Physics - Sound Wave Question with Solution | TestHub

Let the frequency of tuning fork C be $n_C$.

The frequency of tuning fork A is $1.4\%$ more than that of C:
$n_A=n_C+0.014n_C=1.014n_C$

The frequency of tuning fork B is $2.6\%$ less than that of C:
$n_B=n_C-0.026n_C=0.974n_C$

When A and B are sounded together, 10 beats are produced per second:
$n_A-n_B=10$

Substituting the values of $n_A$ and $n_B$:
$1.014n_C-0.974n_C=10$

$0.04n_C=10$

$n_C=\frac{10}{0.04}=250$ Hz

Answer: 250 Hz