Two copper wires of radii r_1 and r_2 (r_1 | vedclass

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(C) The velocity $v$ of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$

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travel faster in the thinner wire

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(C) The velocity $v$ of a transverse wave on a stretched string is given by the formula $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Linear mass density $\mu = \rho \cdot A$,where $\rho$ is the density of the material and $A$ is the cross-sectional area.Since the wires are made of the same material (copper),$\rho$ is constant.The cross-sectional area $A = \pi r^2$,so $\mu = \rho \cdot \pi r^2$.Substituting this into the velocity formula: $v = \sqrt{\frac{T}{\rho \pi r^2}} = \frac{1}{r} \sqrt{\frac{T}{\rho \pi}}$.Since $T$,$\rho$,and $\pi$ are constant,we have $v \propto \frac{1}{r}$.Given $r_1 > r_2$,the thinner wire $(r_2)$ will have a higher velocity compared to the thicker wire $(r_1)$.Therefore,the transverse wave travels faster in the thinner wire.

Two copper wires of radii $r_1$ and $r_2$ $(r_1 > r_2)$ are subjected to the same tension and are plucked. The transverse waves will

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