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

• A. 98
• B. 100
• C. 103
• D. 105

Answer 1

Answer: (C)

Let $n$ be the frequency of fork $C$ then
$n_{A}=n+\frac{3 n}{100}=\frac{103 n}{100}$
and $n_{B}=n-\frac{2 n}{100}=\frac{98}{100}$
but $n_{A}-n_{B}=5 \Rightarrow \frac{5 n}{100}=5$
$\Rightarrow n=100\, Hz$
$\therefore n_{A}=\frac{(103)(100)}{100}=103\, Hz$