$A$ set of $20$ tuning forks is arranged in a series of increasing frequencies. If each fork gives $4 \; Hz$ beats with respect to the preceding fork and the frequency of the last fork is twice the frequency of the first,then the frequency of the last fork is $\dots \; Hz$.

When an air column at $15 ^\circ C$ and a tuning fork are sounded together, $4$ beats per second are produced. The frequency of the fork is less than that of the air column. When the temperature falls to $10 ^\circ C$, the beat frequency decreases by one. The frequency of the fork will be ..... $Hz$ $[V_{sound}$ at $0 ^\circ C = 332, m/s]$.

Two vibrating strings $A$ and $B$ produce beats of frequency $8 \ Hz$. The beat frequency is found to reduce to $4 \ Hz$ if the tension in the string $A$ is slightly reduced. If the original frequency of $A$ is $320 \ Hz$, then the frequency of $B$ is: (in $Hz$)

On sounding tuning fork $A$ with another tuning fork $B$ of frequency $384 \ Hz$,$6$ beats are produced per second. After loading the prongs of $A$ with some wax and then sounding it again with $B$,$4$ beats are produced per second. What is the frequency of the tuning fork $A$ 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:

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce $4 \text{ beats/sec}$. The highest frequency is twice the lowest. The possible highest and lowest frequencies are: