$A$ stretched string resonates with a tuning fork of frequency $512 \; Hz$ when the length of the string is $0.5 \; m$. The length of the string required to vibrate resonantly with a tuning fork of frequency $256 \; Hz$ would be .......... $m$.

Show that when a string fixed at its two ends vibrates in $1$ loop,$2$ loops,$3$ loops and $4$ loops,the frequencies are in the ratio $1 : 2 : 3 : 4$.

$A$ wire of density $8 \times 10^3\,kg/m^3$ is stretched between two clamps $0.5\,m$ apart. The extension developed in the wire is $3.2 \times 10^{-4}\,m$. If Young's modulus $Y = 8 \times 10^{10}\,N/m^2$,the fundamental frequency of vibration in the wire will be $......\,Hz$.

If $n_{1}, n_{2}$ and $n_{3}$ are the fundamental frequencies of three segments into which a string is divided,then the original fundamental frequency $n$ of the string is given by

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

$A$ string fixed at both ends is vibrating in two segments. The wavelength of the corresponding wave is