If two tuning forks $A$ and $B$ are sounded together,they produce $4$ beats per second. $A$ is then slightly loaded with wax,and they produce $2$ beats when sounded again. The frequency of $A$ is $256 \ Hz$. The frequency of $B$ will be: (in $Hz$)

Two tuning forks $A$ and $B$ produce notes of frequencies $256 \ Hz$ and $262 \ Hz$ respectively. An unknown note sounded at the same time as $A$ produces beats. When the same note is sounded with $B$,the beat frequency is twice as large. The unknown frequency could be ... $Hz$

For the formation of beats,two sound notes must have:

$A$ source of frequency $\nu$ gives $5$ beats/second when sounded with a source of frequency $200 \;Hz$. The second harmonic of frequency $2\nu$ of the source gives $10$ beats/second when sounded with a source of frequency $420 \;Hz$. The value of $\nu$ is .... $Hz$.

The displacement at a point due to two waves are $y_1 = 4 \sin(500 \pi t)$ and $y_2 = 2 \sin(506 \pi t)$. The result due to their superposition will be

Two tuning forks have frequencies $256 \ Hz$ $(A)$ and $262 \ Hz$ $(B)$. Tuning fork $A$ produces a certain number of beats per second with an unknown tuning fork. The same unknown tuning fork produces double the number of beats per second with tuning fork $B$. What is the frequency of the unknown tuning fork in $Hz$?