Springs in parallel double effective stiffness

Steps:

• Single spring: F = kx, elastic potential energy E_p = (1/2)kx².
• Two springs in parallel: effective spring constant k_eff = 2k.
• Extension under force F: x' = F / k_eff = F / 2k = x/2.
• Total elastic potential energy: (1/2)k_eff(x')² = (1/2)(2k)(x/2)² = (1/2)kx² = (1/2)E_p.

Why A is correct:

• Parallel combination halves extension and halves stored energy per Hooke's law and energy formula, as stiffness doubles while displacement squares.

Why the others are wrong:

• B: Incorrect extension (should be x/2, not x/3) and energy scaling.
• C: Wrong energy (doubles stiffness halves energy, not same as single spring).
• D: Correct extension but wrong energy (total is half, not same as single).

Final answer: A