Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. 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 :

AIEEEAIEEE 2002Waves

Solution:

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 $=288 \pm 4=292$ 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 $=292 Hz$