Question

Two wires are kept right between the same pair of supports. The tensions in the wires are in the ratio $2:1$, the radii are in the ratio $3:1$ and the densities are in the ratio $1:2$. Find the ratio of their fundamental frequencies.

Answer 1

Step 1: Given

Ratio of tensions in the wire: $T_1:T_2=2:1$

Ratio of radii of the wire: $r_1:r_2=3:1$

Ratio of densities of the wire: ${\rho }_1:{\rho }_2=1:2$

Step 2: Formula Used

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

$\mu =\rho A$

$A=\pi r^2$

Step 3: Find the ratio of frequencies

Find an expression for the ratio of frequencies by using the formulas

$$\begin{gathered} \frac{f_1}{f_2}=\frac{\frac{1}{2l}\sqrt{\frac{T_1}{{\mu }_1}}}{\frac{1}{2l}\sqrt{\frac{T_2}{{\mu }_2}}} \\ =\sqrt{\frac{T_1{\mu }_2}{T_2{\mu }_1}} \\ =\sqrt{\frac{T_1{\rho }_2{A}_2}{T_2{\rho }_1{A}_1}} \\ =\sqrt{\frac{T_1{\rho }_2\pi {r_2}^2}{T_2{\rho }_1\pi {r_1}^2}} \end{gathered}$$

Calculate the ratio of frequencies by substituting the values

$$\begin{gathered} \frac{f_1}{f_2}=\sqrt{\frac{T_1{\rho }_2{r_2}^2}{T_2{\rho }_1{r_1}^2}} \\ =\sqrt{\frac{2\times 2\times 1^2}{1\times 1\times 3^2}} \\ =\sqrt{\frac{4}{9}} \\ =\frac{2}{3} \end{gathered}$$

Hence, the ratio of their fundamental frequencies is $2:3$.