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(C) The fundamental frequency of a sonometer wire is given by $f = \frac{v}{2L} = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since $\mu = A \rho$,we have $f =

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$178.2$

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(C) The fundamental frequency of a sonometer wire is given by $f = \frac{v}{2L} = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$.Since $\mu = A ho$,we have $f = \frac{1}{2L} \sqrt{\frac{T}{A ho}}$.From Young's modulus $Y = \frac{T/A}{\Delta L/L}$,we get $\frac{T}{A} = Y 	imes 	ext{strain} = Y 	imes \frac{\Delta L}{L}$.Substituting this into the frequency formula: $f = \frac{1}{2L} \sqrt{\frac{Y 	imes 	ext{strain}}{ho}}$.Given: $L = 1.5 \ m$,$	ext{strain} = 1\% = 0.01$,$Y = 2.2 	imes 10^{11} \ N/m^2$,$ho = 7.7 	imes 10^3 \ kg/m^3$.$f = \frac{1}{2 	imes 1.5} \sqrt{\frac{2.2 	imes 10^{11} 	imes 0.01}{7.7 	imes 10^3}}$.$f = \frac{1}{3} \sqrt{\frac{2.2 	imes 10^9}{7.7}} = \frac{1}{3} \sqrt{0.2857 	imes 10^9} = \frac{1}{3} \sqrt{28.57 	imes 10^7} \approx \frac{5345}{3} \approx 178.2 \ Hz$.

$A$ sonometer wire of length $1.5 \ m$ is made of steel. The tension in it produces an elastic strain of $1 \%$. What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times 10^3 \ kg/m^3$ and $2.2 \times 10^{11} \ N/m^2$ respectively (in $Hz$)?

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