$A$ 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 remains the same. The frequency of the fork is ............. $Hz$.

$A$ string is under tension of $180 \ N$ and mass per unit length $2 \times 10^{-3} \ kg/m$. It produces two consecutive resonant frequencies with a tuning fork,which are $375 \ Hz$ and $450 \ Hz$. The mass of the string is: (in $g$)

$A$ string in a musical instrument is $50 \,cm$ long and its fundamental frequency is $800 \,Hz$. Keeping the tension applied to the string same, the change in the length to produce a sound note of fundamental frequency $1000 \,Hz$ will be: (in $\,cm$)

$A$ violin emits sound waves of frequency $n_1$ under tension $T$. When tension is increased by $44\%$,keeping the length and mass per unit length constant,the frequency of sound waves becomes $n_2$. The ratio of frequency $n_2$ to frequency $n_1$ is:

Two wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and the radius of the first wire is twice the radius of the second,then the ratio of the length of the first wire to the second wire 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 $1:3:4$?