# How can we calculate entropy produced by plastic deformation in this example?

Suppose we have a massless spring of spring constant $$k$$ attached to a mass $$m$$ at equilibrium position $$x_{0}$$ at temperature $$T$$. The mass may oscillate in one dimension only for simplicity. We choose coordinates so that $$x > x_{0}$$ means the spring is stretched and $$x < x_{0}$$ means the spring is compressed. When the mass is at rest at position $$x$$, the total mechanical energy of the spring is $$E = \frac{1}{2}k(x-x_{0})^{2}.$$

Suppose we overstretch the spring to some position $$x \gg x_{0}$$, causing plastic deformation. I gather that this will shift the equilibrium position to some new value $$x_{1} > x_{0}$$. Then the total mechanical energy shifts: $$E = \frac{1}{2}k(x-x_{0})^{2}\rightarrow E' = \frac{1}{2}k(x-x_{1})^{2} < E.$$ (I'll assume $$k$$ is unchanged for the sake of simplifying the problem.)

Some mechanical energy was lost during the plastic deformation event, causing it to be lost to internal/thermal energy. If I'm not mistaken, this means entropy must have risen in this system. I am wondering if there is a systematic way to calculate this entropy change in the system. Any additional simplifying assumptions can be made as long as they are stated explicitly.