$A$ and $B$ are two wires whose fundamental frequencies are $256 \text{ Hz}$ and $382 \text{ Hz}$ respectively. When the third harmonic of $A$ and the second harmonic of $B$ are sounded together, the number of beats heard in two seconds will be:

$A$ 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 $Hz$)?

$56$ tuning forks are arranged such that each fork produces $4$ beats per second with its previous one. If the frequency of the last fork is twice that of the first, the frequency of the $19^{\text{th}}$ fork is . . . . . . (in $\text{Hz}$)

$A$ 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 .... $cps$.

Why don't we experience beats in case of superposition of two sound waves with a large difference in their frequencies?

$56$ tuning forks are arranged in increasing order of frequencies in a series such that each fork gives $4 \text{ beats per second}$ with the previous one. The frequency of the last fork is the octave of the first. The frequency of the first fork is ..... $Hz$.