question_answer

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  [KCET 1998; AIEEE 2002]

A)            286 cps                                   

B)            292 cps

C)            294 cps                                   

D)            288 cps

Correct Answer: B

Solution :

           nA = Known frequency = 288 cps, nB = ? x = 4 bps, which is decreasing (from 4 to 2) after loading i.e. x¯ Unknown fork is loaded so nB¯ Hence nA ? nB¯ = x¯               Wrong            nB¯ ? nA¯ = x¯             Correct Þ nB = nA + x = 288 + 4 = 292 Hz.