Why is a spring fastest where its force is zero?

Freeze the motion at a quarter cycle

Reset the experiment to a 1.00 kg block, a spring constant of 16 N/m and a release displacement of 20 cm to the right. The block starts from rest. Its elastic energy is ½kA² = 0.320 J. Hooke’s law gives a force of −3.20 N; the minus sign points left, toward equilibrium.

Step by ¼ cycle. The block is at x = 0 and travelling left at 0.800 m/s. The spring stores no energy there; all 0.320 J is kinetic. Step again: at x = −0.20 m the velocity is zero and the force is +3.20 N. The compressed spring stores just as much energy as the stretched spring because elastic energy depends on x².

Change the clock without changing the energy budget

Combining F = −kx with F = ma gives a = −(k/m)x. A cosine released from rest has acceleration equal to −ω²x, so ω = √(k/m) and T = 2π√(m/k). In the default setup T is about 1.571 s. Raise the mass to 4.00 kg with k and A unchanged; the period becomes 3.142 s, but the energy stays 0.320 J.

Reset and move the starting displacement to 10 cm. The period still reads 1.571 s, but the total energy drops to 0.080 J. The energy bar shows the share of each form within a trial. To compare different settings, use the numerical joule readings instead of the bar’s full width.

Check your understanding

With the default mass and 20 cm release, raise the stiffness from 16 to 64 N/m. Predict the new period and the total energy, then check both readings.