Energy in simple harmonic motion

Simple harmonic motion is defined by a restoring force proportional to displacement, F(x) = −mω²x. That force depends on position alone, so the work it does between two points is path-independent and a potential energy exists. Take U = 0 at the equilibrium point and the work you must do against the force to reach x is U(x) = ∫₀ˣ mω²s ds = ½mω²x². The result is a parabola, and its curvature d²U/dx² = mω² is constant — that constant is the entire content of the harmonic assumption, since a well of any other shape would not give a force linear in x. Put numbers on it: m = 0.50 kg and ω = 8.0 rad s⁻¹ give mω² = 32 N m⁻¹, so U(0.20 m) = ½ × 32 × 0.20² = 0.64 J. Dimensions agree, kg × s⁻² × m² = kg m² s⁻² = J.

You do not have to assume that E is conserved; the differential equation says so. Start from d²x/dt² = −ω²x, multiply both sides by m dx/dt, and read each side as a time derivative: m(dx/dt)(d²x/dt²) = d/dt(½mv²) on the left, and −mω²x(dx/dt) = −d/dt(½mω²x²) on the right. Bring them together and d/dt(½mv² + ½mω²x²) = 0, so E = ½mv² + ½mω²x² is constant along every solution. That move — multiply a second-order equation by the velocity and integrate once — is what physicists mean by a first integral, and it is worth more than the answer here: one constant of integration now stands in for one of the two you would otherwise fix from initial conditions. Release the 0.50 kg mass from rest at 0.20 m and E = 0.64 J; launch it through x = 0 at 1.6 m s⁻¹ instead and E = 0.64 J again, so the two motions share an amplitude even though their phase constants differ.

Evaluate the constant where it is easiest. At a turning point the mass is momentarily at rest, v = 0 and x = ±A, so the kinetic term vanishes and E = ½mω²A². Everything else follows from that one line. Inverting it gives the amplitude an energy buys, A = √(2E/(mω²)); evaluating the constant at x = 0 instead, where U vanishes, gives ½mvₘₐₓ² = E and so vₘₐₓ = ωA, the same maximum speed the time-domain solution produces. The quadratic dependence is the headline. Pull the 0.50 kg, 8.0 rad s⁻¹ oscillator to 0.40 m rather than 0.20 m and the store rises from 0.64 J to 2.56 J — four times the energy for twice the displacement, and at exactly the same period, because the amplitude never enters ω. Check the limit: A → 0 sends E → 0, as an oscillator sitting at rest at the bottom of its well must.

Set ½mv² + ½mω²x² = ½mω²A² and solve for the speed: v(x) = ±ω√(A² − x²). This is the practical payoff of the energy route. Asked for the speed at a stated displacement, the time-domain solution would have you invert x = A cos(ωt + φ) for t and substitute; the energy equation answers in one step. Three checks before trusting it: at x = 0 it gives ωA, the known maximum; at x = ±A it gives zero; and rad s⁻¹ × m is m s⁻¹. What it will not give is direction, because squaring destroyed the sign — energy knows speed, phase knows which way. With A = 0.20 m and ω = 8.0 rad s⁻¹, the speed at x = 0.10 m is 8.0√(0.20² − 0.10²) = 1.39 m s⁻¹, which is 0.866 vₘₐₓ while the kinetic share is ¾E = 0.48 J. Kinetic and potential energies are equal not at A/2 but at x = A/√2 = 0.141 m, where v = 1.13 m s⁻¹.

Put x = A cos(ωt + φ) and v = −ωA sin(ωt + φ) back into the two terms. U(t) = ½mω²A²cos²(ωt + φ) and K(t) = ½mω²A²sin²(ωt + φ), and the double-angle identities turn these into U(t) = ½E[1 + cos(2ωt + 2φ)] and K(t) = ½E[1 − cos(2ωt + 2φ)]. Each runs from 0 to E about a mean of ½E at angular frequency 2ω, so the energy curves have period T/2 while the displacement has period T. The reason is that both terms are quadratic: squaring a quantity that changes sign twice per cycle produces a shape that peaks twice per cycle. Their sum is E at every instant, which is why the total never oscillates at all. Averaging over a whole period with (1/T)∫₀ᵀ cos²(ωt + φ) dt = ½ gives ⟨K⟩ = ⟨U⟩ = ½E. For the 0.50 kg oscillator with A = 0.20 m and ω = 8.0 rad s⁻¹, T = 0.785 s, the energy curves repeat every 0.393 s, and each averages 0.32 J.

Two idealisations are hiding in ½mω²A². The first is that the well is exactly quadratic. A real restoring force carries higher terms, U(x) = ½mω²x² + ¼βx⁴ being the first correction for a symmetric well. Energy is still conserved, but E is no longer ½mω²A², the exchange is no longer a pure 2ω sinusoid, and the frequency acquires an amplitude-dependent fractional shift whose leading term is the dimensionless 3βA²/(8mω²) — the tell-tale sign of an anharmonic oscillator, and the reason the harmonic result is always quoted with a small-amplitude caveat. The second is that nothing dissipates. Repeat the first-integral calculation with a drag force −bv added and dE/dt = v(−bv) = −bv² ≤ 0, strictly negative whenever the mass is moving. For light damping E decays as e^(−bt/m) while the amplitude, going as √E, decays as e^(−bt/2m): the store empties at twice the exponential rate of the swing, which is why a struck oscillator falls quiet long before it looks still. Setting b → 0 returns the constant total, which is the correspondence check the damped model must pass.