A wire having a linear mass density 9.0 times | vedclass

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(C) The wave speed $v$ in the wire is given by $v = \sqrt{\frac{T}{\mu}}$.Given $T = 900 \; 	ext{N}$ and $\mu = 9.0 	imes 10^{-4} \; 	ext{kg/m}$.$v

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$10$

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(C) The wave speed $v$ in the wire is given by $v = \sqrt{\frac{T}{\mu}}$.Given $T = 900 \; 	ext{N}$ and $\mu = 9.0 	imes 10^{-4} \; 	ext{kg/m}$.$v = \sqrt{\frac{900}{9.0 	imes 10^{-4}}} = \sqrt{10^6} = 1000 \; 	ext{m/s}$.The resonant frequencies of a string fixed at both ends are given by $f_n = \frac{nv}{2L}$,where $n = 1, 2, 3, \dots$.The difference between two consecutive resonant frequencies is $\Delta f = f_{n+1} - f_n = \frac{(n+1)v}{2L} - \frac{nv}{2L} = \frac{v}{2L}$.Given $\Delta f = 550 \; 	ext{Hz} - 500 \; 	ext{Hz} = 50 \; 	ext{Hz}$.Therefore,$\frac{v}{2L} = 50 \; 	ext{Hz}$.Substituting $v = 1000 \; 	ext{m/s}$,we get $\frac{1000}{2L} = 50$.$2L = \frac{1000}{50} = 20$.$L = 10 \; 	ext{m}$.

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