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Two Wires of Different Densities but Same Area of Cross Section Are Soldered Together at One End and Are Stretched to a Tension T. - Physics

Sum

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.

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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,

\[\nu_1  = 2 \nu_2 \] 

\[ \Rightarrow \sqrt{\left( \frac{T}{a_1 \rho_1} \right)} = 2\sqrt{\left( \frac{T}{a_2 \rho_2} \right)}\] 

\[ \Rightarrow \left( \frac{T}{a_1 \rho_1} \right) = 4  \left( \frac{T}{a_2 \rho_2} \right)\] 

\[ \Rightarrow \frac{\rho_1}{\rho_2} = \frac{1}{4}\] 

\[ \Rightarrow  \rho_1 :  \rho_2    =   1: 4\]

  Is there an error in this question or solution?
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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 15 Wave Motion and Waves on a String
Q 17 | Page 324
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