The fundamental frequency of a sonometer wire is $n$. If the tension is increased $3$ times,the length is increased $3$ times,and the diameter is increased $2$ times,what will be the new frequency?

$A$ 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).

$A$ second harmonic has to be generated in a string of length $l$ stretched between two rigid supports. The points where the string has to be plucked and touched are respectively

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:

$A$ sonometer wire $49 \ cm$ long is in unison with a tuning fork of frequency '$n$'. If the length of the wire is decreased by $1 \ cm$ and it is vibrated with the same tuning fork,$6$ beats are heard per second. The value of '$n$' is (in $Hz$)

Two vibrating strings $A$ and $B$ of the same material but lengths $3 L$ and $2 L$ have radii $2 r$ and $3 r$ respectively. They are stretched under the same tension. String $A$ vibrates in the second overtone and string $B$ in the fundamental mode. The ratio of their frequencies $n_A / n_B$ will be: