The fundamental frequency of a sonometer wire is $n$. If the length and diameter of the wire are doubled while keeping the tension constant,what is the new fundamental frequency?

Two wires made up of the same material are of equal lengths but their radii are in the ratio $1 : 2$. On stretching each of these two strings by the same tension,the ratio between the fundamental frequencies is

$A$ stretched string is vibrating in the second overtone. The number of nodes and antinodes between the ends of the string are respectively:

$A$ sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of $9 \,kg$ is suspended from the wire. When this mass is replaced by a mass $M$, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of $M$ is

Two tuning forks when sounded together produce $4$ beats per second. One of the forks is in unison with $23 \ cm$ length of a sonometer wire and the other with $24 \ cm$ length of the same wire. The frequencies of the two tuning forks are

$A$ tuning fork gives $4$ beats with $50 \ cm$ length of a sonometer wire. If the length of the wire is shortened by $1 \ cm$,the number of beats is still the same. The frequency of the fork is (in $Hz$)