Question

Both the strings shown in the figure are made of same material and have the same cross-section. The pulleys are light. The wave speed of a transverse wave in the string AB is $v_1$ and in CD it is $v_2$ The $v_1/v_2$ is

• A. 1
• B. 2
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is C $\sqrt{2}$

Let the tension in the string CD be=T

Then the tension in the string AB will be=T+T=$2$T

$\therefore$ Velocity of the wave in string CD

$V_2=\sqrt{\frac{T}{\mu }}$

And the velocity of the wave in string AB

$V_1=\sqrt{\frac{2T}{\mu }}=\sqrt{2}{V}_2\Rightarrow [\frac{V_1}{V_2}=\frac{\sqrt{2}}{1}]$

Hence option (C) is correct