The length of a sonometer wire tuned to a frequency of $250 \ Hz$ is $0.60 \ m$. The frequency of the tuning fork with which the vibrating wire will be in tune when the length is made $0.40 \ m$ is .... $Hz$.

Two wires $W_1$ and $W_2$ have the same radius $r$ and respective densities $\rho_1$ and $\rho_2$ such that $\rho_2 = 4\rho_1$. They are joined together at the point $O$,as shown in the figure. The combination is used as a sonometer wire and kept under tension $T$. The point $O$ is midway between the two bridges. When a stationary wave is set up in the composite wire,the joint is found to be a node. The ratio of the number of antinodes formed in $W_1$ to $W_2$ is

$A$ wire of density $9 \times 10^3 \text{ kg/m}^3$ is stretched between two clamps $1 \text{ m}$ apart and is subjected to an extension of $4.9 \times 10^{-4} \text{ m}$. The lowest frequency of transverse vibration in the wire is ..... $\text{Hz}$ $(Y = 9 \times 10^{10} \text{ N/m}^2)$.

$A$ second harmonic has to be generated in a string of length $l$ stretched between two rigid supports. The points where the string has to be plucked and touched are respectively

$A$ sonometer wire $49 \ cm$ long is in unison with a tuning fork of frequency '$n$'. If the length of the wire is decreased by $1 \ cm$ and it is vibrated with the same tuning fork,$6$ beats are heard per second. The value of '$n$' is (in $Hz$)

$A$ 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 $N$)?