A uniform wire of length L,diameter D,and den | vedclass

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(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the linear mass density.Linear ma

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$f \propto \frac{1}{L D}$

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(A) The fundamental frequency of a stretched wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu$ is the linear mass density.Linear mass density $\mu = 	ext{mass per unit length} = 	ext{density} 	imes 	ext{cross-sectional area} = ho 	imes (\pi \frac{D^2}{4})$.Substituting $\mu$ into the frequency formula:$f = \frac{1}{2L} \sqrt{\frac{T}{ho \pi \frac{D^2}{4}}} = \frac{1}{2L} \sqrt{\frac{4T}{\pi ho D^2}} = \frac{1}{2L} \cdot \frac{2}{D} \sqrt{\frac{T}{\pi ho}} = \frac{1}{LD} \sqrt{\frac{T}{\pi ho}}$.Since $T$,$ho$,and $\pi$ are constants,we have $f \propto \frac{1}{LD}$.

$A$ uniform wire of length $L$,diameter $D$,and density $\rho$ is stretched under a tension $T$. The correct relation between its fundamental frequency $f$,the length $L$,and the diameter $D$ is:

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