A tuning fork arrangement (pair) produces 4 beats/sec with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/sec. The frequency of the unknown fork is

  • Option 1)

    286 cps

  • Option 2)

    292\: cps

  • Option 3)

    294 cps

  • Option 4)

    288 cps

 

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.

 

 

 

The wax decreases the frequency of unknown fork. The possible unknown frequencies are (288+4)  cps  and (288-4) cps

Wax reduces 284 cps and so beats should increases. It is not given in the question. This frequency is ruled out. Wax reduced 292 cps and so beats should decrease. It is given that the beats decrease to 2 from 4.

Hence unknown fork has frequency 292 cps.

Correct option is 2.


Option 1)

286 cps

This is an incorrect option.

Option 2)

292\: cps

This is the correct option.

Option 3)

294 cps

This is an incorrect option.

Option 4)

288 cps

This is an incorrect option.

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