Energy in Simple Harmonic Motion

Spec mapping: OCR H556 Module 5.3 — Oscillations (kinetic and potential energy of a simple harmonic oscillator; conservation of total energy in free undamped SHM; graphs of $E_k$Ek​, $E_p$Ep​ and $E_\text{total}$Etotal​ against displacement and time; $E_\text{total} = \frac{1}{2}m\omega^2 A^2 \propto A^2$Etotal​=21​mω2A2∝A2). Refer to the official OCR H556 specification document for exact wording.

Anything that is oscillating back and forth must store and exchange energy between different forms. A mass on a spring swaps kinetic energy (when it is moving fast through equilibrium) for elastic potential energy (when it is momentarily stationary at the extremes). A pendulum swaps kinetic energy for gravitational potential energy. A vibrating atom in a crystal lattice swaps kinetic energy for interatomic-bond potential energy. In every case — in the absence of friction and air resistance — the total mechanical energy remains constant.

This lesson explores the energy of a simple harmonic oscillator: formulae, graphs, conversion between forms, and exam-style worked examples. The key conceptual point: total energy $E_\text{total} = \frac{1}{2}m\omega^2 A^2$Etotal​=21​mω2A2 is constant; kinetic and potential energies vary out of phase, each at frequency $2f$2f (twice the motion's frequency), summing always to the constant total. It is OCR A-Level Physics A Module 5.3 (Oscillations) — the energy chapter that closes the undamped SHM theory and prepares for damping (lesson 6).

---

Total Energy of an SHM Oscillator

From the last lesson, the velocity of a simple harmonic oscillator at displacement $x$x satisfies

$v^2 = \omega^2(A^2 - x^2)$v2=ω2(A2−x2)

Its kinetic energy is therefore

$E_k = \frac{1}{2}mv^2 = \frac{1}{2}m\omega^2(A^2 - x^2)$Ek​=21​mv2=21​mω2(A2−x2)

At the maximum displacement, $x = \pm A$x=±A and $v = 0$v=0, so $E_k = 0$Ek​=0 — the oscillator is momentarily at rest. At equilibrium, $x = 0$x=0 and the speed is maximum, so

$E_{k,\text{max}} = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}mv_\text{max}^2$Ek,max​=21​mω2A2=21​mvmax2​

Because mechanical energy is conserved (no friction, no air resistance, no driving force), the total mechanical energy of the oscillator must be equal to this peak kinetic value — at equilibrium, the entire energy is kinetic:

$E_\text{total} = \frac{1}{2}m\omega^2 A^2$Etotal​=21​mω2A2

This is a beautiful result: for a given oscillator (fixed $m$m and $\omega$ω), the total energy is proportional to the square of the amplitude. Doubling the amplitude quadruples the total energy; the energy of a pendulum swung through $30°$30° is four times that swung through $15°$15°. This $A^2$A2 scaling is the deep reason why each successive cycle of a damped oscillator (lesson 6) loses energy in a regular geometric pattern.

---

Potential Energy

Since total energy is conserved, the potential energy is whatever the total energy is not in kinetic form at that instant:

$E_p = E_\text{total} - E_k = \frac{1}{2}m\omega^2 A^2 - \frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2 x^2$Ep​=Etotal​−Ek​=21​mω2A2−21​mω2(A2−x2)=21​mω2x2

$E_p = \frac{1}{2}m\omega^2 x^2$Ep​=21​mω2x2

So potential energy grows as the square of the displacement from equilibrium. This parabolic "energy well" is the signature of any harmonic system — whether a mass on a spring, a pendulum at small angles, or an atom in a crystal lattice. It is exactly the harmonic approximation discussed in the going-further section of lesson 3: any smooth potential well looks parabolic near its bottom, and the SHM equations capture that universal small-displacement behaviour.

At $x = 0$x=0: $E_p = 0$Ep​=0 (all energy is kinetic).

At $x = \pm A$x=±A: $E_p = \frac{1}{2}m\omega^2 A^2 = E_\text{total}$Ep​=21​mω2A2=Etotal​ (all energy is potential).

At any intermediate $x$x: $E_k + E_p = \frac{1}{2}m\omega^2 A^2$Ek​+Ep​=21​mω2A2 exactly.

Why "Mass times Omega-Squared"?

For a mass-spring with stiffness $k$k, the elastic PE is $\frac{1}{2}kx^2$21​kx2 (Hooke's law, integrated). Using the result from lesson 3 that $\omega^2 = k/m$ω2=k/m, this is identically $\frac{1}{2}m\omega^2 x^2$21​mω2x2 — confirming the SHM-language form. For a pendulum with $\omega^2 = g/L$ω2=g/L, the gravitational PE at angular displacement $\theta$θ is $mgL(1 - \cos\theta) \approx \frac{1}{2}mgL\theta^2 = \frac{1}{2}m\omega^2(L\theta)^2 = \frac{1}{2}m\omega^2 x^2$mgL(1−cosθ)≈21​mgLθ2=21​mω2(Lθ)2=21​mω2x2 in terms of arc-length displacement $x = L\theta$x=Lθ. Same SHM-language form. The formula $E_p = \frac{1}{2}m\omega^2 x^2$Ep​=21​mω2x2 unifies the energy bookkeeping of all SHM systems, irrespective of whether the underlying "spring" is mechanical, gravitational, electrical or otherwise.

---

Graphs of Energy in SHM

Plotting $E_k$Ek​, $E_p$Ep​ and $E_\text{total}$Etotal​ against displacement $x$x is a standard OCR exam skill.

• $E_p = \frac{1}{2}m\omega^2 x^2$Ep​=21​mω2x2 — upward parabola through the origin, equal to $E_\text{total}$Etotal​ at $x = \pm A$x=±A.
• $E_k = \frac{1}{2}m\omega^2(A^2 - x^2)$Ek​=21​mω2(A2−x2) — downward parabola, equal to $E_\text{total}$Etotal​ at $x = 0$x=0, zero at $x = \pm A$x=±A.
• $E_\text{total}$Etotal​ — horizontal line at height $\frac{1}{2}m\omega^2 A^2$21​mω2A2.

The two parabolae mirror each other about the horizontal line $E = \frac{1}{4}m\omega^2 A^2$E=41​mω2A2 (half of $E_\text{total}$Etotal​), and at every $x$x their sum adds up to the flat total-energy line.

Against Time

Against time the picture is different but equally instructive. Using $x = A\cos(\omega t)$x=Acos(ωt):

• $E_p(t) = \frac{1}{2}m\omega^2 A^2 \cos^2(\omega t)$Ep​(t)=21​mω2A2cos2(ωt)
• $E_k(t) = \frac{1}{2}m\omega^2 A^2 \sin^2(\omega t)$Ek​(t)=21​mω2A2sin2(ωt)
• $E_\text{total}(t) = \frac{1}{2}m\omega^2 A^2$Etotal​(t)=21​mω2A2 (constant)

Because $\sin^2 + \cos^2 = 1$sin2+cos2=1, the sum is still a flat line. Both $E_k$Ek​ and $E_p$Ep​ oscillate with a period of $T/2$T/2 — exactly half the period of the motion itself. Every cycle the oscillator passes through equilibrium twice (so KE reaches its maximum twice) and through each extreme twice (so PE reaches its maximum twice). The frequency of either energy variation is $2f$2f, twice the motion's frequency.

Exam Tip: A classic OCR question: "What is the frequency of the kinetic energy variation?" The answer is $2f$2f, twice the frequency of the motion, because KE peaks twice per cycle (once at each pass through equilibrium). Similarly for PE (peaks at each extreme). The total energy is constant, so it has no frequency at all — or equivalently, frequency zero.

This $2f$2f-doubling is also why a low-frequency seismic wave can resonantly drive a high-frequency vibrational mode in a building (parametric resonance) — a phenomenon investigated in earthquake engineering.

---

Worked Example 1 — Energy of a Mass on a Spring

A $0.40$0.40 kg mass on a spring oscillates with amplitude $0.050$0.050 m at angular frequency $12$12 rad s$^{-1}$−1. Calculate (a) the total energy and (b) the kinetic energy when the mass is $0.030$0.030 m from equilibrium.

(a) Total energy:

$\begin{aligned} E_\text{total} &= \frac{1}{2}m\omega^2 A^2 \\ &= 0.5 \times 0.40 \times 12^2 \times 0.050^2 \\ &= 0.5 \times 0.40 \times 144 \times 0.00250 \\ &= 0.0720 \text{ J} \end{aligned}$Etotal​​=21​mω2A2=0.5×0.40×122×0.0502=0.5×0.40×144×0.00250=0.0720 J​

(b) Kinetic energy at $x = 0.030$x=0.030 m:

$\begin{aligned} E_k &= \frac{1}{2}m\omega^2(A^2 - x^2) \\ &= 0.5 \times 0.40 \times 144 \times (0.00250 - 0.000900) \\ &= 28.8 \times 0.001600 \\ &= 0.0461 \text{ J} \end{aligned}$Ek​​=21​mω2(A2−x2)=0.5×0.40×144×(0.00250−0.000900)=28.8×0.001600=0.0461 J​

Sanity check. Potential energy should be $E_\text{total} - E_k = 0.0720 - 0.0461 = 0.0259$Etotal​−Ek​=0.0720−0.0461=0.0259 J. Direct calculation: $E_p = \frac{1}{2}m\omega^2 x^2 = 28.8 \times 0.000900 = 0.0259$Ep​=21​mω2x2=28.8×0.000900=0.0259 J. ✓

This is the bread-and-butter SHM-energy calculation, and it appears regularly on OCR papers. The technique is always: compute $E_\text{total}$Etotal​ from the amplitude, then either subtract from total or compute directly from displacement.

---

Worked Example 2 — Where is $E_k = E_p$Ek​=Ep​?

At what displacement is the kinetic energy of a simple harmonic oscillator equal to its potential energy?

Setting $E_k = E_p$Ek​=Ep​:

$\begin{aligned} \frac{1}{2}m\omega^2(A^2 - x^2) &= \frac{1}{2}m\omega^2 x^2 \\ A^2 - x^2 &= x^2 \\ A^2 &= 2x^2 \\ x &= \pm \frac{A}{\sqrt{2}} \approx \pm 0.707\,A \end{aligned}$21​mω2(A2−x2)A2−x2A2x​=21​mω2x2=x2=2x2=±2​A​≈±0.707A​

So when the oscillator is about $70.7\%$70.7% of the way out toward its extreme, its energy is split evenly between kinetic and potential. At that instant it is still moving at about $70.7\%$70.7% of its maximum speed (since $v^2 = \omega^2(A^2 - x^2) = \omega^2 A^2/2$v2=ω2(A2−x2)=ω2A2/2, giving $v = v_\text{max}/\sqrt{2}$v=vmax​/2​).

This is a popular OCR short question. The answer $A/\sqrt{2}$A/2​ should be memorised — and the result holds regardless of $m$m or $\omega$ω.

---

Worked Example 3 — Pendulum at the Bottom of its Swing

A pendulum of length $0.80$0.80 m swings with an amplitude (angular) of $0.060$0.060 rad. The bob has mass $0.15$0.15 kg. Take $g = 9.81$g=9.81 m s$^{-2}$−2 and assume small-angle SHM.

Step 1 — Maximum height above the lowest point. Geometry: when the string makes angle $\theta$θ with the vertical, the bob is at height $h = L(1 - \cos\theta)$h=L(1−cosθ) above its equilibrium position. For small $\theta$θ, $\cos\theta \approx 1 - \theta^2/2$cosθ≈1−θ2/2, so

$h_\text{max} \approx \frac{1}{2}L\theta_\text{max}^2 = 0.5 \times 0.80 \times (0.060)^2 = 1.44 \times 10^{-3} \text{ m}$hmax​≈21​Lθmax2​=0.5×0.80×(0.060)2=1.44×10−3 m

Step 2 — Gravitational PE at the extreme.

$E_p = mgh_\text{max} = 0.15 \times 9.81 \times 1.44 \times 10^{-3} = 2.12 \times 10^{-3} \text{ J}$Ep​=mghmax​=0.15×9.81×1.44×10−3=2.12×10−3 J

Step 3 — KE at equilibrium (by energy conservation). At the lowest point all the energy is kinetic:

$\begin{aligned} \frac{1}{2}mv^2 &= 2.12 \times 10^{-3} \\ v &= \sqrt{\frac{2 \times 2.12 \times 10^{-3}}{0.15}} = \sqrt{0.02827} = 0.168 \text{ m s}^{-1} \end{aligned}$21​mv2v​=2.12×10−3=0.152×2.12×10−3​​=0.02827​=0.168 m s−1​

Cross-check using $v_\text{max} = A\omega$vmax​=Aω. Here $A = L\theta_\text{max} = 0.80 \times 0.060 = 0.048$A=Lθmax​=0.80×0.060=0.048 m and $\omega = \sqrt{g/L} = \sqrt{9.81/0.80} = 3.50$ω=g/L​=9.81/0.80​=3.50 rad s$^{-1}$−1. So $v_\text{max} = 0.048 \times 3.50 = 0.168$vmax​=0.048×3.50=0.168 m s$^{-1}$−1. ✓ Two routes, same answer — a good confidence check.

The bob swings through equilibrium at about $17$17 cm s$^{-1}$−1. This is a tiny speed — but a $0.15$0.15 kg bob carries momentum $p = 0.025$p=0.025 kg m s$^{-1}$−1, enough to register clearly on a light gate.

---

Worked Example 4 — Energy Loss from a Damped Oscillator

A damped oscillator loses $10\%$10% of its total energy each complete cycle. What fraction of its original amplitude remains after one cycle?

Because $E \propto A^2$E∝A2, if $E$E falls to $0.90\,E_0$0.90E0​ then

$\frac{A}{A_0} = \sqrt{\frac{E}{E_0}} = \sqrt{0.90} = 0.949$A0​A​=E0​E​​=0.90​=0.949

So the amplitude drops by about $5\%$5% per cycle. The energy drops twice as fast (proportionally) as the amplitude — a useful rule of thumb. After $n$n cycles, the amplitude is $A_0 \times 0.949^n$A0​×0.949n and the energy is $E_0 \times 0.90^n$E0​×0.90n, both decaying geometrically.

This is a preview of lesson 6: damped oscillators decay exponentially in amplitude (and twice as fast in energy), as a consequence of $E \propto A^2$E∝A2.

---

Worked Example 5 — Total Energy and Maximum Speed

A loudspeaker cone of effective mass $0.50$0.50 g oscillates at $1000$1000 Hz with amplitude $0.20$0.20 mm. Find (a) the maximum speed of the cone, (b) the total mechanical energy.

(a) $\omega = 2\pi f = 2\pi \times 1000 = 6283$ω=2πf=2π×1000=6283 rad s$^{-1}$−1. Then $v_\text{max} = A\omega = 2.0 \times 10^{-4} \times 6283 = 1.26$vmax​=Aω=2.0×10−4×6283=1.26 m s$^{-1}$−1 — a substantial speed for a tiny amplitude.

(b) $E_\text{total} = \frac{1}{2}mv_\text{max}^2 = 0.5 \times 5.0 \times 10^{-4} \times (1.26)^2 = 3.97 \times 10^{-4}$Etotal​=21​mvmax2​=0.5×5.0×10−4×(1.26)2=3.97×10−4 J $= 0.397$=0.397 mJ.

Alternatively $E_\text{total} = \frac{1}{2}m\omega^2 A^2 = 0.5 \times 5.0 \times 10^{-4} \times (6283)^2 \times (2.0 \times 10^{-4})^2 = 3.95 \times 10^{-4}$Etotal​=21​mω2A2=0.5×5.0×10−4×(6283)2×(2.0×10−4)2=3.95×10−4 J (rounding agrees). The total kinetic-plus-potential energy of the cone is under a milli-joule, but it is continuously being supplied by the amplifier and dissipated to the air as sound waves.

---

Conservation of Energy — The Big Picture

Energy conservation is the reason simple harmonic motion is so widespread. Any system with a stable equilibrium has a potential-energy curve with a minimum, and near the minimum any smooth curve looks parabolic. If the curve is parabolic, the restoring force is linear, and the motion is simple harmonic with conserved $E = \frac{1}{2}m\omega^2 A^2$E=21​mω2A2.

In other words, SHM is what you get whenever a system is disturbed by a small amount from stable equilibrium. This is why everything from a violin string to a carbon dioxide molecule in the atmosphere obeys (at least approximately) the SHM equations — and why $E \propto A^2$E∝A2 is one of the most ubiquitous proportionalities in all of physics. It applies to:

• Mechanical oscillators (mass-spring, pendulum, vibrating ruler).
• Sound waves: intensity $\propto A^2$∝A2.
• Electromagnetic waves: intensity $\propto E_0^2$∝E02​ (where $E_0$E0​ is electric field amplitude).
• Water waves: energy density $\propto A^2$∝A2.
• Quantum mechanics: probability density $\propto |\psi|^2$∝∣ψ∣2.

The squared-amplitude rule is the energetic signature of harmonic motion in every branch of physics.

---

Synoptic Links