If x is doubled in a spring with potential energy U_s = (1/2) k x^2, how does U_s change?

If x is doubled in a spring with potential energy U_s = (1/2) k x^2, how does U_s change?

• A. Remains the same
• B. Increases by factor of 4
• C. Increases by factor of 2
• D. Increases by factor of 8

Answer: (B)

For a spring, the potential energy scales with the square of the displacement from equilibrium. If you double the displacement, you plug into U_s = (1/2) k x^2: U_s' = (1/2) k (2x)^2 = (1/2) k · 4x^2 = 4 · (1/2) k x^2 = 4 U_s. So the potential energy becomes four times larger. The squared dependence means doubling x fourfolds the energy, not just doubling it or keeping it the same.

For a spring, the potential energy scales with the square of the displacement from equilibrium. If you double the displacement, you plug into U_s = (1/2) k x^2:

U_s' = (1/2) k (2x)^2 = (1/2) k · 4x^2 = 4 · (1/2) k x^2 = 4 U_s.

So the potential energy becomes four times larger. The squared dependence means doubling x fourfolds the energy, not just doubling it or keeping it the same.