question_answer

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce 4 beats/sec. The highest frequency is twice of the lowest. Possible highest and the lowest frequencies are                                                                    [MP PMT 1990; MHCET 2002]

A)            80 and 40                               

B)            100 and 50

C)            44 and 22                               

D)            72 and 36

Correct Answer: D

Solution :

                   Using nLast = nFirst + (N ? 1)x where N = Number of tuning fork in series           x = beat frequency between two successive forks Þ 2n = n + (10 ? 1) ´ 4 Þ n = 36 Hz \ nFirst = 36 Hz and nLast = 2 ´ nFirst = 72 Hz