Question

A metallic wire is suspended by suspending weight to it. If $\alpha $is the longitudinal strain and $Y$is the Young’s Modulus of elasticity, show that the elastic potential energy per unit volume is given by $\dfrac{{Y{\alpha ^2}}}{2}$.

Answer 1

Hint: This question is from the chapter elasticity. We have to calculate the expression for elastic potential energy. For this we have to use the formula of work done in stretching a wire. Using the formula, we will calculate the elastic potential energy per unit volume.

Complete step by step answer:The elastic potential energy stored in a wire is equal to the work done in stretching a wire to that length. Therefore, if we find the work done it will be equal to the potential energy of the wire and then we will find the potential energy per unit volume.
The work done in stretching a wire is given by the expression:
\[W = \int\limits_0^l {F \cdot dl} \]
The force acting on a stretched string is given by:
\[F = \dfrac{{YAl}}{L}\]
Putting this value in equation of Work done, we get:
\[W = \int\limits_0^l {\dfrac{{YAl}}{L} \cdot dl} \], if we have to integrate the work done in stretching the string to length $l$.
Integrating we get:
\[\begin{array}{l}
W = \dfrac{{YA}}{L}\int\limits_0^l l \\
W = \dfrac{{YA}}{L}{\left[ {\dfrac{{{l^2}}}{2}} \right]_0}^l\\
W = \dfrac{{YA{l^2}}}{{2L}}
\end{array}\]
Now to add terms of volume and stress and strain we have to multiply and divide $L$ in the above equation.
$W = \dfrac{{YAL{l^2}}}{{2{L^2}}}$,
We will separate the terms:
$\begin{array}{l}
W = \dfrac{Y}{2}\left[ {AL} \right]{\left[ {\dfrac{l}{L}} \right]^2}\\
W = \dfrac{Y}{2} \times V \times {(strain)^2}
\end{array}$
We know this is the formula of potential energy and we have to calculate the formula of potential energy per unit volume.
The strain is $\alpha $ and we put this value in the derived equation:
$\begin{array}{l}
W = \dfrac{Y}{2} \times V \times {\alpha ^2}\\
\dfrac{W}{V} = \dfrac{1}{2}Y{\alpha ^2}
\end{array}$
The above expression is of potential energy per unit volume.
Hence, we have proved the required condition which needed to be proved.

Note:In questions like this we have to perform integration therefore the students should know that the basic integration formulas. Students should remember the relations because without knowing the relations, such questions cannot be solved.