Two tuning forks $A$ and $B$ vibrating simultaneously produce $5$ beats. The frequency of $B$ is $512 \ Hz$. It is observed that if one arm of $A$ is filed,the number of beats increases. The frequency of $A$ is: (in $Hz$)

The string of a violin emits a note of $205 \,Hz$ at its correct tension. The string is tightened slightly and then it produces six beats in two seconds with a tuning fork of frequency $205 \,Hz$. The frequency of the note emitted by the taut string is .......... $Hz$.

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 $Hz$)

$A$ string under a tension of $129.6 \ N$ produces $10 \ beats/s$ when it is vibrated along with a tuning fork. When the tension in the string is increased to $160 \ N$,it sounds in unison with the same tuning fork. Calculate the fundamental frequency of the tuning fork in $Hz$.

When two tuning forks (fork $1$ and fork $2$) are sounded together,$4$ beats per second are heard. Now some tape is attached to the prong of fork $2$. When the tuning forks are sounded again,$6$ beats per second are heard. If the frequency of fork $1$ is $200 \,Hz$,then the original frequency of fork $2$ is ........... $Hz$.

$50$ tuning forks are arranged in increasing order of their frequencies such that each gives $4 \, \text{beats/sec}$ with its previous tuning fork. If the frequency of the last fork is the octave of the first, then the frequency of the first tuning fork is ... $Hz$.