Two wires of different densities but same area of cross section are 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.
Solution
Let:
m = Mass per unit length of the first wire
a = Area of the cross section
ρ = Density of the wire
T = Tension
Let the velocity of the first string be v1.
Thus, we have:
\[\nu_1 = \sqrt{\left( \frac{T}{m_1} \right)}\]
The mass per unit length can be given as
\[m_1 = \left( \frac{\rho_1 a_1 I_1}{I_1} \right) = \rho_1 a_1 \]
\[ \Rightarrow \nu_1 = \sqrt{\left( \frac{T}{\rho_1 a_1} \right)} . . . (1)\]
Let the velocity of the first string be v2.
Thus, we have:
\[\nu_2 = \sqrt{\left( \frac{T}{m_2} \right)}\]
\[ \Rightarrow \nu_2 = \sqrt{\left( \frac{T}{\rho_2 a_2} \right)} . . . (2)\]
Given,
\[\nu_1 = 2 \nu_2 \]
\[ \Rightarrow \sqrt{\left( \frac{T}{a_1 \rho_1} \right)} = 2\sqrt{\left( \frac{T}{a_2 \rho_2} \right)}\]
\[ \Rightarrow \left( \frac{T}{a_1 \rho_1} \right) = 4 \left( \frac{T}{a_2 \rho_2} \right)\]
\[ \Rightarrow \frac{\rho_1}{\rho_2} = \frac{1}{4}\]
\[ \Rightarrow \rho_1 : \rho_2 = 1: 4\]
