A wire of density $9 imes 10^{3}\ kgm^{- 3}$ is stretched between two clamps 1m part and is subjected to an extension of$4.9 imes 10^{- 4}m$. The lowest frequency of transverse vibration in the wire is $Y = 9 imes 10^{10}Nm^{- 2})\ $

Question

A wire of density $9 	imes 10^{3}\ kgm^{- 3}$ is stretched between two clamps 1m part and is subjected to an extension of$4.9 	imes 10^{- 4}m$. The lowest frequency of transverse vibration in the wire is $Y = 9 	imes 10^{10}Nm^{- 2})\ $

Dr. Abubacker · Accepted Answer

(b) Given :$l = 4.9 	imes 10^{- 4}m$
From the formula
$Y = \frac{stress}{strain} = \frac{F/\pi r^{2}}{l/L}$
Or
$\frac{F}{\pi r^{2}} = Y\frac{l}{L}$
=$9 	imes 10^{10} 	imes \frac{{{\ \ \ \ 4.9 	imes 10}^{10}}^{-}}{4}\ \ $
=44.1$	imes 10^{6}$
$V = \frac{1}{2L}\sqrt{\left( \frac{T}{m} ight)}$
$= \frac{1}{2L}\sqrt{\left( \frac{F}{\pi r^{2}d} ight)} = \frac{1}{2}\sqrt{\left( \frac{F/\pi r^{2}}{d} ight)}\ $
$= \frac{1}{2} 	imes \sqrt{\left( \frac{44.1 	imes 10^{6}}{9 	imes 10^{3}} ight)}$
$= \frac{1}{2}\sqrt{\left( 4.9 	imes 10^{3} ight)} = \frac{1}{2}\sqrt{\left( 49 	imes 10^{2} ight)}$
$= \frac{7 	imes 10}{2} = 35\ Hz$

Q. A wire of density $9 \times 10^{3}\ kgm^{- 3}$ is stretched between two clamps 1m part and is subjected to an extension of$4.9 \times 10^{- 4}m$. The lowest frequency of transverse vibration in the wire is $Y = 9 \times 10^{10}Nm^{- 2})\ $

Explanation

Reviewed by Dr. Abubacker

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