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:

$A$ tuning fork arrangement (pair) produces $4 \, beats/sec$ with one fork of frequency $288 \, cps$. $A$ little wax is placed on the unknown fork and it then produces $2 \, beats/sec$. The frequency of the unknown fork is .... $cps$.

$A$ set of $28$ tuning forks is arranged in an increasing order of frequencies. Each fork produces '$x$' beats per second with the preceding fork and the last fork is an octave of the first. If the frequency of the $12^{\text{th}}$ fork is $152 \text{ Hz}$,the value of '$x$' (number of beats per second) is:

Three sound waves of equal amplitudes have frequencies $(n - 1)$,$n$,and $(n + 1)$. They superimpose to produce beats. The number of beats produced per second is:

Two tuning forks have frequencies $380 \, Hz$ and $384 \, Hz$ respectively. When they are sounded together,they produce $4 \, beats$ per second. After hearing the maximum sound,how long will it take to hear the minimum sound?

Two tuning forks $A$ and $B$ are sounded together giving rise to $8$ beats in $2$ s. When fork $A$ is loaded with wax,the beat frequency is reduced to $4$ beats in $2$ s. If the original frequency of tuning fork $B$ is $380$ Hz,then the original frequency of tuning fork $A$ is . . . . . . Hz.