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  • Solve! 56 tuning forks are so arranged in series, that each fork gives 4 beats/second with the previous one. If the frequency of the last fork is 3 times that of the first, then the frequency of the first fork will be

56 tuning forks are so arranged in series, that each fork gives 4 beats/second with the previous one. If the frequency of the last fork is 3 times that of the first, then the frequency of the first fork will be

  • Option 1)

    55 Hz

  • Option 2)

    110 Hz

  • Option 3)

    75 Hz

  • Option 4)

    220 Hz

 

Answers (1)

As we learnt in 

Beat Frequency -

\Delta \nu = \left | \nu _{1}-\nu _{2} \right |

module of \left ( \nu _{1} -\nu _{2}\right )

 

- wherein

Where \nu _{1} \: and\: \nu _{2} are frequency of two wave differ slightly  in value of frequency.

 

Let frequency of first fork is \nu. Frequency of forks are

\nu+4,\:\nu+8,\:\nu+12--------------\:+\nu+4\times 55

According to Question

\nu+220\:=\:3\nu\:=>\nu=110\:Hz


Option 1)

55 Hz

This option is incorrect.

Option 2)

110 Hz

This option is correct.

Option 3)

75 Hz

This option is incorrect.

Option 4)

220 Hz

This option is incorrect.

Posted by

Vakul

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