A wire of density 9 times 10^3 ,kg/m^3 is str | vedclass

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(C) The Young's modulus $Y$ is given by $Y = \frac{T/A}{\Delta \ell / \ell}$,where $T$ is tension,$A$ is cross-sectional area,$\Delta \ell$ is extensi

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$35$

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(C) The Young's modulus $Y$ is given by $Y = \frac{T/A}{\Delta \ell / \ell}$,where $T$ is tension,$A$ is cross-sectional area,$\Delta \ell$ is extension,and $\ell$ is length.$T = \frac{Y A \Delta \ell}{\ell}$.The velocity of transverse waves in the wire is $v = \sqrt{\frac{T}{\mu}}$,where $\mu = A ho$ is the mass per unit length and $ho$ is the density.$v = \sqrt{\frac{Y A \Delta \ell}{\ell A ho}} = \sqrt{\frac{Y \Delta \ell}{\ell ho}}$.Substituting the given values: $v = \sqrt{\frac{9 	imes 10^{10} 	imes 4.9 	imes 10^{-4}}{1 	imes 9 	imes 10^3}} = \sqrt{\frac{44.1 	imes 10^6}{9 	imes 10^3}} = \sqrt{4.9 	imes 10^3} = \sqrt{4900} = 70 \,m/s$.The lowest frequency (fundamental frequency) is $f = \frac{v}{2\ell} = \frac{70}{2 	imes 1} = 35 \,Hz$.

$A$ wire of density $9 \times 10^3 \,kg/m^3$ is stretched between two clamps $1 \,m$ apart and is subjected to an extension of $4.9 \times 10^{-4} \,m$. What will be the lowest frequency of the transverse vibrations in the wire in $Hz$ $[Y = 9 \times 10^{10} \,N/m^2]$?

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