$A$ uniform wire of length $L$,diameter $D$,and density $\rho$ is stretched under a tension $T$. The correct relation between its fundamental frequency $f$,the length $L$,and the diameter $D$ is:

$A$ stretched wire of length $260 \ cm$ is set into vibrations. It is divided into three segments whose frequencies are in the ratio $2:3:4$. Their lengths must be

$A$ 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:

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:

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

$A$ sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of $9 \ kg$ is suspended from the wire. When this mass is replaced by a mass $M$,the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of $M$ is ... $kg$.