Derive an expression for elastic potential energy per unit volume stored for the wire is 1/2 xx stress xx strain.

Derive an expression for elastic potential energy per unit volume stor

1.	Derive an expression for elastic potential energy per unit volume stored for the wire is 1/2 xx stress xx strain.
Answer» Solution :Young MODULUS `Y= ( F//A)/(x//L)` where x = length of wire increased `therefore F = ((YA)/(L ))x` Now, work done to be done to increae length dx, `dW = Fdx` `= ((YA)/(L)) x dx` `therefore `Work to be done to increase the length l against RESTORING FORCE, `W = int _(0) ^(l) ((YA)/( L ))d dx` `= (AY)/(2L) l ^(2)` Which is the elastic potential energy stored in wire. `therefore U= (AY)/(2L) l ^(2)` `therefore ` Energy stored per unit volume `= (U )/(LA)` `=1/2 (Yl ^(2))/(L ^(2))` `=1/2 xx (("stress")/("STRAIN")) xx ("strain") ^(2)` `=1/2 xx ` stress `xx` strain