A metal wire of linear mass density of 9.8 , | vedclass

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(B) In the condition of resonance,the frequency of the alternating current $(n)$ is equal to the fundamental frequency of the vibrating wire.The funda

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$50$

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(B) In the condition of resonance,the frequency of the alternating current $(n)$ is equal to the fundamental frequency of the vibrating wire.The fundamental frequency of a stretched string is given by $n = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$,where $l$ is the length,$T$ is the tension,and $\mu$ is the linear mass density.Given:Length $l = 1 \, m$Tension $T = 10 \, kg 	ext{-} wt = 10 	imes 9.8 \, N = 98 \, N$Linear mass density $\mu = 9.8 \, g/m = 9.8 	imes 10^{-3} \, kg/m$Substituting the values:$n = \frac{1}{2 	imes 1} \sqrt{\frac{98}{9.8 	imes 10^{-3}}}$$n = \frac{1}{2} \sqrt{10000}$$n = \frac{1}{2} 	imes 100 = 50 \, Hz$.

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