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

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Question

$Y = \frac{{T/\pi {r^2}}}{{\Delta \ell /\ell }} = \frac{{T\ell }}{{{\pi ^2}\Delta \ell }} \Rightarrow T = \frac{{Y\pi {r^2}\Delta \ell }}{\ell }$

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Vedclass · Accepted Answer

$35$

Vedclass · Answer

$Y = \frac{{T/\pi {r^2}}}{{\Delta \ell /\ell }} = \frac{{T\ell }}{{{\pi ^2}\Delta \ell }} \Rightarrow T = \frac{{Y\pi {r^2}\Delta \ell }}{\ell }$

$V=\sqrt{\frac{T}{m}}=\sqrt{\frac{Y \pi r^{2} \Delta \ell}{\pi r^{2} d \ell}}=\sqrt{\frac{Y \Delta \ell}{\ell d}}=70 \mathrm{\,m} / \mathrm{s}$

Lowest Frequency $=\frac{V}{2 \ell}=\frac{70}{2}=35 \mathrm{\,Hz}$

A wire of density $9 \times 10^3 \,kg/m^3$ is stretched between two clamps one meter 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 ... $Hz$ $[Y = 9 \times 10^{10} \,N/m^2]$ ?

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