Beats are produced by frequencies $v_1$ and $v_2$ $(v_1 > v_2)$. The duration of time between two successive minima is

$A$ tuning fork $A$,marked $512 \ Hz$,produces $5$ beats per second when sounded with another unmarked tuning fork $B$. If $A$ is loaded with wax,the number of beats is again $5$ per second. What is the frequency of the tuning fork $B$ when not loaded (in $Hz$)?

The displacement equations of sound waves produced by two sources are given by $y_1 = 5 \sin(400 \pi t)$ and $y_2 = 8 \sin(408 \pi t)$, where $t$ is time in seconds. If the waves are produced simultaneously, the number of beats produced per minute is

$A$ tuning fork gives $5$ beats with another tuning fork of frequency $100\,Hz.$ When the first tuning fork is loaded with wax,the number of beats remains unchanged. What is the frequency of the first tuning fork in $Hz$?

Each of the two strings of length $51.6 \, cm$ and $49.1 \, cm$ are tensioned separately by $20 \, N$ force. The mass per unit length of both strings is the same and equal to $1 \, g/m$. When both strings vibrate simultaneously,the number of beats is:

$25$ tuning forks are arranged in series in the order of decreasing frequency. Any two successive forks produce $3 \, Hz$ beats. If the frequency of the first tuning fork is the octave of the last fork,then the frequency of the $21^{st}$ fork is .... $Hz$