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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,

$\nu_1 = 2 \nu_2$

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

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

$\Rightarrow \frac{\rho_1}{\rho_2} = \frac{1}{4}$

$\Rightarrow \rho_1 : \rho_2 = 1: 4$

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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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