Lesson 7 of 9 · 7 min
Energy in SHM
NCERT §13.7
The rig's block races through the middle and stops dead at the ends. Its speed keeps changing, yet the spring loses nothing. Where does the energy go at the ends?
The lesson in notes
In short
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.
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Energy graphs for a spring oscillator
Khan Academy · English · Lecture · Open on YouTube