two strings A and B made of the same material have equal lengths. the cross sectional area of A is half that of B while the tension on A is twice that on B. the ratio of the velocities of transverse waves in A and B is? (ans : 2:1)

Dear Student,

Please find below the solution to the asked query:

Velocity of transverse wave in a stretched string is,

$$\begin{gathered} V=\sqrt{\frac{T}{\mu }} =\sqrt{\frac{Tl}{m}}=\sqrt{\frac{Tl}{\rho Al}} \\ \Rightarrow V=\sqrt{\frac{T}{\rho A}} \end{gathered}$$

So, the ratio of the velocities of the transverse waves is,

$\frac{V_A}{V_B}=\frac{\sqrt{{\displaystyle \frac{T_A}{{\rho }_A{A}_A}}}}{\sqrt{{\displaystyle \frac{T_B}{{\rho }_B{A}_B}}}}=\sqrt{\frac{T_A}{T_B}}\sqrt{\frac{A_B}{A_A}}\sqrt{\frac{{\rho }_B}{{\rho }_A}}$

Given that the two strings are of same material. So, the densities are equal. Cross sectional area of string A is half of that of string B. The tension in string A is twice of that of string B.

Therefore,

$$\begin{gathered} \frac{V_A}{V_B}=\sqrt{\frac{T_A}{T_B}}\sqrt{\frac{A_B}{A_A}}\sqrt{\frac{{\rho }_B}{{\rho }_A}}=\sqrt{\frac{2}{1}}\sqrt{\frac{2}{1}}\sqrt{\frac{1}{1}}=\sqrt{\frac{4}{1}} \\ \Rightarrow \frac{V_A}{V_B}=\frac{2}{1} \end{gathered}$$

Hope this information will clear your doubts about the topic.

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