$A$ sonometer wire is in unison with a tuning fork when it is stretched by weight $w$ and the corresponding resonating length is $L_1$. If the weight is reduced to $\frac{w}{4}$,the corresponding resonating length becomes $L_2$. The ratio $\frac{L_1}{L_2}$ is:

$A$ tuning fork vibrating with a sonometer having $20 \ cm$ wire produces $5$ beats per second. The beat frequency does not change if the length of the wire is changed to $21 \ cm$. The frequency of the tuning fork (in Hertz) must be

$A$ 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:

$A$ 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 $kg$)

If the frequency of vibrations of a string is to be increased by a factor of two,then the tension in the string must be made:

When a stretched wire of fundamental frequency $f$ is divided into three segments,the fundamental frequencies of these three segments are $f_1, f_2$ and $f_3$ respectively. Then the relation among $f_1, f_2, f_3$ and $f$ is (Assume tension is constant).