In case of an ideal spring, we take the reference point (where potential energy is assumed to be zero) at extension of `x_(0)` instead of taking at natural length. Then potential energy when spring is extended by `x` is `:`

Options

A. `(1)/(2)k(x-x_(0))^(2)`
B. `(1)/(2)k(x^(2)-x_(0)^(2))`
C. `(1)/(2)k(x^(2)-2x x_(0))`
D. `(1)/(2)k(2x x_(0)-x^(2))`

Answer

B

Explanation

`DeltaU=U_(x)=U_(x_(0))=(1)/(2)kx^(2)-(1)/(2)kx_(0)^(2)`