 340 m/s. Actual frequency of note = n = 540 Hz

To Find: Apparent frequency when approaching and receding   =?

Solution:

Ans: Apparent frequency when the two engines are approaching each other = 684 Hz and the apparent frequency when the two engines are receding each other = 426.3 Hz

Example – 08:

A train blows the whistle of frequency 640 Hz in air. Find the difference in apparent frequencies of the whistle for a stationary observer when the train moves towards and away from the observer with the speed 72 km/hr. The speed of sound in air is 340 m/s.

Given: Actual frequency of the note = 640 Hz, Speed of listener V_L =0, Velocity of sound in air = v = 340 m/s. Speed of the source = 72 km/h = 72 × 5/18 = 20 m/s.

To Find: n_{a(approaching)}  –  n_a(receding) =?

Solution:

n_{a(approaching)}  –  n_a(receding) =680 – 604.4 = 75.6 Hz

Ans:  The difference in apparent frequencies of the whistle for a stationary observer 75.6 Hz

Example – 09:

A stationary source produces a note of frequency 350 Hz. An observer in a car moving towards the source measures the frequency of sound at 370 Hz. Find the speed of the car. What will be the frequency of sound as measured by the observer in the car if the car moves away from the source at the same speed? Assume speed of sound = 340 m/s.

Given: Source is stationary V_S =0, Velocity of sound in air = v = 340 m/s. Actual frequency of note = n = 350 Hz, Apparent frequency of note when approaching = n_a = 370 Hz

To Find: Speed of car = V_L = ? apparent frequency when receding   =?

Solution:

Ans: Speed of the car is 19.43 m/s and the apparent frequency heard during receding is 330 Hz.

Example – 10:

The speed limit for a vehicle on a road is 100 km/h. A policeman detects a drop of 20% in the pitch of the horn of a car as it passes him. Is the policeman justified in punishing the car driver for crossing the speed limit? The velocity of sound in air = 340 m/s.

Given: Allowed speed = 100 km/h = 100 × 5/18 = 27.78 m/s, Speed of listener  V_L =0, Velocity of sound in air = v = 340 m/s. Drop in frequency when approaching and receding = 20 % = 0.2 i.e. n_a(receding) = n_{a(approaching)}  – 20% n_{a(approaching)}  = 0.8  n_{a(approaching)}

To Find: Justification for policeman

Solution:

Ans: The speed of the car is 37.77 m/s which is more than the allowed speed of 27.78 m/s. Hence policeman is justified in his action.

Example – 11:   

The speed limit for a vehicle on road is 120 km/ hr. A policeman detects a drop of 10 % in the pitch of horn of a car as it passes him. Is the policeman justified in punishing the car driver for crossing the speed limit? Given velocity of sound = 340 m/s.

Given: Allowed speed = 120 km/h = 120 × 5/18 = 33.33 m/s, Speed of listener  V_L =0, Velocity of sound in air = v = 340 m/s. Drop in frequency when approaching and receding = 10 % = 0.1 i.e. n_a(receding) = n_{a(approaching)}  – 10% n_{a(approaching)}  = 0.9  n_{a(approaching)}

To Find: Justification for policeman

Solution:

Ans: Now the speed of the car is 17.89 m/s which is less than the allowed speed of 33.33 m/s. Hence policeman is not justified in his action.

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Science > Physics > Wave Motion > Numerical Problems on Dopler Effect

The post Numerical Problems on Dopler Effect appeared first on The Fact Factor.

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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In this article, we shall study the reflection of waves at the boundary of the denser medium and rarer medium. Wave in a medium may be defined as the disturbance moving through the medium without change of form. e.g. ripple in water formed due to dropping a stone in water.

Reflection of a Wave:

When a wave travelling in a homogeneous medium meets a boundary of some other medium, it has a tendency to travel in the opposite direction. This phenomenon is called the reflection of the wave (or energy). Both light waves (transverse waves) and sound waves (longitudinal waves) exhibit this phenomenon. As the wavelength of a light wave is very very small of order 10^-10 m, hence it can be reflected from a small surface. While the wavelength of a sound wave is large of order 10^-3 m, hence it requires a large surface for reflection.

The light waves (transverse waves) and sound waves (longitudinal waves) both obey the laws of reflection. Echo is an example of the reflection of sound.

Reflection of Sound Waves from Curved Surface:

Reflection of sound waves from a curved surface can be demonstrated by an arrangement as shown in the figure. The concave reflectors are arranged parallel to each other such that they have the common axis and kept apart. A clock is placed at the focus F₁ of reflector M₁ and a funnel tube is arranged at the focus F₂ of reflector M₂. Sound waves diverge from the clock and are incident on reflector M₁ and after reflection move parallel to the common axis. They are incident on reflector M2₂ and after reflection gets converged at the focus F₂ of reflector M₂ and can be heard through funnel tube.

Reflection of Longitudinal Waves (Sound Waves) 

A progressive wave is a wave in which the medium particles are vibrating in the direction parallel to the direction of propagation of the wave is called a longitudinal wave. e.g. sound wave. Longitudinal waves propagate in a form of compressions and rarefactions.

At a Boundary of a Denser Medium:  

Suppose that longitudinal waves in air (sound waves) are incident normally on a rigid wall. When a compression strikes the wall, it exerts a force on the wall. But as the wall is rigid, it exerts an equal and opposite force on the layer of air in compression and thus pushes the compression in the backward direction. Thus a compression travelling towards the right is reflected as a compression travelling towards the left.  The displacement of a medium particle in the reflected wave is in the opposite direction to the displacement of the particle in incident wave. Thus, there is a phase difference of π radian or 180° between the incident wave and the reflected wave. Similarly, incident rarefaction is reflected as a rarefaction.

At a Boundary of a Rarer Medium:

Suppose that a longitudinal wave travelling in a denser medium is incident at the boundary of a rarer medium. If compression in the incident wave strikes the surface of separation (e.g. air at the open end of a pipe) then due to the high pressure of compression, the surface of separation is pushed back. As the particles of air at the open end of a pipe are free to move, the surrounding air goes away quickly and compression is converted into a rarefaction before the wave is reflected. Thus, the compression after reflection at a rarer medium returns as a rarefaction.  Thus no phase change takes place when a longitudinal wave is reflected from the surface of a rarer medium. Similarly, incident rarefaction is reflected as a compression.

Reflection of Transverse Waves:

A progressive wave is a wave in which the medium particles are vibrating in the direction perpendicular to the direction of propagation of the wave is called a transverse wave. e.g. the wave produced in a rope by tying the rope at one end to a rigid wall and jerked at the other end. A transverse wave propagates in the form of crests and troughs.

At a Boundary of a Denser Medium:

Suppose that one end of a string is rigidly fixed to a wall at P, Transverse waves are produced in the string and they travel towards the wall. When the crest reaches the fixed end P, it exerts an upward force on the support but the support is not free to move. By Newton’s third law of motion, it exerts an equal and opposite reaction on the string and sends the pulse backwards. This is the reflected wave. Since the displacement at P is zero, the upward motion is cancelled by a downward motion.  Hence a trough is formed on reflection Thus after reflection at a denser medium, a crest returns as a trough i.e. there is a phase change of π radian or 180° between the incident wave and the reflected wave. Similarly, a trough is reflected as a crest.

At a Boundary of a Rarer Medium:

If a crest of a transverse wave, in a denser medium, is incident at a point of a rarer medium, as the particles of rarer medium are free to vibrate, it will displace the particle at the point of incidence in the upward direction and after reflection from the rarer medium, a crest returns as a crest or a trough returns as a trough. Thus there is no change of a phase when a transverse wave is reflected from a rarer medium.

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Example – 01:

The equation of transverse simple harmonic progressive wave is y = 3 sin 2π(t/0.04 – x/40), where the length is expressed in cm and the time in seconds. Calculate the wavelength, frequency, amplitude and the speed of the wave.

Given: Equation of the wave y =3 sin 2π(t/0.04 – x/40) cm

To Find: Wavelength = λ =? Frequency = n =? Amplitude = a =? speed of wave = v =?

Solution:

Equation of given wave is

Comparing with

Amplitude = a = 3 cm, Period = T = 0.04 s, Wavelength = λ = 40 cm

We have n = 1/T = 1/0.04 = 25 Hz

Now, v = n λ = 25 × 40 = 1000 cm/s

Ans: Wavelength = 40 cm, Frequency = 25 Hz, Amplitude = 3 cm, speed of wave = 1000 cm/s = 10 m/s

Example – 02:

The equation of a certain sound wave (simple harmonic progressive wave) is given by y = 0.05 sin 10π(t/0.025 – x/8.5), where x and y are in meters and t is in seconds. What are the (1) amplitude (2) frequency (3) wavelength of the wave? What is the velocity and direction of propagation of the wave?

Given: Equation of the wave y = 0.05 sin 10π(t/0.025 – x/8.5) m

To Find: Amplitude = a =? Frequency = n =?,  Wavelength = λ = ?,  speed of wave = v =? and direction = ?

Solution:

Equation of given wave is

Comparing with

Amplitude = a = 0.05 m, Period = T = 0.005 s, Wavelength = λ = 1.7 m

We have n = 1/T = 1/0.005 = 200 Hz

Now, v = n λ = 200 × 1.7 = 340 m/s

The term – x/1.7 shows that the wave is moving in positive direction of x-axis

Ans: Amplitude =0.05 m, Frequency = 20 Hz,  Wavelength = 1.7 m,

Speed of wave = v = 340 m/s and direction = in positive direction of x-axis

Example-03:

The equation of a wave can be represented by y = 0.02 sin 2π /0.5 (320t – x) where x and y are in metres and t is in seconds. Find the amplitude, frequency, wavelength, and velocity of propagation of the wave.

Given: Equation of the wave y = y = 0.02 sin 2π /0.5 (320t – x) m

Solution:

To Find: Amplitude = a =? Frequency = n =? Wavelength = λ = ?  velocity of wave = v =?

Comparing with

Amplitude = a = 0.02 m, Frequency = n = 640 Hz, Wavelength = λ = 0.5 m

Now, v = n λ = 640 × 0.5 = 320 m/s

Ans: Amplitude =0.02 m, Frequency =640 Hz, Wavelength = 0.5 m, velocity of wave = 320 m/s

Example – 04:

A simple harmonic progressive wave of amplitude 5 cm and frequency 5 Hz is traveling along the positive X-direction with a speed of 40 cm/s. Calculate (1) the displacement at x = 38cm and t = 1 second. (2) the phase difference between two points in the path of the wave separated by a distance of 0.8 cm. (3) the phase difference between two positions of a particle at an interval of 0.01 s.

Given: Amplitude = a = 5 cm, Frequency = n = 5 Hz, Velocity = v = 40 cm/s, Direction = + X-axis.

To Find: 1) Displacement = y =? when x = 38cm and t = 1 second. 2) ∅ =? when x = 0.8 cm. 3) ∅ = ? when t = 0.01 s.

Solution:

We have v = n λ

∴ λ = v/n  =40/5 = 8m

The equation of progressive wave is in the form

1) Displacement = y =? when x = 38cm and t = 1 second.

2) ∅ = ? when x = 0.8 cm.

3) ∅ = ? when t = 0.01 s.

Example – 05:

The equation of progressive wave is y= 0.01 sin 2π (2t – 0.01x) when all quantities are expressed in SI units. Calculate (a) frequency of the wave. (b) the phase difference between two positions of the same particle at a time interval of 0.25 s. (c) phase difference at a given instant of time between two particles 50 m apart.

Given: Equation of wave y= 0.01 sin 2π (2t – 0.01x) m

To Find: 1) Frequency = n =?  2) ∅ =? when t = 0.25 s. 2) ∅ = ? when x = 50 m.

Solution:

Equation of wave y= 0.01 sin 2π (2t – 0.01x) m

Comparing with

Amplitude = a = 0.01 m, Frequency = n = 2 Hz, Wavelength = λ = 100 m

1) ∅ = ? when t = 0.25 s.

 2) ∅ = ? when x = 50 m.

Example – 06:

The equation of a simple harmonic progressive wave is given by y= 0.002 sin 2π(5t – x/12) where all the quantities are in S.I. units. Calculate the displacement of the particle at a distance of 5 m from the origin after 0.2 s.

Given: Equation of wave y= 0.002 sin 2π(5t – x/12) m

To Find: Displacement of particle = ?.

Solution:

Equation of wave y= 0.002 sin 2π(5t – x/12) m

∴   y = 0.002 sin 2π(5(0.2) – 5/12)

∴   y = 0.002 sin 2π(1 – 5/12)

∴   y = 0.002 sin 2π(7/12)

∴   y = 0.002 sin 7π/6

∴   y = 0.002 sin (6π + π)/6

∴   y = 0.002 sin (π + π/6)

∴   y = – 0.002 sin (π/6)

∴   y = – 0.002 × ½ = – 0.001 m

Ans: Displacement of partiocle = -0.001 m

Example – 07:

The equation of simple harmonic progressive wave of a source is y = 6 sin300πt cm. Write down the equation of the wave. Find the displacement, velocity and acceleration of a point 1.5 m from the source at the instant t = 0.01 s after the start of oscillations. The velocity of propagation of waves is 300 m/s.

Given: Equation of source is y = 6 sin300πt cm, v = 300 m/s

To Find: 1) displacement = y =?, Velocity of point = v = ?,  Acceleration of point = ?, at x = 1.5 m and t = 0.01 s.

Solution:

Equation of source is y = 6 sin300πt cm

Comparing with

y = a sin 2πn t cm and y = a sin ω t cm

Amplitude = a = 6 cm = 0.06 m, 2πn = 300 π, n = 150 Hz, ω = 300 π rad/s

We have v = n λ

∴ λ = v/n  = 300/150 = 2 m

Hence equation of the wave is

∴   y= 0.06 sin 2π(150t – x/2)

∴   y = 0.06 sin 2π(150(0.01) – 1.5/2)

∴ y = 0.06 sin 2π(1.5 – 0.75)

∴ y = 0.06 sin 2π(0.75)

∴ y = 0.06 sin 2π(3/4)

∴ y = 0.06 sin (3π/2)

∴  y = 0.06 (-1) = – 0.06 m

Velocity of particle is given by

Magnitude of the acceleration of particle is given by

f = ω²y = (300 π)² × 0.06 = 5.33 × 10⁴ m/s²

Example – 08:

The equation of simple harmonic oscillations of a source is y = 10 sin 20πt cm. Find the displacement from the position of equilibrium, the velocity and acceleration of point 10 m away from the source 3 seconds after oscillations begin. The velocity of propagation of waves 200 m/s.

Given: Equation of source is y = 10 sin 20πt cm, v = 200 m/s

To Find: 1) displacement = y =?, Velocity of point = v = ?,  Acceleration of point = ?, at x = 10 m and t = 3 s.

Solution:

Equation of source is y = 10 sin 20πt cm

Comparing with

y = a sin 2πn t cm and y = a sin ω t cm

Amplitude = a = 10 cm = 0.1 m, 2πn = 20 π, n = 10 Hz, ω = 20 π rad/s

We have v = n λ

∴ λ = v/n  = 200/10 = 20 m

Hence equation of the wave is

∴   y = 0.1 sin 2π(10t – x/20)

∴ y = 0.1  sin 2π(10(3) – 10/20)

∴ y = 0.1 sin 2π(30 – 1/2)

∴ y = 0.1 sin 2π(59/2)

∴ y = 0.1 sin 59π

∴ y = 0.1 (0) = 0 m

Magnitude of the velocity of particle is given by

Magnitude of the acceleration of particle is given by

f = ω²y = (20 π)² × 0 = 0  m/s²

Example – 09:

A transverse simple harmonic progressive wave of amplitude 0.01 m and frequency 500 Hz is traveling along a stretched string with a speed of 200 m/s. Find the displacement of the particle at a distance of 0.7 m from the origin and after 0.01 s.

Given: Amplitude = a = 0.01 m, frequency = n = 500 Hz, Velocity of wave = v = 200 m/s

To Find: 1) displacement = y =? at x = 0.7 m and t = 0.01 s.

Solution:

We have v = n λ

∴ λ = v/n  = 200/500 = 0.4 m

Hence equation of the wave is

∴   y = 0.01 sin 2π(500t – x/0.4)

∴   y = 0.01 sin 2π(500(0.01) – 0.7/0.4)

∴   y = 0.01 sin 2π(5 – 7/4)

∴   y = 0.01 sin 2π(13/4)

∴   y = 0.01 sin (13π/2)

∴   y = 0.01 sin (12π + π)/2

∴   y = 0.01 sin (6π + π/2)

∴   y = 0.01 sin (π/2)  = 0.01 × 1 = 0.01 m

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In this article, we shall study the concept of a simple harmonic progressive wave, its characteristics and its equation.

Wave in a medium may be defined as the disturbance moving through the medium without change of form. e.g. ripple in water formed due to dropping a stone in water. Wave motion is a mode of transfer of energy or a form of disturbance travelling through an elastic medium due to the repeated oscillations of the particles of the medium about their mean positions.

When a stone is dropped in water, it transfers part of its kinetic energy to the particles of water with which it comes into contact. These particles are set into vibration. Now, these vibrating particles give their energy to the neighbouring particles. Thus neighbouring particles are set into vibrations.  Thus each particle receives energy from the previous particle and passes it to the next particle. Thus wave transfer energy from one point to another.

Simple Harmonic Progressive Wave:

A simple harmonic progressive wave is a wave which continuously advances in a given direction without change of form and the particles of the medium perform simple harmonic motion about their mean position with the same amplitude and period, when the waves pass over them.

Expression for One Dimensional Simple Harmonic Progressive Wave:

Consider a simple, harmonic progressive wave travelling along the positive direction of the x-axis.  Let the vibration of the particles of the medium be parallel or perpendicular to the x-axis.  At any instant t, the displacement of the particles of the medium situated at the origin is given by,

y    =    a sin ωt

Where a is the amplitude and 2π/ω is the period of oscillation.

Now consider a particle of the medium situated at a point P at a distance of x from O. This particle also performs S.H.M. with the same amplitude and period, however, the disturbance from O will reach P only after some time.  In other words, the particle at P lags behind the particle at O in phase.  Hence the displacement of the particle at P at the instant t is given by

y    =    a sin (ωt – ∅).

Where, ∅ is the phase difference between O and P.

The angle ∅ depends on the distance OP. In wave motion, two particles separated by a distance λ differ in phase by 2π radians.  In other words, the path difference λ corresponds to a phase difference of 2p radians.  Therefore, the path difference x will correspond to a phase difference of ∅ = 2πx / λ. Hence, the displacement of the particle at P is given by,

This is the equation of a one dimensional simple harmonic progressive wave travelling along the positive x-axis in terms of the period of the wave.

If n is the frequency of oscillation of the particle, then T = 1/n, substituting in equation (1) we get

This is an expression for a simple harmonic progressive wave in terms of frequency of a wave.

If v is the velocity of the wave in the medium,

This is an expression for a simple harmonic progressive wave in terms of velocity of a wave. Equation (2) and (3) are two different forms of the equation of a one dimensional simple harmonic progressive wave travelling along the positive x-axis.

Equation of Simple Harmonic Progressive Wave Travelling in the Negative X – Direction:

Characteristics of a Simple Harmonic Progressive Wave:

• All particles of medium perform S.H.M. when the wave passes through the medium.
• All particles vibrate with the same amplitude.
• All particles vibrate with the same frequency.
• The propagation of the simple harmonic progressive wave is a doubly periodic phenomenon, periodic in time and periodic in space.
• Energy is transmitted through the medium.

Numerical Problems on Formation of Equation of Simple Harmonic Progressive Wave:

Example – 01:

The equation of simple harmonic progressive wave from a source is y =15 sin 100πt. Find the equation of the wave generated if it propagates along the + X-axis with a velocity of 300 m/s.

Given: Equation of source y =15 sin 100πt, Direction = + X-axis, Velocity of wave v = 300 m/s.

To Find: Equation of the wave =?

Solution:

Equation of Source wave is y =15 sin 100πt

Comparing with y =15 sin 2πnt

a = 15 m and 2πn = 100 π

∴ n = 50 Hz

Now, v = nλ

∴ λ = v/n = 300/50 = 6 m

The equation of simple harmonic progressive wave is given by

Example – 02:

Write down the equation of a transverse wave travelling along a stretched string. Given: amplitude 3 m, wavelength = 40 m and frequency = 25 Hz.

Given: Amplitude = a = 3 m, wavelength = λ = 40m, Frequency n = 25 Hz

To Find: Equation of the wave =?

Solution:

The equation of progressive wave is in the form

Example – 03:

Write down the equation of a wave moving in the positive direction of the x-axis and of amplitude 0.05 m and period 0.04 s traveling along a stretched string with a velocity 12.5 m/s.

Given: Amplitude = a = 0.05 m, Period = T = 0.04 s, Velocity = v = 12.5 m/s, Direction = + X-axis.

To Find: Equation of the wave =?

Solution:

We have v = n λ

∴  λ = v/n  = v T = 12.5 × 0.04 = 0.5 m

The equation of progressive wave is in the form

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