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

Question

A tuning fork arrangement (pair) produces 4 beats/s with one fork of frequency 288 cps. A little wax is placed on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is : (A) 286 cps (B) 292 cps (C) 294 cps (D) 288 cps

Tardigrade Admin · Accepted Answer

The tuning fork of frequency 288 Hz is producing 4 beats/s with the unknown tuning fork i. e., the frequency difference between them is 4 .Therefore, the frequency of unknown tuning fork =288 ± 4=292 or 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 Hz

Q. A tuning fork arrangement (pair) produces $4$ beats/s with one fork of frequency $288 c p s$ . A little wax is placed on the unknown fork and it then produces $2$ beats/s. The frequency of the unknown fork is :

$286 c p s$

$292 c p s$

$294 c p s$

$288 c p s$

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

286cps

292cps

294cps

288cps

Q. A tuning fork arrangement (pair) produces $4$ beats/s with one fork of frequency $288 c p s$ . A little wax is placed on the unknown fork and it then produces $2$ beats/s. The frequency of the unknown fork is :

$286 c p s$

$292 c p s$

$294 c p s$

$288 c p s$