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

Options

A. 286 cps
B. 284 cps
C. 292 cps
D. 290 cps

Answer

C

Explanation

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