A steel wire of length 1.25 m is stretched b | vedclass

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Question

(C) Given: Length of wire $L = 1.25 \ m$,strain $\epsilon = 0.14 \% = 0.0014$,density $ho = 7.7 	imes 10^3 \ kg \ m^{-3}$,Young's modulus $Y = 2.2

Vedclass · Accepted Answer

$80$

Vedclass · Answer

(C) Given: Length of wire $L = 1.25 \ m$,strain $\epsilon = 0.14 \% = 0.0014$,density $ho = 7.7 	imes 10^3 \ kg \ m^{-3}$,Young's modulus $Y = 2.2 	imes 10^{11} \ N \ m^{-2}$.Fundamental frequency $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = ho A$.Since $Y = \frac{T/A}{\epsilon}$,we have $\frac{T}{A} = Y \epsilon = 2.2 	imes 10^{11} 	imes 0.0014 = 3.08 	imes 10^8 \ N \ m^{-2}$.Substituting $\frac{T}{\mu} = \frac{T}{ho A} = \frac{Y \epsilon}{ho}$ into the frequency formula:$f = \frac{1}{2L} \sqrt{\frac{Y \epsilon}{ho}} = \frac{1}{2 	imes 1.25} \sqrt{\frac{3.08 	imes 10^8}{7.7 	imes 10^3}} = \frac{1}{2.5} \sqrt{0.4 	imes 10^5} = \frac{1}{2.5} \sqrt{40000} = \frac{200}{2.5} = 80 \ Hz$.

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