$A$ 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 $Hz$.

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 $\,Hz$)

$A$ wire of length $2L$ is made by joining two wires $A$ and $B$ of same lengths but different radii $r$ and $2r$ and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire $A$ is $p$ and that in $B$ is $q$,then the ratio $p : q$ is

The length of the wire shown in the figure between the pulleys is $1.5 \, m$ and its mass is $12.0 \, g$. The frequency of vibration with which the wire vibrates in three loops,forming an antinode at the midpoint of the wire,is: (Given $g = 9.8 \, m/s^2$)

$A$ stretched string of $1 \ m$ length and mass $5 \times 10^{-4} \ kg$ is having a tension of $20 \ N$. If it is plucked at $25 \ cm$ from one end,then it will vibrate with a frequency of ... $Hz$.