Energy in Simple Harmonic Motion

The energy ledger of the harmonic oscillator is now precision technology. Every MEMS gyroscope and timing resonator in a phone is a micromachined oscillator whose femtojoule energy budget must be topped up against dissipation each cycle; the figure of merit quoted on every datasheet, the quality factor $Q = \omega E/P_{\mathrm{loss}}$, is nothing but this article's conserved $E$ divided by the energy lost per radian. The same bookkeeping runs the LIGO and Virgo mirror suspensions, engineered as pendulums with $Q$ in the hundreds of millions precisely so that the thermal energy $k_B T_{\mathrm{emp}}$ assigned to each mode by equipartition — the $\langle U\rangle = \tfrac12 k_B T_{\mathrm{emp}}$ of equation (6) — stays spread thinly enough in frequency not to drown a gravitational wave. The sharpest 2026 frontier is the bottom of the energy ladder itself. Optomechanical and superconducting-circuit experiments now routinely cool nanogram-scale oscillators below one phonon of average excitation, exchange energy with them one quantum $\hbar\omega$ at a time, and verify that the classical phase-space ellipse of equation (8) dissolves into the quantum ground state of area $\sim\hbar$ — the zero-point energy $\tfrac12\hbar\omega$ sitting exactly where the classical ledger would read $E = 0$. The action $J = 2\pi E/\omega$ of derivation C is the bridge: the classical area, measured in units of Planck's constant, literally counts the phonons.

Watch a child on a swing at the two special moments. At the highest point of the arc she hangs motionless for an instant — all the energy is stored as height, none as motion. Sweeping through the bottom she is fastest — all the energy is motion, none is storage. In between, the books are split. The total never changes; it only migrates between the two reservoirs, twice each way every full swing. A frictionless mass on a spring does exactly the same thing with "height" replaced by "stretch": the energy ledger of any harmonic oscillator has just these two columns, and the sum of each row is the same number. The cleanest way to see the whole story at once is the energy diagram: draw the potential well $U(x) = \tfrac12 kx^2$ and rule a horizontal line at the total energy $E$. The motion is confined to where the line sits above the parabola; the line meets the curve exactly at the turning points $x = \pm A$, and at any intermediate position the vertical gap between the line and the parabola is the kinetic energy.

Now let time run. Both reservoirs breathe sinusoidally, but because each depends on the square of a sinusoid, each completes a full fill-and-empty cycle in half a period. Watch only the kinetic energy and you see it peak twice per swing — once passing through equilibrium moving right, once moving left. The two curves are mirror images pivoting about the half-energy line, and their sum is a dead-flat constant:

Why does the total depend on the amplitude squared? Because the well is quadratic: push twice as far and the force you fight grows twice as large along a path twice as long, costing four times the work. That square law is why doubling a pendulum's swing quadruples the energy you must supply, and why a loudspeaker cone driven at twice the excursion radiates four times the acoustic power. Read in reverse, it is a diagnostic: measure how stored energy scales with amplitude and you learn whether the underlying well is really quadratic.

Everything in this article flows from one bookkeeping identity. The kinetic energy of the mass is $\tfrac12 m\dot x^2$; the work stored against the restoring force $F = -kx$ is the potential energy $U = \tfrac12 kx^2$. The work–energy theorem then guarantees that their sum is a constant of the motion:

where: $E$ — total mechanical energy (joules); $m$ — the oscillating mass; $\dot x$ — its velocity; $k$ — the spring constant or effective stiffness; $x$ — displacement from equilibrium. This single line already is the dynamics: differentiate it and Newton's law $m\ddot x = -kx$ falls out; integrate it and the full motion follows. Conservation holds because $F = -kx$ is conservative — no friction, no drive. The cleanest way to evaluate the constant is to catch the oscillator where one reservoir is empty. At a turning point $x = \pm A$ the mass is momentarily at rest, so the whole budget is potential:

where: $A$ — the amplitude, the maximum displacement. Energy depends only on stiffness and amplitude — not on the phase, not on the mass, not on the instant you look. A stiffer spring or a wider swing means more energy, quadratically. Catch it instead at equilibrium, where the spring is relaxed and the budget is all motion. The speed there is the maximum speed $v_{\max} = \omega A$, so the same constant reads

where: $\omega = \sqrt{k/m}$ — the angular frequency. Equations (2) and (3) agree precisely because $m\omega^2 = k$; checking that equality is the fastest sanity test of any SHM energy calculation. Form (3) travels better than (2): it survives into systems where "$k$" is hidden (pendulums, circuits, molecules) but $\omega$ and $A$ are measurable. Now substitute the full solution $x = A\cos(\omega t + \varphi)$, $\dot x = -\omega A\sin(\omega t + \varphi)$ to see how the fixed total is split at every instant:

where: $\varphi$ — the phase constant set by initial conditions. The two shares are $\sin^2$ and $\cos^2$ of the same angle, so their sum collapses to $\tfrac12 kA^2$ by the Pythagorean identity — conservation made visible. Each share, being the square of a sinusoid, never goes negative and repeats twice as fast as the motion itself. The half-angle identity $\sin^2\theta = \tfrac12(1 - \cos 2\theta)$ makes the doubled tempo explicit:

where: the argument $2(\omega t + \varphi)$ announces that energy oscillates at $2\omega$ — kinetic energy peaks both times the mass crosses equilibrium each period, moving right and moving left. The constant offset $\tfrac14 kA^2$ is the mean value, which is exactly half of $E$: averaging the cosine away over a full cycle,

where: $\langle\cdot\rangle$ — the average over one period. On time-average the two reservoirs hold equal shares — the virial theorem specialized to a quadratic potential ($\langle T\rangle = \tfrac{n}{2}\langle U\rangle$ for $U \propto x^n$, with $n = 2$). This equality is the classical seed of equipartition: couple the oscillator to a heat bath and each quadratic term claims $\tfrac12 k_B T_{\mathrm{emp}}$ of thermal energy. For problem-solving, the most-used consequence of (1) is what you get by solving it for the velocity at a given position:

where: the sign tracks the direction of travel. "How fast is the mass at $x$?" needs no differential equation — read the kinetic energy off the diagram as $E - U(x)$ and convert. Note the shape: $v$ is maximal at $x = 0$, falls off as a quarter-circle, and hits zero exactly at $\pm A$. At half-amplitude the speed is still $\sqrt{3}/2 \approx 87\%$ of maximum — the oscillator hurries through the middle and lingers near the edges.

Which equation answers which question

There is a second geometry in which the energy budget becomes a picture: plot momentum against position. Squaring and adding $x = A\cos(\omega t + \varphi)$ and $p = m\dot x = -m\omega A\sin(\omega t + \varphi)$ eliminates time entirely, leaving the orbit

where: $p$ — momentum. The state of the oscillator circulates clockwise around an ellipse with semi-axes $A$ and $m\omega A$; conservation of $E$ is the geometric statement that the point never leaves its ellipse. Larger energy, larger ellipse — and it is the area that keeps the score.

1 · A mass on a spring, audited three ways (confidence builder)

A $0.50\ \mathrm{kg}$ mass on a spring with $k = 200\ \mathrm{N/m}$ oscillates with amplitude $A = 0.10\ \mathrm{m}$. Find the total energy, the maximum speed, and the speed and kinetic energy at $x = A/2$.

Step 1. The angular frequency: $\omega = \sqrt{k/m} = \sqrt{200/0.50} = 20\ \mathrm{rad/s}$.

Step 2. Total energy from the amplitude, equation (2): $E = \tfrac12 kA^2 = \tfrac12(200)(0.10)^2 = \boxed{1.0\ \mathrm{J}}$. Cross-check via (3): $\tfrac12 m\omega^2 A^2 = \tfrac12(0.50)(400)(0.01) = 1.0\ \mathrm{J}$. ✓

Step 3. Maximum speed at equilibrium: $v_{\max} = \omega A = 20 \times 0.10 = \boxed{2.0\ \mathrm{m/s}}$.

Step 4. Speed at half amplitude from equation (7): $v(A/2) = \omega\sqrt{A^2 - A^2/4} = 20\sqrt{0.0100 - 0.0025} = 20 \times 0.0866 = \boxed{1.73\ \mathrm{m/s}}$ — still $87\%$ of $v_{\max}$, though the mass is halfway out.

Step 5. Kinetic energy there, two ways. From the ledger: $T = E - U = 1.0 - \tfrac12(200)(0.05)^2 = 1.0 - 0.25 = \boxed{0.75\ \mathrm{J}}$. Directly: $\tfrac12 mv^2 = \tfrac12(0.50)(1.73)^2 = 0.75\ \mathrm{J}$. ✓ At half the displacement the potential holds only a quarter of the energy — the quadratic well at work.

2 · An AFM cantilever against the thermal noise floor (exam level)

The tip of an atomic-force-microscope cantilever behaves as a harmonic oscillator with stiffness $k = 40\ \mathrm{N/m}$ and resonance frequency $f = 300\ \mathrm{kHz}$. It is driven at amplitude $A = 10\ \mathrm{nm}$ at room temperature ($k_B T_{\mathrm{emp}} = 4.14\times10^{-21}\ \mathrm{J}$). Find the stored energy, compare it with thermal energy, and estimate the cantilever's thermal jitter when the drive is off.

Step 1. Stored energy from equation (2): $E = \tfrac12 kA^2 = \tfrac12(40)(10^{-8})^2 = \boxed{2.0\times10^{-15}\ \mathrm{J} = 2.0\ \mathrm{fJ}}$. Femtojoules sound tiny — the comparison in Step 2 is what gives the number meaning.

Step 2. Compare with the thermal scale: $E/k_B T_{\mathrm{emp}} = 2.0\times10^{-15}/4.14\times10^{-21} \approx \boxed{4.8\times10^{5}}$. The driven motion carries half a million times the thermal energy — which is exactly why a 10 nm AFM scan is crisp rather than fuzzy.

Step 3. Thermal jitter with the drive off. Equipartition — the thermodynamic echo of equation (6) — assigns the potential term its share: $\langle\tfrac12 kx^2\rangle = \tfrac12 k_B T_{\mathrm{emp}}$, so $x_{\mathrm{rms}} = \sqrt{k_B T_{\mathrm{emp}}/k} = \sqrt{4.14\times10^{-21}/40} = \boxed{1.0\times10^{-11}\ \mathrm{m} \approx 10\ \mathrm{pm}}$ — a thousandth of the driven amplitude, and the fundamental resolution floor of the instrument. (Consistency: scaling the driven case by amplitude, $E \propto A^2$ predicts energy ratio $(10\ \mathrm{nm}/10\ \mathrm{pm})^2 = 10^6$ — within a factor of 2 of Step 2, the factor being $\langle U\rangle = E/2$.)

Step 4. For flavor, the effective mass: $m_{\mathrm{eff}} = k/\omega^2 = 40/(2\pi \times 3\times10^5)^2 = 1.1\times10^{-11}\ \mathrm{kg} \approx 11\ \mathrm{ng}$, giving $v_{\max} = \omega A = 1.9\times10^{-2}\ \mathrm{m/s}$ — a nanogram-scale mass crossing equilibrium at two centimetres per second, 300,000 times a second.

References. Morin, Introduction to Classical Mechanics, ch. 4 (Oscillations) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §6.1 · Landau & Lifshitz, Mechanics, Vol. 1, §21–22 · Taylor, Classical Mechanics, ch. 5 · Rayleigh, The Theory of Sound, Vol. 1, ch. 4 (energy methods).

Provenance. Standard SHM energetics. A verifier should: (1) confirm $\mathrm{d}E/\mathrm{d}t=0$ from $m\ddot x+kx=0$; (2) check $E=\tfrac12 kA^2=\tfrac12 m\omega^2A^2$ via $\omega^2=k/m$; (3) verify $\langle\sin^2\rangle=\langle\cos^2\rangle=\tfrac12$ and the resulting equal averages; (4) reproduce $T_{\text{period}}=2\pi\sqrt{m/k}$ from the energy integral; (5) recompute the ellipse area $J=2\pi E/\omega$. Sources: Morin ch. 4, Goldstein §6.1 & §10, Landau–Lifshitz Vol. 1 §21–22 & §49, Taylor ch. 5. Worked-example numbers on this page ($E = 1.0$ J, $v_{\max} = 2.0$ m/s, $v(A/2) = 1.73$ m/s for the spring; $E = 2.0$ fJ, $E/k_BT_{\mathrm{emp}} = 4.8\times10^5$, $x_{\mathrm{rms}} \approx 10$ pm for the cantilever) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Simple Harmonic Motion · Amplitude, Phase, and Angular Frequency · Small Oscillations About Equilibrium. Leads to: Damped Harmonic Oscillator · Driven Oscillations · Resonance · Anharmonic Pendulum and Amplitude Dependence.