The wire of density 9 X 10-3 kg cm-3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4.9 X 10-4 . The lowest frequency of the transverse vibration in the wire is (Young's modulus of wire Y=9 X 10 10 Nm-2 ) (to the nearest integer),.........
Option: 1 3
Option: 2 35
Option: 3 49
Option: 4 135

Answers (1)

Fundamental frequency in the string is given as

\begin{array}{l} f=\frac{1}{2 \ell} \sqrt{\frac{T}{\mu}}=\frac{1}{2 \ell} \sqrt{\frac{T}{\rho A}}=\frac{1}{2 \ell} \frac{\sqrt{Y \Delta \ell}}{\rho \ell} \\ f=\frac{1}{2 \ell} \sqrt{\frac{Y \Delta \ell}{\rho \ell}} \\ \quad\left(Using \ \ \frac{\Delta \ell}{\ell}=4.9 \times 10^{-4}\right) \\ \\ f=\frac{1}{2} \sqrt{\frac{9 \times 10^{10} \times 4.9 \times 10^{-4}}{9 \times 10^{-3} \times 10^{-6}}}=35 \end{array}

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