# Hooke's Law

## Problems from IIT JEE

Problem (IIT JEE 1996): The extension in a string, obeying Hooke's law, is $x$. The speed of transverse wave in the stretched string is $v$. If the extension in the string is increased to $1.5 x$, the speed of transverse wave will be,

1. $1.22 v$
2. $0.61 v$
3. $1.50 v$
4. $0.75 v$

Solution: The velocity of transverse wave in a string under tension $T$ and mass per unit length $\mu$ is given by, \begin{align} \label{yxa:eqn:1} v=\sqrt{T/\mu}. \end{align} The tension in the string is related to its extension $x$ by Hooke's law, \begin{align} \label{yxa:eqn:2} T=kx, \end{align} where $k$ is a constant. The tension is $kx$ in the first case and $1.5kx$ in the second case. Substitute $T=1.5kx$ in first equation to get velocity in the second case, \begin{align} v^\prime=\sqrt{1.5kx/\mu}=\sqrt{1.5}\,\sqrt{kx/\mu}=1.22v.\nonumber \end{align} We have assumed $x$ to be much smaller than the string length to keep $\mu$ constant.