The transverse displacement in a stretched string is given by $y = 0.06 \sin \left( \frac{2\pi}{3}x \right) \cos (120\pi t)$,where $x$ and $y$ are in $m$ and $t$ is in $s$. The length of the string is $1.5 \, m$ and its mass is $3.0 \times 10^{-2} \, kg$. The tension in the string is ..... $N$.

$A$ $1 \, cm$ long string vibrates with a fundamental frequency of $256 \, Hz$. If the length is reduced to $\frac{1}{4} \, cm$ keeping the tension unaltered,the new fundamental frequency will be: (in $, Hz$)

Two identical strings $X$ and $Z$ made of the same material have tensions $T_{x}$ and $T_{z}$ in them. If their fundamental frequencies are $450\, Hz$ and $300\, Hz$,respectively,then the ratio $T_{x} / T_{z}$ is:

If the length of a stretched string is reduced by $40 \%$ and the tension is increased by $44 \%$,then the ratio of the final to the initial frequencies of the stretched string is:

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 wires of same material are vibrating under the same tension. If the $1^{\text{st}}$ overtone of the $1^{\text{st}}$ wire is equal to the $2^{\text{nd}}$ overtone of the $2^{\text{nd}}$ wire and the radius of the $1^{\text{st}}$ wire is twice the radius of the $2^{\text{nd}}$ wire,find the ratio of the length of the $1^{\text{st}}$ wire to that of the $2^{\text{nd}}$ wire.