Physics - Wave on String Question with Solution | TestHub

The formula for frequency of a sonometer is $\mathrm{f}=\frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\pi \rho {\mathrm{D}}^2}}$ Here $l$ is length, $\mathrm{T}$ is tension, $\mathrm{D}$ is diameter and $\rho$ is density. The frequency of both the wires is same. The frequency of the wire $\mathrm{A}$ is ${\mathrm{f}}_{\mathrm{A}}=\frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\pi {\rho }_1{\mathrm{D}}^2}}$ The frequency of the wire $\mathrm{B}$ is $f_{\mathrm{B}}=\frac{1}{2l}\sqrt{\frac{2\mathrm{T}}{\pi {\rho }_2(2\mathrm{D}{)}^2}}$ Equating both the frequencies $\begin{array}{rl}& \frac{1}{2l}\sqrt{\frac{\mathrm{T}}{\pi \rho {\mathrm{D}}^2}}=\frac{1}{2l}\sqrt{\frac{2\mathrm{T}}{\pi {\rho }_2(2\mathrm{D}{)}^2}}\\ & \sqrt{\frac{1}{{\rho }_1}}=\sqrt{\frac{1}{2{\rho }_2}}\\ & \frac{1}{{\rho }_1}=\frac{1}{2{\rho }_2}\\ & \therefore {\rho }_2=\frac{{\rho }_1}{2}\end{array}$