A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is $4\ .9\ imes \ 10^{- 7\ }m^{2\ .}$ If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140\ rad\ s^{- 1}.\ $If the Young’s modulus of the material of the wire is $n\ imes 10^{9}\ {Nm}^{- 2\ \ }$, the value of n is

Question

A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is $4\ .9\ 	imes \ 10^{- 7\ }m^{2\ .}$ If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140\ rad\ s^{- 1}.\ $If the Young’s modulus of the material of the wire is $n\ 	imes 10^{9}\ {Nm}^{- 2\ \ }$, the value of n is

Dr. Abubacker · Accepted Answer

(a) $\omega = \sqrt{\frac{K}{m} =}\ \sqrt{\frac{YA}{lm}}$
$= \sqrt{\frac{{(n 	imes 10}^{9})\ (4.9 	imes 10^{- 7})}{1 	imes 0.1}}$
Given,$\omega = 140\ rad\ s^{- 1}\ in\ above\ equation,\ we\ get,$
$n = 4$

Explanation

Reviewed by Dr. Abubacker

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