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 $N$)?

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:

$A$ wire of length $0.3\,m$,stretched between rigid supports,has its $n^{\text{th}}$ and $(n+1)^{\text{th}}$ harmonics at $400\,Hz$ and $450\,Hz$,respectively. If the tension in the string is $2700\,N$,its linear mass density is ......... $kg/m$.

$A$ string $A$ has twice the length,twice the diameter,twice the tension,and twice the density of another string $B$. The overtone of $A$ which will have the same fundamental frequency as that of $B$ is:

When a string of length $l$ is divided into three segments of length $l_1, l_2$ and $l_3$,the fundamental frequencies of the three segments are $n_1, n_2$ and $n_3$ respectively. The original fundamental frequency $n$ of the string is:

$A$ wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $25 \ Hz$. The mass of the wire is $2 \ g$ and its linear mass density is $4 \times 10^{-3} \ kg/m$. What is the tension in the string (in $N$)?