Question

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 folk is:

• A. 286 cps
• B. 292 cps
• C. 294 cps
• D. 288 cps

Answer 1

Answer: (B)

292 cps

Answer 2

The tuning fork of frequency 288 Hz is producing 4 beats/sec with the unknown tuning fork i.e., the frequency difference between them is 4. Therefore, the frequency of unknown tuning fork $=288\pm 4$ $=292\text{ }or\text{ }284$ On placing a little wax on unknown tuning fork, its frequency decreases but now the number of beats produced per second is 2 i.e. the frequency difference now decreases. It is possible only when before placing the wax, the frequency of unknown fork is greater than the frequency of given tuning fork. Hence, the frequency of unknown tuning fork $=292\text{ }Hz$