Question

When two tuning forks $A$ and $B$ are sounded together, $4$ beats per second are heard. The frequency of the fork $B$ is $384Hz$. When one of the prongs of the fork $A$ is filed and sounded with $B$, then beat frequency increases, then frequency of the fork $A$ is

• A. $380Hz$
• B. $388Hz$
• C. $379Hz$
• D. $389Hz$

Answer 1

The correct option is B $388Hz$

Beat frequency $b=|{\nu }_A-{\nu }_B|$

As on filing the fork A, beat frequency increases. Thus fork A has greater frequency than fork B as on filing, frequency of fork increases.

$⟹$ $b={\nu }_A-{\nu }_B$

$\therefore$ $4={\nu }_A-384$ $⟹{\nu }_A=388$ $Hz$