Physics - Sound Wave Question with Solution | TestHub

Given $n_1=1.04n_2$ and $n_3=0.99n_2$.

The beat frequency between A and C is given by $|n_1-n_3|=5$.

Substituting the given relations:

$|1.04n_2-0.99n_2|=5$

$0.05n_2=5$

$n_2=\frac{5}{0.05}=100\text{ Hz}$

Calculating $n_1$ and $n_3$:

$n_1=1.04\times 100=104\text{ Hz}$

$n_3=0.99\times 100=99\text{ Hz}$

The frequencies of forks A, B, and C are $104\text{ Hz}$, $100\text{ Hz}$, and $99\text{ Hz}$ respectively.

Answer: 104 Hz, 100 Hz, 99 Hz