$A$ uniform string of length $L$ and mass $M$ is fixed at both ends while it is subject to a tension $T$. It can vibrate at frequencies $(v)$ given by the formula (where $n=1, 2, 3, \ldots$):

$A$ stone is hung in air from a wire which is stretched over a sonometer. The bridges of the sonometer are $L \, cm$ apart when the wire is in unison with a tuning fork of frequency $N$. When the stone is completely immersed in water,the length between the bridges is $l \, cm$ for re-establishing unison. The specific gravity of the material of the stone is:

$A$ string vibrates with a frequency of $200 \ Hz$. When its length is doubled and tension is altered,it begins to vibrate with a frequency of $300 \ Hz$. The ratio of the new tension to the original tension is

The frequency of transverse vibrations in a stretched string is $200 \ Hz$. If the tension is increased four times and the length is reduced to one-fourth the original value,the frequency of vibration will be .... $Hz$.

$A$ string is stretched between fixed points separated by $75.0\, cm$. 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 .... $Hz$

The tension in a piano wire is $10 \ N$. What should be the tension in the wire to produce a note of double the frequency (in $N$)?