Speed of a transverse wave on a straight wire | vedclass

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(C) The speed of a transverse wave on a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass de

Vedclass · Accepted Answer

$0.03$

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(C) The speed of a transverse wave on a stretched wire is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Given: $v = 90\; ms^{-1}$,$m = 6.0 	imes 10^{-3}\; kg$,$L = 0.6\; m$,$A = 1.0 	imes 10^{-6}\; m^{2}$,$Y = 16 	imes 10^{11}\; Nm^{-2}$.Linear mass density $\mu = \frac{m}{L} = \frac{6.0 	imes 10^{-3}}{0.6} = 10^{-2}\; kg/m$.From $v = \sqrt{\frac{T}{\mu}}$,we have $T = v^{2} \mu = (90)^{2} 	imes 10^{-2} = 8100 	imes 10^{-2} = 81\; N$.Young's modulus $Y = \frac{T/A}{\Delta L/L}$,so $\Delta L = \frac{T L}{Y A}$.Substituting the values: $\Delta L = \frac{81 	imes 0.6}{16 	imes 10^{11} 	imes 1.0 	imes 10^{-6}} = \frac{48.6}{16 	imes 10^{5}} = 3.0375 	imes 10^{-5}\; m \approx 0.03\; mm$.

Speed of a transverse wave on a straight wire (mass $6.0\; g$,length $60\; cm$,and area of cross-section $1.0\; mm^{2}$) is $90\; ms^{-1}$. If the Young's modulus of the wire is $16 \times 10^{11}\; Nm^{-2}$,the extension of the wire over its natural length is: (in $; mm$)

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