Question

Two wires of different densities but same area of cross section is soldered together at one end and are stretched to a tension $T$. The velocity of a transverse wave in the first wire is double of that in the second wire. Find the ratio of the density of the first wire to that of the second wire.

Answer 1

Step 1: Given

Tension in the wires: $T_1=T_2=T$

Area of cross section of both wires: $A_1=A_2=A$

Ratio of velocities of waves: $\frac{v_1}{v_2}=\frac{2}{1}$

Step 2: Formula Used

$v=\sqrt{\frac{T}{m}}=\sqrt{\frac{T}{\rho A}}\left(m=\rho A\right)$ [mass per unit length id product of area and density]

Step 3: Calculate the ratio of densities by dividing the velocities and substituting the values

$$\begin{gathered} \frac{v_1}{v_2}=\frac{\sqrt{\frac{T_1}{{\rho }_1{A}_1}}}{\sqrt{\frac{T_2}{{\rho }_2{A}_2}}} \\ \Rightarrow \frac{2}{1}=\frac{\sqrt{\frac{T}{{\rho }_{1}A}}}{\sqrt{\frac{T}{{\rho }_{2}A}}} \\ \Rightarrow \frac{2}{1}=\sqrt{\frac{{\rho }_2}{{\rho }_1}} \\ \Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{1}{4} \end{gathered}$$

Hence, the ratio of densities is $1:4$.