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.
\[Frequency,  f = \frac{1}{LD}\sqrt{\frac{T}{\pi\rho}}\]

\[\Rightarrow  f_1  = \frac{1}{L D_1}\sqrt{\frac{T_1}{\pi \rho_1}}\] 

\[ \Rightarrow  f_2  = \frac{1}{L D_2}\sqrt{\frac{T_2}{\pi \rho_2}}\]v

\[\therefore   \frac{f_1}{f_2} = \frac{L D_2}{L D_1}\sqrt{\frac{T_1}{T_2}}\sqrt{\frac{\pi \rho_2}{\pi \rho_1}}\] 

\[ \Rightarrow \frac{f_1}{f_2} = \frac{1}{3}\sqrt{\frac{2}{1} \times \frac{2}{1}}\] 

\[ \Rightarrow    f_1 :  f_2  = 2: 3\]