wave_mechanics Beats Question 40
Question: 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]
Options:
A) 286 cps
B) 292 cps
C) 294 cps
D) 288 cps
Show Answer
Answer:
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
Therefore nB = nA + x = 288 + 4 = 292 Hz.
BETA
