Question

Two wires are kept tight 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

Given:
The tensions in the two wires are in the ratio of 2:1.
$\Rightarrow \frac{T_1}{T_2}=2$
Ratio of the radii is 3:1.
$\Rightarrow \frac{r_1}{r_2}=3=\frac{D_1}{D_2}$
Density in the ratios of 1:2.
$\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{1}{2}$
Let the length of the wire be L.
$\mathrm{Frequency,}f=\frac{1}{LD}\sqrt{\frac{T}{\pi \rho }}$
$$\begin{gathered} \Rightarrow f_1=\frac{1}{L{D}_1}\sqrt{\frac{T_1}{\pi {\rho }_1}} \\ \Rightarrow f_2=\frac{1}{L{D}_2}\sqrt{\frac{T_2}{\pi {\rho }_2}} \end{gathered}$$
$$\begin{gathered} \therefore \frac{f_1}{f_2}=\frac{L{D}_2}{L{D}_1}\sqrt{\frac{T_1}{T_2}}\sqrt{\frac{\pi {\rho }_2}{\pi {\rho }_1}} \\ \Rightarrow \frac{f_1}{f_2}=\frac{1}{3}\sqrt{\frac{2}{1}\times \frac{2}{1}} \\ \Rightarrow f_1:f_2=2:3 \end{gathered}$$