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

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(B) The fundamental frequency $f$ of a stretched wire is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\

Vedclass · Accepted Answer

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

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(B) The fundamental frequency $f$ of a stretched wire is given by the formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Linear mass density $\mu$ is defined as mass per unit length,which can be expressed in terms of density $\rho$ and cross-sectional area $A$ as $\mu = A \rho$.Since the wire has a circular cross-section with diameter $D$,the area $A = \pi \frac{D^2}{4}$.Substituting this into the expression for $\mu$,we get $\mu = \frac{\pi D^2 \rho}{4}$.Now,substitute $\mu$ back into the frequency formula: $f = \frac{1}{2L} \sqrt{\frac{T}{\frac{\pi D^2 \rho}{4}}} = \frac{1}{2L} \sqrt{\frac{4T}{\pi D^2 \rho}} = \frac{1}{2L} \cdot \frac{2}{D} \sqrt{\frac{T}{\pi \rho}} = \frac{1}{LD} \sqrt{\frac{T}{\pi \rho}}$.From this expression,it is clear that $f \propto \frac{1}{LD}$.

$A$ uniform wire of length $L$,diameter $D$,and density $\rho$ is stretched by a tension $T$. The frequency $f$ of the wire is proportional to:

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