Two vibrating strings A and B of the same mat | vedclass

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(B) The frequency of a vibrating string is given by $n = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \rho \pi r^2$ is the linear mass density.Sinc

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$3: 1$

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(B) The frequency of a vibrating string is given by $n = \frac{p}{2L} \sqrt{\frac{T}{\mu}}$,where $\mu = \rho \pi r^2$ is the linear mass density.Since the material is the same,density $\rho$ is constant. Given tension $T$ is also constant,the frequency is $n \propto \frac{p}{Lr}$.For string $A$: Length $L_A = 3L$,radius $r_A = 2r$,and it vibrates in the second overtone,so $p_A = 3$.Thus,$n_A \propto \frac{3}{(3L)(2r)} = \frac{1}{2Lr}$.For string $B$: Length $L_B = 2L$,radius $r_B = 3r$,and it vibrates in the fundamental mode,so $p_B = 1$.Thus,$n_B \propto \frac{1}{(2L)(3r)} = \frac{1}{6Lr}$.The ratio is $\frac{n_A}{n_B} = \frac{1/2Lr}{1/6Lr} = \frac{6}{2} = 3$.Therefore,the ratio $n_A / n_B = 3: 1$.

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