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In simple harmonic motion the mechanical energy is conserved — the centripetal-style restoring force is conservative, no work is done by external forces, and the kinetic and potential energies merely exchange. The shape of that exchange, its symmetry, and its frequency turn out to be richer than they first appear. The kinetic energy oscillates at twice the frequency of the displacement; the total energy scales with the square of the amplitude; and the phase relationships between x, v and a — separated by precisely π/2 in successive derivatives — are exam currency at A-Level. This lesson establishes those relationships precisely.
Spec mapping: This lesson sits under AQA 7408 section 3.6.1.2. It covers the total energy E = ½mω²A² of an SHM oscillator, the kinetic-potential split Eₖ = ½mω²(A² − x²) and Eₚ = ½mω²x², the doubling of the energy-oscillation frequency relative to the displacement frequency, and the phase relationships between displacement, velocity and acceleration (x and v are π/2 out of phase; x and a are π out of phase). Maximum and minimum speeds and accelerations are reviewed in this energy context. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.4 (energy in mechanics): the work-energy theorem and conservative-force framework underpinning SHM energy bookkeeping are first developed here.
- Section 3.6.1.1 (circular motion): the reference-circle construction extends to energy — the constant total energy of an SHM oscillator corresponds to the constant kinetic energy of the reference point in uniform circular motion.
- Section 3.5 (capacitors and inductors): in an LC oscillator, the electric-field energy ½(Q²/C) in the capacitor and the magnetic-field energy ½LI² in the inductor exchange at twice the oscillation frequency — the direct electromagnetic analogue of the SHM Eₖ/Eₚ exchange.
A simple harmonic oscillator possesses two reservoirs of mechanical energy: kinetic energy associated with the speed of the moving mass, and potential energy stored in whatever provides the restoring force (a spring's elastic energy, gravity's gravitational PE, an electric-field energy). In an undamped SHM system, these two add to a constant.
Using the velocity-displacement identity from the previous lesson, v² = ω²(A² − x²):
Eₖ = ½mv² = ½mω²(A² − x²)
Eₖ is maximum at x = 0 (equilibrium), where Eₖ_max = ½mω²A². Eₖ is zero at x = ±A (the extremes), where the body is momentarily at rest.
The work done against the restoring force in displacing the body from 0 to x is
W = ∫₀ˣ kx' dx' = ½kx² = ½(mω²)x² = ½mω²x².
This work is stored as elastic (or equivalent) potential energy, so
Eₚ = ½mω²x²
Eₚ is zero at x = 0 (equilibrium) and maximum at x = ±A (the extremes), where Eₚ_max = ½mω²A².
Adding:
E_total = Eₖ + Eₚ = ½mω²(A² − x²) + ½mω²x² = ½mω²A² − ½mω²x² + ½mω²x² = ½mω²A².
The total energy is constant — it does not depend on displacement, and it does not depend on time. The two reservoirs continuously exchange, but their sum is fixed by the initial conditions.
E_total = ½mω²A² = ½kA² = ½m(v_max)²
The three forms are equivalent: the first uses ω, the second uses the spring constant k = mω² (for a mass-spring system), the third uses v_max = ωA.
A 0.30 kg mass attached to a spring of constant k = 75 N m⁻¹ oscillates with amplitude 0.080 m. Calculate (a) the total energy of the oscillation, (b) the kinetic energy at x = 0.050 m, (c) the speed at x = 0.050 m.
(a) E_total = ½kA² = ½ × 75 × (0.080)² = 37.5 × 0.0064 = 0.24 J.
(b) Eₚ at x = 0.050 m: Eₚ = ½kx² = ½ × 75 × (0.050)² = 0.0938 J.
Eₖ at x = 0.050 m: Eₖ = E_total − Eₚ = 0.24 − 0.0938 = 0.146 J.
(c) Eₖ = ½mv² → v = √(2Eₖ/m) = √(2 × 0.146/0.30) = √0.974 = 0.987 m s⁻¹.
Cross-check: ω² = k/m = 75/0.30 = 250 → ω = 15.81 rad s⁻¹. v = ω√(A² − x²) = 15.81 × √(0.0064 − 0.0025) = 15.81 × √0.0039 = 15.81 × 0.0625 = 0.988 m s⁻¹. ✓ (small rounding).
Plotting Eₖ and Eₚ against displacement x reveals two parabolas:
The two parabolas cross at Eₖ = Eₚ = ¼kA². Setting Eₚ = ¼kA² gives ½kx² = ¼kA² → x² = A²/2 → x = A/√2 ≈ 0.707A. At this displacement the energy is split equally between kinetic and potential. This crossover point is a routine exam question.
Equipartition point. Eₖ = Eₚ at x = ±A/√2.
graph TD
A["x = −A (extreme)"] -->|"Eₖ = 0<br/>Eₚ = ½kA²"| B["x = 0 (equilibrium)"]
B -->|"Eₖ = ½kA²<br/>Eₚ = 0"| C["x = +A (extreme)"]
C -->|"Eₖ = 0<br/>Eₚ = ½kA²"| B
D["At x = ±A/√2:<br/>Eₖ = Eₚ = ¼kA²<br/>(equipartition)"]
style A fill:#e74c3c,color:#fff
style B fill:#27ae60,color:#fff
style C fill:#e74c3c,color:#fff
style D fill:#3498db,color:#fff
The displacement varies as x(t) = A cos(ωt + φ). The two energies as functions of time follow directly:
Eₚ(t) = ½mω²A² cos²(ωt + φ)
Eₖ(t) = ½mω²A² sin²(ωt + φ)
Both are squared trigonometric functions, oscillating between 0 and ½mω²A². Their sum is ½mω²A²(cos² + sin²) = ½mω²A², constant. The exchange in time is perfectly out of phase: when Eₚ is at its maximum, Eₖ is at its minimum, and vice versa.
A crucial observation: cos² θ = ½(1 + cos 2θ) and sin² θ = ½(1 − cos 2θ). So:
Eₚ(t) = ¼mω²A²[1 + cos(2ωt + 2φ)]
Eₖ(t) = ¼mω²A²[1 − cos(2ωt + 2φ)]
Both Eₖ and Eₚ oscillate at angular frequency 2ω, i.e. at twice the frequency of the displacement. The doubling has a clear physical reason: as the body sweeps from x = −A to x = +A in half a period, the kinetic energy rises from zero (at −A) to a maximum (at x = 0) and back to zero (at +A). Then in the second half-period, exactly the same Eₖ profile repeats as the body returns from +A through 0 back to −A. The body distinguishes its two extremes (sign of x); the energy does not (sign of x² is the same).
In summary:
The displacement-time graph has period T. The energy-time graphs (Eₖ-t and Eₚ-t) have period T/2.
This is one of the most testable observations in the SHM topic — students who can articulate the frequency-doubling routinely earn marks that students who cannot do not.
A mass-spring system has frequency f = 4.0 Hz of displacement oscillation. State (a) the period of the displacement, (b) the frequency of the kinetic-energy oscillation, (c) the period of the kinetic-energy oscillation.
(a) T = 1/f = 0.250 s.
(b) Energy oscillates at twice the displacement frequency: f_E = 2f = 8.0 Hz.
(c) T_E = 1/f_E = 0.125 s (or equivalently T/2).
The three kinematic descriptions of SHM are:
| Quantity | Expression | Magnitude peaks when |
|---|---|---|
| Displacement | x = A cos(ωt + φ) | x = ±A (cos = ±1) |
| Velocity | v = −Aω sin(ωt + φ) | x = 0 (sin = ±1) |
| Acceleration | a = −Aω² cos(ωt + φ) | x = ±A (cos = ±1) |
The phase relationships between them are clean:
The cleanest way to remember these is via the reference circle. On a reference circle, four quantities sit at four equally spaced angular positions: position (radial), velocity (90° ahead, tangential), acceleration (180° from position, centripetal-inward), and the velocity's own derivative (270° ahead). The successive π/2 phase lags reproduce as derivatives in time.
If displacement is a cosine starting at +A (φ = 0):
The acceleration is always proportional to the negative of the displacement: a = −ω²x. This is the defining equation of SHM, and on an a–x plot it appears as a straight line through the origin with negative gradient. The gradient is −ω².
graph LR
A["x(t) = A cos(ωt)<br/>(cosine, peak at t=0)"] --> B["v(t) = −Aω sin(ωt)<br/>(neg. sine, peak at t=T/4)"]
B --> C["a(t) = −Aω² cos(ωt)<br/>(neg. cosine, peak at t=T/2)"]
A -.-> D["x and v: π/2 phase difference"]
B -.-> E["v and a: π/2 phase difference"]
A -.-> F["x and a: π phase difference<br/>(antiphase)"]
C -.-> F
style A fill:#3498db,color:#fff
style B fill:#27ae60,color:#fff
style C fill:#e67e22,color:#fff
style F fill:#e74c3c,color:#fff
Bringing the energy and phase discussion together:
| Position | Displacement | Velocity | Acceleration | Eₖ | Eₚ |
|---|---|---|---|---|---|
| Centre | x = 0 | v = ±Aω (max) | a = 0 | max | 0 |
| Extremes | x = ±A | v = 0 | a = ∓Aω² (max magnitude) | 0 | max |
| Half-amplitude | x = ±A/2 | v = ±Aω√(3)/2 ≈ 0.866Aω | a = ∓ω²A/2 | (3/4)E_total | (1/4)E_total |
| Equipartition | x = ±A/√2 ≈ 0.707A | v = ±Aω/√2 ≈ 0.707Aω | a = ∓ω²A/√2 | E_total/2 | E_total/2 |
The position–velocity–acceleration interplay tells the whole story of the motion. At the centre, the body is moving fastest and accelerating nowhere — pure kinetic energy. At the extremes, it is stationary but accelerating most strongly — pure potential energy. Everywhere in between, the totals exchange while their sum stays constant.
An SHM oscillator has amplitude 0.10 m and angular frequency 20 rad s⁻¹. At what displacement is the kinetic energy equal to the potential energy?
Set Eₖ = Eₚ. ½mω²(A² − x²) = ½mω²x² → A² − x² = x² → x² = A²/2 → x = ±A/√2 = ±0.10/√2 = ±0.0707 m.
At this displacement, each form holds half the total energy: ¼mω²A² = ¼ × m × 400 × 0.01 = m. So Eₖ = Eₚ = m × (units of J kg⁻¹), or with m = 0.2 kg, each equals 0.2 J of a total 0.4 J.
Specimen question modelled on the AQA paper format. Nine marks.
A pendulum bob of mass 0.040 kg oscillates with amplitude 0.020 m about its equilibrium position, performing simple harmonic motion at frequency 0.50 Hz.
(a) Calculate (i) the angular frequency ω, (ii) the total energy of the oscillation. (3 marks)
(b) Sketch on the same axes against time t (for two complete periods) (i) the displacement x, (ii) the kinetic energy Eₖ. Describe the phase and frequency relationship between the two. (3 marks)
(c) Calculate the displacement at which the kinetic energy equals the potential energy and explain the physical significance of this displacement. (3 marks)
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