$A$ tuning fork produces $5 \, Hz$ beats per second when the length of a sonometer wire is either $1 \, m$ or $1.05 \, m$. Calculate the frequency of the tuning fork in $Hz$.

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

When tension $T$ is applied to a sonometer wire of length $l$, it vibrates with the fundamental frequency $n$. Keeping the setup same, when the tension is increased by $8 \,N$, the fundamental frequency becomes three times the earlier. The initial tension applied to the wire was: (in $\,N$)

Two uniform wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and the radius of the first wire is twice the radius of the second wire,then the ratio of the lengths of the first wire to the second wire is:

$A$ 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 $\,kg$)

When a stretched wire of fundamental frequency $f$ is divided into three segments,the fundamental frequencies of these three segments are $f_1, f_2$ and $f_3$ respectively. Then the relation among $f_1, f_2, f_3$ and $f$ is (Assume tension is constant).