A wire of density 9 times 10^{-3} ,kg, cm^{-3 | vedclass

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(A) Given: Density $ho = 9 	imes 10^{-3} \,kg/cm^3 = 9 	imes 10^{-3} / 10^{-6} \,kg/m^3 = 9000 \,kg/m^3$.Length $L = 1 \,m$.Strain $= 4.9 	imes 1

Vedclass · Accepted Answer

$35$

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(A) Given: Density $ho = 9 	imes 10^{-3} \,kg/cm^3 = 9 	imes 10^{-3} / 10^{-6} \,kg/m^3 = 9000 \,kg/m^3$.Length $L = 1 \,m$.Strain $= 4.9 	imes 10^{-4}$.Young's modulus $Y = 9 	imes 10^{10} \,N/m^2$.For the lowest frequency (fundamental mode),$L = \lambda / 2$,so $\lambda = 2L = 2 \,m$.The frequency $f$ is given by $f = v / \lambda = (1 / \lambda) \sqrt{T / \mu}$,where $T$ is tension and $\mu$ is mass per unit length.Since $Y = (T/A) / 	ext{strain}$,we have $T = Y \cdot A \cdot 	ext{strain}$.Also,$\mu = m/L = (ho \cdot V) / L = (ho \cdot A \cdot L) / L = ho \cdot A$.Substituting these into the frequency formula:$f = (1 / 2L) \sqrt{(Y \cdot A \cdot 	ext{strain}) / (ho \cdot A)} = (1 / 2L) \sqrt{(Y \cdot 	ext{strain}) / ho}$.Substituting the values:$f = (1 / 2) \sqrt{(9 	imes 10^{10} 	imes 4.9 	imes 10^{-4}) / 9000} = (1 / 2) \sqrt{(9 	imes 10^6 	imes 4.9) / 9000} = (1 / 2) \sqrt{10^3 	imes 4.9} = (1 / 2) \sqrt{4900} = 70 / 2 = 35 \,Hz$.

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