Energy in Simple Harmonic Motion
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Direct answer
Total energy in SHM is a fixed account, E = ½kA² = ½mω²A², sloshing between kinetic at the centre and potential at the extremes. At displacement x the split is exact: KE = ½k(A² − x²), PE = ½kx², and the two are equal only at x = A/√2, where each holds half of E. At the mean position the speed peaks at v_max = Aω; at the extremes the body stops and all the energy is potential. Averaged over time, kinetic and potential each carry exactly E/2 — a result NEET asks straight. The quadratic dependence on x means doubling the amplitude quadruples the energy, and every energy-versus-displacement graph is a parabola.
What you must remember
- Total energy: E = ½kA² = ½mω²A²; proportional to the square of amplitude, independent of time in ideal SHM.
- Position formulas: KE = ½k(A² − x²); PE = ½kx²; their sum is E at every point of the motion.
- The crossover: KE = PE at x = A/√2 ≈ 0.707A, each then equal to E/2; at x = A/2 the split is KE = 3E/4, PE = E/4.
- Speed relation: v = ω√(A² − x²); v_max = Aω at x = 0; v = 0 at x = ±A.
- Time averages: over one full cycle, ⟨KE⟩ = ⟨PE⟩ = E/2 = ¼kA² — each form, on average, holds half the total.
- Graph signatures: KE and PE versus x are parabolas opening opposite ways; versus time, each oscillates between 0 and E at twice the frequency of the motion.
- Energy shortcut: equating ½mv²(max) with ½kA² recovers v_max = Aω without solving any equation of motion — exam craft worth practising.
- Pendulum cousin: for a small-angle pendulum PE is gravitational, mgh with h ≈ Lθ²/2, but the E ∝ A² bookkeeping is identical.
Following the energy through one oscillation
Take k = 50 N/m, m = 0.2 kg, A = 10 cm. Total energy E = ½ × 50 × 0.01 = 0.25 J. At x = 5 cm, PE = ½ × 50 × 0.0025 = 0.0625 J, exactly E/4, leaving KE = 0.1875 J = 3E/4 — and the speed formula agrees: ω = √(k/m) = √250 ≈ 15.8 rad/s, v = 15.8 × √(0.01 − 0.0025) ≈ 1.37 m/s, giving KE = ½ × 0.2 × 1.37² ≈ 0.19 J. At x = A/√2 ≈ 7.07 cm the accounts balance at 0.125 J each. The quadratic bookkeeping explains the shape of every graph: quarter energy at half amplitude, not half energy — the mistake the distractors are always built on. One oscillation, three checkpoints, no formula beyond ½kx² and its complement.
Where NEET sets the trap
The champion trap is the crossover question: "at what displacement is KE equal to PE?" with options A/2 and A/√2 both present; the quadratic split, not linear intuition, decides. The second is amplitude scaling: doubling A quadruples E, and the option "doubles" never leaves the list. Averages trip the unwary — the time-average of each form is E/2, but a subtler question averaged over positions gives ⟨PE⟩ = E/3 and ⟨KE⟩ = 2E/3 (the oscillator lingers near the extremes), a distinction tougher papers have exploited. Graph items hand you a KE-versus-t curve and ask for the period: the energy curve completes two cycles per mechanical cycle. Assertion-reason close the set: "KE is maximum at the extremes" (false) and "total energy is constant in ideal SHM" (true).
Frequently asked questions
At what displacement is kinetic energy equal to potential energy in SHM?
At x = A/√2 ≈ 0.707A, where each equals half the total — not at A/2, where PE is only E/4.
How does SHM energy depend on amplitude?
E = ½kA² = ½mω²A² — quadratic: doubling the amplitude multiplies the energy fourfold.
What are the average kinetic and potential energies over one cycle?
Each equals half the total, ⟨KE⟩ = ⟨PE⟩ = ¼kA², because the exchange between the two forms is symmetric over a period.
What is the speed at the mean position?
v_max = ωA; the centre holds all the energy kinetically, which is why ½mω²A² = ½mv²(max) is the standard shortcut.
At what frequency does the kinetic energy oscillate?
At twice the oscillator's frequency — KE peaks twice per cycle (once per pass through the centre), completing two full variations per mechanical oscillation.