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Total Energy Is Constant at (1/2)kA²: Split It into Kinetic and Potential

In SHM the total energy stays fixed and sloshes between kinetic and potential. Energy questions are answered by knowing the three values: total (1/2) k A², potential (1/2) k x² and kinetic (1/2) k (A² - x²).

NSEJS · Oscillations · 7 linked questions · 100% coverage

Know this first

Motion and force questions become simpler when direction, starting condition, time and the acting forces are marked first.

Distance · Length of the path travelledDisplacement · Straight-line change in positionResultant force · Net force including direction

Study deeper: NCERT Science · Oscillations

The three energies

E = (1/2) k A² = (1/2) m ω² A², U = (1/2) k x² and K = (1/2) k (A² - x²)

Kinetic energy is greatest at the mean position and zero at the extremes, with the potential energy the other way round.

If K = K0 cos² ωt, the maximum kinetic energy K0 equals the total energy and the maximum potential energy

Over a full cycle, <K> = <U> = (1/4) k A², each half of the total. The kinetic energy is half its maximum when (1/2) k (A² - x²) = (1/4) k A², so x = A/sqrt(2), which the particle reaches T/8 after leaving the mean position.

Reading U against x

On a plot of U against x for SHM,

U = (1/2) k x²

If U = 1 J at x = 0.2 m, then k = 50 N/m, and for a 4 kg mass T = 2π sqrt(4/50) = 2π sqrt(2)/5 s

The force is F = -dU/dx, so U = 8 x² gives F = -16 x, which is 32 (in the same unit system) at x = -2: with CGS units this is 32 dyne, and the mass does not enter.

Energy bookkeeping with a reference level

Add the potential energy at the mean position (set by the reference) to the maximum kinetic energy to get the total. With a total energy of 9 J and a potential energy of 5 J at the mean position, the kinetic energy at the mean position is 4 J. If the potential energy at the mean is 3 J and the mean (time-averaged) kinetic energy is 4 J, the maximum kinetic energy is 8 J since the average is half the maximum, so the total is 3 + 8 = 11 J.

OlympIQ exam lens

✓E = (1/2) k A²; U = (1/2) k x²; K = (1/2) k (A² - x²)
✓Time averages: <K> = <U> = E/2
✓K = E/2 at x = A/sqrt(2), reached T/8 after the mean position
✓F = -dU/dx; read k from a point on the U-x parabola
✓Average kinetic energy over a cycle is half the maximum kinetic energy

OlympIQ trap

Do not call the mean (time-averaged) kinetic energy the maximum. The average over a cycle is half of the maximum, so a question that gives the average must be doubled before it is added to the potential energy at the mean position.

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