wave_mechanics critical_thinking Question 24
Question: 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]
Options:
A) 80 and 40
B) 100 and 50
C) 44 and 22
D) 72 and 36
Show Answer
Answer:
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
Therefore 2n = n + (10 ? 1) ´ 4
Therefore n = 36 Hz \ nFirst = 36 Hz and nLast = 2 ´ nFirst = 72 Hz
BETA
