The frequency of a sonometer wire is $n$. If its tension is increased $4$ times and its length is doubled,then the new frequency will be

The transverse displacement of a string (clamped at its both ends) is given by $y(x,t) = 0.6 \sin \left( \frac{2\pi}{3}x \right) \cos (120\pi t)$,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 will be .... $N$.

$A$ 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 ......... $kg/m$.

In a sonometer wire,the tension is maintained by suspending a $50.7 \, kg$ mass from the free end of the wire. The suspended mass has a volume of $0.0075 \, m^3$. The fundamental frequency of the wire is $260 \, Hz$. If the suspended mass is completely submerged in water,the fundamental frequency will become .... $Hz$ (take $g = 10 \, ms^{-2}$)

$A$ 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 $N$)?

$A$ sonometer wire resonates with $4$ antinodes between two bridges for a given tuning fork, when $1 \,kg$ mass is suspended from the wire. Using the same fork, when mass $M$ is suspended, the wire resonates producing $2$ antinodes between the two bridges (the distance between the two bridges remains the same). The value of $M$ is: (in $\,kg$)