$A$ uniform string is vibrating with a fundamental frequency '$n$'. If the radius and length of the string are both doubled while keeping the tension constant,then the new frequency of vibration is:

Two uniform stretched steel strings $A$ and $B$ are vibrating under the same tension. The first overtone of $A$ is equal to the second overtone of $B$. If the radius of $A$ is twice that of $B$,then the ratio of the lengths of the strings $l_A/l_B$ is:

If the length of a stretched string is shortened by $40 \%$ and the tension is increased by $44 \%$, then the ratio of the final and initial fundamental frequencies is :

$A$ sonometer wire of resonating length $90 \ cm$ has a fundamental frequency of $400 \ Hz$ when kept under some tension. The resonating length of the wire with a fundamental frequency of $600 \ Hz$ under the same tension is . . . . . . $cm$.

Two strings $A$ and $B$ of lengths $L_A = 80 \text{ cm}$ and $L_B = x \text{ cm}$ respectively are used separately in a sonometer. The ratio of their densities $(d_A / d_B)$ is $0.81$. The diameter of $B$ is one-half that of $A$. If the strings have the same tension and fundamental frequency, the value of $x$ is:

$A$ 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 $1:3:4$?