# 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,

- $1.22 v$
- $0.61 v$
- $1.50 v$
- $0.75 v$

**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.