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

Answer 1

Let $n$ be the frequency of fork $C$, then
$n_A=n+\frac{3n}{100}=\frac{103n}{100}\mathrm{and}{n}_B=n-\frac{2n}{100}=\frac{98n}{100}$
but $n_A-n_B=5\Rightarrow \frac{5n}{100}=5$$\Rightarrow n=100Hz\therefore n_A=\frac{\left(103\right)\left(100\right)}{100}=103Hz$