Question

When a uniform metallic wire is stretched the lateral strain produced in it β .If σ and Y are the Poisson 's' ratio and Young's modulus for wire,then elastic potential energy density of wire is

Answer 1

Given :

Lateral strain =

$\beta$

Poisson's ratio =

$\sigma$

Young's modulus =

$\gamma$

To find :

Potential energy density of wire.

Solution :

The formula of Potential energy density =

$\frac{1}{2} \times lateral \: strain \times stress$

(equation 1)

We know that

Young's modulus :

$\gamma = \frac{stress}{lateral \: strain} \\ so \\ stress = \gamma \times lateral \: strain$

(equation 2)

Putting the value of stress from equation 2 to equation 1.

Potential energy density =

$\frac{1}{2} \times \gamma \times {(lateral \: strain)}^{2} = \frac{1}{2} \times \gamma \times { \beta }^{2}$

Let longitudinal strain =

$\alpha$

Putting Longitudinal strain in numerator and denominator.

Potential energy density=

$= \frac{1}{2} \times \gamma \times { \beta } \times \frac{ \beta }{ \alpha } \times \alpha$

We know that

Poisson's ratio =

$\sigma = \frac{ \beta }{ \alpha }$

So, the value of potential energy density =

$\bf{\frac{1}{2}\times \gamma \times \beta \times \sigma \times \alpha} = \frac{1}{2} \alpha \beta \gamma \sigma$