$A$ string $A$ has twice the length,twice the diameter,twice the tension,and twice the density of another string $B$. The overtone of $A$ which will have the same fundamental frequency as that of $B$ is:

$A$ string is divided into three segments such that the segments possess fundamental frequencies in the ratio $1: 2: 3$. Then,the lengths of the segments are in the ratio:

If ${n_1}, {n_2}, {n_3}, \dots$ are the frequencies of segments of a stretched string,the frequency $n$ of the string is given by

$A$ segment of wire vibrates with a fundamental frequency of $450 \,Hz$ under a tension of $9 \,kg-wt$. The tension at which the fundamental frequency of the same wire becomes $900 \,Hz$ is:

$A$ wire of density $8 \times 10^3\,kg/m^3$ is stretched between two clamps $0.5\,m$ apart. The extension developed in the wire is $3.2 \times 10^{-4}\,m$. If Young's modulus $Y = 8 \times 10^{10}\,N/m^2$,the fundamental frequency of vibration in the wire will be $......\,Hz$.

$A$ wire of length $2L$ is made by joining two wires $A$ and $B$ of same lengths but different radii $r$ and $2r$ and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire $A$ is $p$ and that in $B$ is $q$,then the ratio $p : q$ is