Question

Both the string, shown in figure, are made of same material and have same cross-section. The pulleys are light. The wave speed of a transverse wave in the string AB is ${\upsilon }_1$ and in CD it is ${\upsilon }_2$. Then ${\upsilon }_1/{\upsilon }_2$ is

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

Answer 1

The correct option is D $1/\sqrt{2}$

according to Free body diagram as shown in figure,

by equilibrium condition we can write,

$T_{AB}+T_{AB}=T_{CD}$

$\Rightarrow 2T_{AB}=T_{CD}$

$\frac{T_{AB}}{T_{CD}}=\frac{1}{2}$ ................equation(1)

as we know that wave speed of transverse wave in string is given by,

$v=\sqrt{\frac{T}{\mu }}$ (Where T=tension in string, $\mu$=mass per unit length of string)

SInce, here $\mu$ is constant,

So,

$v\propto \sqrt{T}$

$\Rightarrow \frac{V_1}{V_2}=\sqrt{\frac{T_{AB}}{T_{CD}}}$

from equation(1),

$\Rightarrow \frac{V_1}{V_2}=\sqrt{\frac{1}{2}}$