Energy in SHM: where K equals U.

K and U versus x (two parabolas adding to a flat E) and versus t (period T/2), plus a v-x phase plot that is an ellipse (a circle when v is scaled by ω).

Find the displacement at which K = U (A/√2), then read the frequency of U's oscillation (2ν).

K and U oscillate with period T; zero velocity means zero acceleration.

• Kinetic energy K = ½mv² = ½mω²A² sin²(ωt + φ) = ½kA² sin²(ωt + φ). It is zero at the extremes and largest at the mean position.
• The spring force F = −kx is conservative, with potential energy U = ½kx² = ½kA² cos²(ωt + φ). It is zero at the mean position and largest at the extremes.
• Adding them, E = K + U = ½kA², because sin² + cos² = 1. The total mechanical energy stays constant in time, as it must under a conservative force.
• K and U are each periodic with period T/2, not T: each reaches its peak twice in every oscillation, since the sign of v and of x does not matter.
• In terms of displacement, U = ½kx² and K = ½k(A² − x²): two parabolas that always add up to the flat line E = ½kA².
• Both K and U are never negative. K cannot be, since it depends on v²; U is made non-negative by choosing its zero at the mean position.
• Worked example: a 1 kg block on a spring of 50 N m⁻¹, pulled 10 cm from equilibrium and released. ω = √(50/1) = 7.07 rad s⁻¹. At 5 cm from the mean position cos(7.07t) = 0.5 and v = 0.1 × 7.07 × 0.866 ≈ 0.61 m s⁻¹, so K ≈ 0.19 J, U = ½ × 50 × 0.05² = 0.0625 J and E = 0.25 J, equal to ½ × 50 × 0.1².

Teaching this idea? Put the simulation on your own page or LMS with this code.