41 forks are so arranged that each produces 5 beats/s when sounded with its near fork. If the frequency of last fork is double the frequency of first fork, then the frequencies of the first and last fork are respectively

Question

41 forks are so arranged that each produces 5 beats/s when sounded with its near fork. If the frequency of last fork is double the frequency of first fork, then the frequencies of the first and last fork are respectively (A) 200, 400 (B) 205 , 410 (C) 195, 390 (D) 100, 200

Tardigrade Admin · Accepted Answer

Each fork produces 5 beat with its near fork. So, total number of beats =40 × 5=200 If n1 and n2 are the frequencies of first and last fork then n2-n1=200 ldots (i) By the question n2=2 n1 ldots (ii) Solving Eqs. (i) and (ii), we get n1=200 Hz , n2=400 Hz

Q. 41 forks are so arranged that each produces 5 beats/s when sounded with its near fork. If the frequency of last fork is double the frequency of first fork, then the frequencies of the first and last fork are respectively

200, 400

205 , 410

195, 390

100, 200

Each fork produces 5 beat with its near fork. So, total number of beats =40×5=200 If n1​ and n2​ are the frequencies of first and last fork then n2​−n1​=200… (i) By the question n2​=2n1​… (ii) Solving Eqs. (i) and (ii), we get n1​=200Hz,n2​=400Hz

Q. $41$ forks are so arranged that each produces $5$ beats/s when sounded with its near fork. If the frequency of last fork is double the frequency of first fork, then the frequencies of the first and last fork are respectively