What is the potential energy of a spring at the equilibrium position x = 0?

What is the potential energy of a spring at the equilibrium position x = 0?

• A. 0
• B. (1/2) k x^2 with x>0
• C. Maximum
• D. Negative

Answer: (A)

The potential energy of a spring depends on how far it is displaced from the point where it’s neither stretched nor compressed. For an ideal spring, U = (1/2) k x^2, with x measured from equilibrium. At the equilibrium position, x = 0, so U = (1/2) k (0)^2 = 0. Energy stored is zero when there is no displacement, and it grows as you move away from equilibrium, since it’s proportional to x^2 and cannot be negative. The other descriptions don’t fit this specific situation: the energy is not a nonzero value at equilibrium, it isn’t negative, and the “with x>0” phrasing doesn’t apply at x = 0.

The potential energy of a spring depends on how far it is displaced from the point where it’s neither stretched nor compressed. For an ideal spring, U = (1/2) k x^2, with x measured from equilibrium. At the equilibrium position, x = 0, so U = (1/2) k (0)^2 = 0. Energy stored is zero when there is no displacement, and it grows as you move away from equilibrium, since it’s proportional to x^2 and cannot be negative. The other descriptions don’t fit this specific situation: the energy is not a nonzero value at equilibrium, it isn’t negative, and the “with x>0” phrasing doesn’t apply at x = 0.