wave_mechanics Beats Question 46
Question: The frequency of tuning forks A and B are respectively 3% more and 2% less than the frequency of tuning fork C. When A and B are simultaneously excited, 5 beats per second are produced. Then the frequency of the tuning fork ‘A’ (in Hz) is
[EAMCET 2001]
Options:
A) 98
B) 100
C) 103
D) 105
Show Answer
Answer:
Correct Answer: C
Solution:
Let n be the frequency of fork C then $ n _{A}=n+\frac{3n}{100}=\frac{103n}{100} $ and $ n _{B}=n-\frac{2n}{100}=\frac{98}{100} $ but $ n _{A}-n _{B}=5 $
Therefore $ \frac{5n}{100}=5 $
Therefore $ n=100Hz $ \ $ n _{A}=\frac{(103)(100)}{100}=103Hz $
BETA
