A wire of density 8 times 10^3,kg/m^3 is stre | vedclass

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(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Here,$T$ is the tension and $\mu$ is the linear

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Here,$T$ is the tension and $\mu$ is the linear mass density.We know that $T = Y A \frac{\Delta L}{L}$ and $\mu = ho A$,where $A$ is the cross-sectional area.Substituting these into the frequency formula:$f = \frac{1}{2L} \sqrt{\frac{Y A \Delta L / L}{ho A}} = \frac{1}{2L} \sqrt{\frac{Y \Delta L}{ho L}}$.Given values: $L = 0.5\,m$,$\Delta L = 3.2 	imes 10^{-4}\,m$,$Y = 8 	imes 10^{10}\,N/m^2$,$ho = 8 	imes 10^3\,kg/m^3$.$f = \frac{1}{2 	imes 0.5} \sqrt{\frac{8 	imes 10^{10} 	imes 3.2 	imes 10^{-4}}{8 	imes 10^3 	imes 0.5}}$.$f = 1 	imes \sqrt{\frac{25.6 	imes 10^6}{4 	imes 10^3}} = \sqrt{6.4 	imes 10^3} = \sqrt{6400} = 80\,Hz$.

$A$ wire of density $8 \times 10^3\,kg/m^3$ is stretched between two clamps $0.5\,m$ apart. The extension developed in the wire is $3.2 \times 10^{-4}\,m$. If Young's modulus $Y = 8 \times 10^{10}\,N/m^2$,the fundamental frequency of vibration in the wire will be $......\,Hz$.

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