In order to double the frequency of the fundamental note emitted by a stretched string,the length is reduced to $\frac{3}{4}$ of the original length and the tension is changed. The factor by which the tension is to be changed is:

Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$,the ratio of the lengths of the strings is :

The length of a sonometer wire tuned to a frequency of $250 \ Hz$ is $0.60 \ m$. The frequency of the tuning fork with which the vibrating wire will be in tune when the length is made $0.40 \ m$ is .... $Hz$.

$A$ 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 ... $kg$.

The fundamental frequency of a sonometer wire is $n$. If the length,tension,and diameter of the wire are all tripled,what is the new fundamental frequency?

$A$ sonometer wire of length $114\, cm$ is fixed at both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio $1 : 3 : 4$?