Two vibrating metal wires of the same material of length 'L' and '3L' and have radii '2r' and 'r' respectively. They are stretched under same tension. Both the

Question

Two vibrating metal wires of the same material of length 'L' and '3L' and have radii '2r' and 'r' respectively. They are stretched under same tension. Both the wires vibrate in their fundamental modes. The ratio n 2 ​ n 1 ​ ​ is [ n 1 ​ and n 2 ​ are the frequencies of first and second wire respectively]

TestHub · Accepted Answer

The fundamental frequency of a stretched wire is given by $n = \frac{1}{2 L} \frac{T}{μ}$

The linear mass density $μ = π r^{2} ρ$ , where $ρ$ is the density of the material.

Substituting $μ$ , we get $n = \frac{1}{2 L} \frac{T}{π r ^{2} ρ} = \frac{1}{2 L r} \frac{T}{π ρ}$

Since the tension $T$ and density $ρ$ are the same for both wires, $n \propto \frac{1}{L r}$

For the first wire, $L_{1} = L$ and $r_{1} = 2 r$ , so $n_{1} \propto \frac{1}{L ( 2 r )} = \frac{1}{2 L r}$

For the second wire, $L_{2} = 3 L$ and $r_{2} = r$ , so $n_{2} \propto \frac{1}{( 3 L ) r} = \frac{1}{3 L r}$

Taking the ratio of the frequencies:

$\frac{n _{1}}{n _{2}} = \frac{\frac{1}{2 L r}}{\frac{1}{3 L r}} = \frac{3}{2} = 1.5$

The ratio is $1.5 : 1$ .

Answer: $1.5 : 1$

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