The frequency of a sonometer wire is $n$. If its tension is increased $4$ times and its length is doubled,then the new frequency will be

Two wires are fixed in a sonometer. Their tensions are in the ratio $8:1$. The lengths are in the ratio $36:35$. The diameters are in the ratio $4:1$. Densities of the materials are in the ratio $1:2$. If the lower frequency in the setting is $360 \ Hz$,the beat frequency when the two wires are sounded together is: (in $Hz$)

Two strings of the same material having lengths $L$ and $2L$,and radii $2r$ and $r$ respectively,are vibrating in the fundamental mode. The tension applied to both strings is the same. The ratio of their respective fundamental frequencies is:

The length of the wire shown in the figure between the pulleys is $1.5 \, m$ and its mass is $12.0 \, g$. The frequency of vibration with which the wire vibrates in three loops,forming an antinode at the midpoint of the wire,is: (Given $g = 9.8 \, m/s^2$)

$A$ string is divided into three segments such that the segments possess fundamental frequencies in the ratio $1: 2: 3$. Then,the lengths of the segments are in the ratio:

$A$ wire of length $L$,diameter $d$,and density of material $\rho$ is under tension $T$,having a fundamental frequency of vibration $n_A$. Another wire of length $2L$,tension $2T$,density $2\rho$,and diameter $3d$ has a fundamental frequency of vibration $n_B$. The ratio $n_B : n_A$ is