Two sound sources,when sounded simultaneously,produce four beats in $0.25 \ s$. The difference in their frequencies must be: (in $Hz$)

$41$ tuning forks are arranged in increasing order of frequency such that each produces $5 \text{ beats/second}$ with the next tuning fork. If the frequency of the last tuning fork is double that of the frequency of the first fork,then the frequency of the first and last fork is:

When a tuning fork $A$ of unknown frequency is sounded with another tuning fork $B$ of frequency $256 \ Hz$,then $3$ beats per second are observed. After that,$A$ is loaded with wax and sounded again,and $3$ beats per second are observed. The frequency of the tuning fork $A$ is ..... $Hz$.

Two sources of sound placed close to each other are emitting progressive waves given by $y_1 = 4 \sin(600\pi t)$ and $y_2 = 5 \sin(608\pi t)$. An observer located near these two sources of sound will hear:

The frequency of a tuning fork is $n \ Hz$ and the velocity of sound in air is $V \ m/s$. When the tuning fork completes $x$ vibrations,the distance traveled by the wave is:

When two tuning forks (fork $1$ and fork $2$) are sounded simultaneously,$4$ beats per second are heard. Now,some tape is attached to the prong of fork $2$. When the tuning forks are sounded again,$6$ beats per second are heard. If the frequency of fork $1$ is $200 \, Hz$,then what was the original frequency of fork $2$ (in $, Hz$)?