When the area of cross-section of a stretched wire is halved and tension is doubled, the speed of…
When the area of cross-section of a stretched wire is halved and tension is doubled, the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then, $k$
1
4
2
8
Solution
Speed of transverse wave in stretched wire is given as
$
v=\sqrt{\frac{T}{m}}
$
where, $T=$ tension in the wire
and $\quad m=$ mass per unit length of wire.
$
\begin{aligned}
& \therefore \quad v=\sqrt{\frac{T}{M / l}} \quad\left[\because m=\frac{M}{l}\right] \\
& =\sqrt{\frac{T l}{M}}=\sqrt{\frac{T l}{V \cdot \rho}} \\
& {[\because M=V \cdot \rho]} \\
& =\sqrt{\frac{T l}{A \cdot l \cdot \rho}} \\
& {[\because V=A \cdot l]} \\
& v=\sqrt{\frac{T}{A \rho}} \Rightarrow v \propto \sqrt{\frac{T}{A}} \\
& \Rightarrow \quad \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1} \times \frac{A_1}{A_2}} \\
& \text { Given, } \quad A_2=\frac{A_1}{2}, T_2=2 T_1 \\
&
\end{aligned}
$
$\therefore$ From Eq. (i), we get
$
\begin{aligned}
& \frac{v_2}{v_1}=\sqrt{\frac{2 T_1}{T_1} \times \frac{A_1}{\frac{A_1}{2}}}=\sqrt{4} \\
& v_2=2 v_1 \Rightarrow K=2
\end{aligned}
$