If the length of a stretched string is shortened by $x \%$ and the tension is increased by $44 \%$,then the ratio of the initial and final fundamental frequencies is $1:2$. The value of '$x$' is:

$A$ sonometer wire resonates with a given tuning fork,forming a standing wave with $5$ antinodes between two bridges when a mass of $9 \ kg$ is suspended from the wire. When a mass '$m$' is suspended from the wire,with the same tuning fork and the same length between the two bridges,$3$ antinodes are formed. The value of mass '$m$' is: (in $kg$)

$A$ sonometer wire has a length of $114 \ cm$, between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $cm$) whose fundamental frequencies are in the ratio $1:3:4$?

$A$ stretched string is divided into three segments of lengths $50\,cm$,$40\,cm$,and $10\,cm$ with the help of bridges. Their vibrations will have frequencies in the ratio:

The length of a sonometer wire is $0.75\, m$ and density is $9 \times 10^3\, kg/m^3$. It can bear a stress of $8.1 \times 10^8\, N/m^2$ without exceeding the elastic limit. What is the fundamental frequency that can be produced in the wire in $Hz$?

$A$ sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same,the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?