Four wires of identical length,diameter,and material are stretched on a sonometer. If the ratio of their tensions is $1 : 4 : 9 : 16$,then the ratio of their fundamental frequencies is:

Calculate the frequency of the second harmonic formed on a string of length $0.5 \ m$ and mass $2 \times 10^{-4} \ kg$ when stretched with a tension of $20 \ N$.

$A$ 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 $\,kg$)

The fundamental frequency of a sonometer wire is $50 \text{ Hz}$ for a given length and tension. If the length is increased by $25\%$ while keeping the tension constant, what is the percentage change in the frequency of the second harmonic?

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 $N$)?

$A$ string is stretched between two rigid supports separated by $75 \,cm$. There are no resonant frequencies between $420 \,Hz$ and $315 \,Hz$. The lowest resonant frequency for the string is (in $\,Hz$)