When two waves of almost equal frequencies $v_1$ and $v_2$ reach a point simultaneously,the time interval between successive maxima is

Two tuning forks with frequencies $380 \,Hz$ and $384 \,Hz$ are sounded together. They produce $4$ beats per second. What is the time interval between the maximum sound and the next minimum sound?

$A$ tuning fork of unknown frequency produces $3 \, \text{beats/sec}$ with a standard tuning fork of frequency $384 \, \text{Hz}$. The beat frequency decreases when a small piece of wax is placed on the prong of the unknown tuning fork. The frequency of the unknown tuning fork is .... $\text{Hz}$

Two waves $Y_1 = 0.25 \sin(316t)$ and $Y_2 = 0.25 \sin(310t)$ are propagating along the same direction. The number of beats produced per second is:

When two tuning forks $A$ and $B$ are sounded together, $4$ beats per second are heard. The frequency of the fork $B$ is $384 \,Hz$. When one of the prongs of the fork $A$ is filed and sounded with $B$, the beat frequency increases. Then the frequency of the fork $A$ is: (in $\,Hz$)

The sound waves of wavelengths $5 \,m$ and $6 \,m$ produce $30$ beats in $3 \,s$. The velocity of sound is (in $\,m/s$)