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(B) The frequency of vibration $n$ must be the same for both segments of the wire.The frequency of a stretched wire is given by $n = \frac{p}{2L} \sqr

Vedclass · Accepted Answer

$1:2$

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(B) The frequency of vibration $n$ must be the same for both segments of the wire.The frequency of a stretched wire is given by $n = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \pi r^2 \rho$ is the linear mass density.For the first wire (radius $r$,length $L$): $n = \frac{p_1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}}$.For the second wire (radius $2r$,length $L$): $n = \frac{p_2}{2L} \sqrt{\frac{T}{\pi (2r)^2 \rho}} = \frac{p_2}{2L} \sqrt{\frac{T}{4 \pi r^2 \rho}} = \frac{p_2}{4L} \sqrt{\frac{T}{\pi r^2 \rho}}$.Equating the frequencies: $\frac{p_1}{2L} \sqrt{\frac{T}{\pi r^2 \rho}} = \frac{p_2}{4L} \sqrt{\frac{T}{\pi r^2 \rho}}$.Simplifying,we get $\frac{p_1}{2} = \frac{p_2}{4}$,which implies $\frac{p_1}{p_2} = \frac{2}{4} = \frac{1}{2}$.

Two wires of the same material with radii $r$ and $2r$ respectively are welded together end to end. The combination is then used as a sonometer wire under tension $T$. The joint is kept midway between the two bridges. The ratio of the number of loops in the wires,given that the joint is a node,is:

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