$A$ wire of length $0.4 \,m$ stretched at both ends vibrates $250$ times per second. If the length of the wire is increased by $0.1 \,m$ and the stretching force is reduced to $1/4$ of its original value, then the new frequency is (in $\,Hz$)

The length of a sonometer wire $AB$ is $110\; cm$. Where should the two bridges be placed from $A$ to divide the wire into $3$ segments whose fundamental frequencies are in the ratio of $1:2:3$?

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?

$A$ steel wire of length $1 \ m$ and mass $0.1 \ kg$ and having a uniform cross-sectional area of $10^{-6} \ m^2$ is rigidly fixed at both ends. The temperature of the wire is lowered by $20^{\circ} C$. If the wire is vibrating in its fundamental mode,find the frequency (in $Hz$).
$(Y_{\text{steel}} = 2 \times 10^{11} \ N/m^2, \alpha_{\text{steel}} = 1.21 \times 10^{-5} /^{\circ} C)$

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

$A$ 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 ..... $\text{Hz}$ $(Y = 9 \times 10^{10} \text{ N/m}^2)$.