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
$n_A=n+\frac{3n}{100}=\frac{103n}{100}$
and $n_B=n-\frac{2n}{100}=\frac{98}{100}$
but $n_A-n_B=5\Rightarrow \frac{5n}{100}=5$
$\Rightarrow n=100Hz$
$\therefore n_A=\frac{(103)(100)}{100}=103Hz$