A metallic wire of length L, area of cross-section A is suspended by attaching some wieght to it. If `alpha` is the longitudinal strain and Y is Young's modulus, find the ratio between elastic potential energy and the elastic energy density.

Explanation

Elastic potential energy
` = 1/2 xx stress xx strai n xx volume`
`= 1/2 Y xx (strai n)^(2) xx volumw`
Elastic energy density
`=("elastic potential energy")/("volume of wire")`
`=1/2 Y xx (strai n)^(2)`
`:. ("elasticity pontential energy" )/("elastic energy density" )`
= volume of the wire = AL .