The fundamental frequency of a sonometer wire is $n$. If the length and diameter of the wire are doubled while keeping the tension constant,what is the new fundamental frequency?

$A$ wire of length $2L$ is made by joining two wires $A$ and $B$ of same lengths but different radii $r$ and $2r$ and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire $A$ is $p$ and that in $B$ is $q$,then the ratio $p : q$ is

Two uniform 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 wire,then the ratio of the lengths of the first wire to the second wire is:

The fundamental frequency of a transverse wave of a stretched string subjected to a tension $T_1$ is $300 \ Hz$. If the length of the string is doubled and subjected to a tension of $T_2$,the fundamental frequency of the transverse wave in the string becomes $100 \ Hz$. Then $T_2: T_1=$ (Linear density of the string is constant)

$A$ string is stretched between two fixed points separated by $75\,cm$. It is observed to have resonant frequencies of $420\,Hz$ and $315\,Hz$. There are no other resonant frequencies between these two. Then,the lowest resonant frequency for this string is ..... $Hz$.

$A$ wire of density $9 \times 10^{-3} \,kg\, cm^{-3}$ is stretched between two clamps $1 \,m$ apart. The resulting strain in the wire is $4.9 \times 10^{-4}$. The lowest frequency of the transverse vibrations in the wire is......$Hz$ (Young's modulus of wire $Y = 9 \times 10^{10} \,N m^{-2}$),(to the nearest integer).