Unlike a laboratory sonometer,a stringed instrument is seldom plucked in the middle. Supposing a sitar string is plucked at about $\frac{1}{4}$th of its length from the end. The most prominent harmonic would be

$A$ metallic wire of length $L$ is fixed between two rigid supports. If the wire is cooled through a temperature difference $\Delta T$ ($Y =$ Young's modulus,$\rho =$ density,$\alpha =$ coefficient of linear expansion),then the frequency of transverse vibration is proportional to:

Two uniform stretched steel strings $A$ and $B$ are vibrating under the same tension. The first overtone of $A$ is equal to the second overtone of $B$. If the radius of $A$ is twice that of $B$,then the ratio of the lengths of the strings $l_A/l_B$ is:

$A$ 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 $Hz$).
$(Y_{\text{steel}} = 2 \times 10^{11} \ N/m^2, \alpha_{\text{steel}} = 1.21 \times 10^{-5} /^{\circ} C)$

$A$ steel wire of length $1.25 \ m$ is stretched between two rigid supports. The tension in the wire produces an elastic strain of $0.14 \%$. The fundamental frequency of the wire is (Density and Young's modulus of steel are $7.7 \times 10^3 \ kg \ m^{-3}$ and $2.2 \times 10^{11} \ N \ m^{-2}$ respectively). (in $Hz$)

$A$ 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 ... $Hz$.