When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the…
When two tuning forks $A$ and $B$ are sounded together, 4 beats per second are heard. The frequency of the fork $B$ is $384\text{ Hz}$. When one of the prongs of the fork $A$ is filled and sounded with $B$, the beat frequency increases, then the frequency of the fork $A$ is [KCET 2015]
$379\text{ Hz}$
$380\text{ Hz}$
$389\text{ Hz}$
$388\text{ Hz}$
Solution
Given, frequency of fork $B$, $\nu_B = 384\text{ Hz}$
Beat frequency, $\nu_A - \nu_B = 4$, so $\nu_A = 388\text{ or } 380$
$\because$ When one of prong of $A$ is filled, then beat frequency increase.
$\therefore$ Frequency of fork $A$, $\nu_A = 388\text{ Hz}$