Beats due to Sounding of Tuning Forks

Physics Chemistry Biology Mathematics
Science > Physics > Wave Motion > You are Here

Example – 1:

  • 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 = nD= 340 Hz

Number of beats heard = 8 per second

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



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

When prongs of 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 can not 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 – 2:

  • 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 = nC= 512 Hz

Number of beats heard = 4 per second



∴  Frequency of Tuning fork B before filing the prongs = nB= (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 can not 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 – 3:

  • 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 = nA= 480 Hz

Number of beats heard = 4 per second

∴  Frequency of Tuning fork B before filing the prongs = nB= (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 = nB= (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 – 4:

  • 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 = nC= 512 Hz

Number of beats heard = 8 per second

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

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

When prongs of 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 can not take values of 506 Hz and 520 Hz and hence the frequency of B can not be 504 Hz

Hence 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 – 5:

  • 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 = nC= 384 Hz

Number of beats heard = 4 per second

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

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

When prongs of 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 – 6:

  • 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 = nA= 256 Hz

Number of beats heard = 5 per second

∴  Frequency of Tuning fork B before loading with wax = nB= (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 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 – 7:

  • 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 – 8:

  • A set of 8 tuning forks is arranged in 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 nth fork in ascending series is given by

nn = 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 – 9:

  • 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 nth fork in ascending series is given by

nn = 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 22nd 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 nth fork in descending series is given by

nn = 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 22nd 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 22nd 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 heardwith consecutive tuning forks  = X = 5 per second

Frequency of nth fork in descending series is given by

nn = 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



Science > Physics > Wave Motion > You are Here
Physics Chemistry Biology Mathematics

3 Comments

  1. The answer given in example 6 is not related to the question

    • Thank you for your reporting. We always appreciate suggestions and corrections from you. Corrections are made to the problem. All the best for your studies and bright future.

  2. hamedullah

    Thank you…..very helpfull article

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