Energy in SHM

Energy is central to understanding SHM. During each oscillation, energy is continuously exchanged between kinetic and potential forms, but the total mechanical energy remains constant (in the absence of damping). This energy analysis provides deep insight into why the motion behaves as it does and gives us another powerful equation for solving problems.

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Kinetic Energy in SHM

The kinetic energy of an oscillating object at any displacement x is:

KE = ½mv²

Using the velocity-displacement relation v² = ω²(A² − x²), we can substitute:

KE = ½mω²(A² − x²)

Key features:

• KE is maximum at the equilibrium position (x = 0): KE_max = ½mω²A²
• KE is zero at maximum displacement (x = ±A): the object is momentarily stationary
• KE varies as a function of displacement — it decreases as the object moves away from equilibrium

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Potential Energy in SHM

The potential energy stored in the system at displacement x is:

PE = ½mω²x²

For a mass-spring system, this is elastic potential energy stored in the stretched or compressed spring (PE = ½kx², and since ω² = k/m, we get ½kx² = ½mω²x²). For a pendulum, it is gravitational potential energy.

Key features:

• PE is zero at the equilibrium position (x = 0)
• PE is maximum at maximum displacement (x = ±A): PE_max = ½mω²A²
• PE increases as the object moves away from equilibrium

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Total Energy

The total mechanical energy at any point is:

E_total = KE + PE = ½mω²(A² − x²) + ½mω²x² = ½mω²A²

This is a constant — it does not depend on x or t. The total energy depends only on:

• The mass m
• The angular frequency ω (which depends on the system parameters)
• The amplitude A

This means that doubling the amplitude quadruples the total energy (since E ∝ A²).

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Energy Exchange During One Complete Oscillation

flowchart LR
    A["x = +A\nKE = 0\nPE = max\nv = 0"] -->|"PE converts to KE\nObject accelerates"| B["x = 0\nKE = max\nPE = 0\nv = v_max"]
    B -->|"KE converts to PE\nObject decelerates"| C["x = −A\nKE = 0\nPE = max\nv = 0"]
    C -->|"PE converts to KE\nObject accelerates"| D["x = 0\nKE = max\nPE = 0\nv = −v_max"]
    D -->|"KE converts to PE\nObject decelerates"| A

This continuous, lossless interchange between KE and PE is what sustains the oscillation indefinitely in the idealised, undamped case. Energy is never created or destroyed — it simply changes form.

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Energy-Displacement Graphs

Plotting KE, PE, and total energy against displacement produces one of the most important diagrams in SHM:

• KE is an inverted parabola, maximum at x = 0 and zero at x = ±A
• PE is an upward parabola, zero at x = 0 and maximum at x = ±A
• Total energy is a horizontal straight line at E = ½mω²A²

The KE and PE curves intersect at x = ±A/√2. At these points, the kinetic and potential energies are equal, each being half the total energy.

At every displacement, KE + PE = constant. Energy flows continuously between the two forms as the object oscillates.

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Energy-Time Graphs

If x = A cos(ωt), then:

• KE = ½mω²A² sin²(ωt)
• PE = ½mω²A² cos²(ωt)

Both KE and PE oscillate with twice the frequency of the displacement (frequency 2f, period T/2). This makes sense: the object passes through equilibrium twice per cycle (once in each direction), so KE reaches its maximum twice per cycle.

The KE and PE curves are sinusoidal, always positive (energy cannot be negative), and always sum to the constant total energy.

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Energy Equations Summary

Quantity	Formula	Maximum value	Where maximum occurs
Kinetic energy	½mω²(A² − x²)	½mω²A²	At equilibrium (x = 0)
Potential energy	½mω²x²	½mω²A²	At extremes (x = ±A)
Total energy	½mω²A²	Constant	Everywhere
For mass-spring	E = ½kA²	—	—

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Effect of Amplitude on Energy

Since E_total = ½mω²A², the total energy is proportional to A². This has practical consequences:

• If you double the amplitude, the total energy quadruples
• If you halve the amplitude, the total energy is reduced to one quarter
• The maximum speed also increases with amplitude: v_max = ωA

Amplitude	Total Energy	Maximum Speed
A	E	v_max
2A	4E	2v_max
A/2	E/4	v_max/2
3A	9E	3v_max

In real systems with damping, the amplitude gradually decreases, meaning the total energy gradually decreases. We will study damping in a later lesson.

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Worked Example 1

A 0.3 kg mass on a spring (k = 48 N m⁻¹) oscillates with amplitude 0.05 m.

Find: (a) the total energy, (b) the KE when x = 0.03 m, (c) the speed when x = 0.03 m, (d) the displacement when KE = PE.

Solution:

ω² = k/m = 48/0.3 = 160 s⁻²

(a) E_total = ½kA² = ½ × 48 × 0.05² = 0.060 J

(b) PE at x = 0.03 m: PE = ½kx² = ½ × 48 × 0.03² = 0.0216 J KE = E_total − PE = 0.060 − 0.0216 = 0.0384 J

(c) KE = ½mv², so v = √(2 × KE/m) = √(2 × 0.0384/0.3) = √(0.256) = 0.506 m s⁻¹

(d) When KE = PE, each equals E_total/2 = 0.030 J PE = ½kx² = 0.030, so x² = 0.060/48 = 0.00125 x = ±0.0354 m = ±A/√2 ✓

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Worked Example 2

A mass-spring system has a total oscillation energy of 0.50 J when the amplitude is 0.10 m. Calculate the energy when the amplitude decreases to 0.07 m due to damping, and find the energy lost.

Solution:

E ∝ A², so E₁/E₂ = A₁²/A₂²

E₂ = E₁ × (A₂/A₁)² = 0.50 × (0.07/0.10)² = 0.50 × 0.49 = 0.245 J

Energy lost = 0.50 − 0.245 = 0.255 J

This energy has been transferred to the surroundings as thermal energy through the damping mechanism.

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Worked Example 3

An oscillator has mass 0.25 kg and oscillates with amplitude 0.08 m and frequency 5 Hz. At what displacement is 75% of the total energy stored as kinetic energy?

Solution:

If KE = 0.75 × E_total, then PE = 0.25 × E_total.

PE/E_total = ½mω²x² / ½mω²A² = x²/A² = 0.25

x² = 0.25 × A² = 0.25 × 0.08² = 0.25 × 0.0064 = 0.0016

x = ±0.04 m = ±A/2

So when the displacement is half the amplitude, 75% of the energy is kinetic and 25% is potential.

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Deriving v = ±ω√(A² − x²) from Energy Conservation

The velocity-displacement equation can be derived elegantly from energy conservation:

At any displacement x: ½mv² + ½mω²x² = ½mω²A²

Dividing through by ½m: v² + ω²x² = ω²A²

Rearranging: v² = ω²(A² − x²)

Taking the square root: v = ±ω√(A² − x²)

This derivation demonstrates the deep connection between energy conservation and the kinematics of SHM.

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Common Exam Mistakes with Energy in SHM

Mistake	Correction
"Doubling amplitude doubles energy"	E ∝ A², so doubling A gives 4× energy
Drawing energy-time graphs at frequency f	Energy oscillates at 2f, not f
Writing negative kinetic energy	KE is always ≥ 0
Forgetting PE exists at equilibrium for a vertical spring	The equilibrium shifts, but we measure PE from new equilibrium where PE = 0
Assuming total energy changes during oscillation	E_total is constant (undamped)

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Real-World Application: Energy Harvesting from Vibrations

Piezoelectric energy harvesters convert mechanical vibrations into electrical energy using the SHM energy principle. A small mass on a cantilever beam oscillates when subjected to ambient vibrations (from machinery, footsteps, or traffic). The total energy ½mω²A² determines the maximum electrical power that can be extracted. Engineers design the harvester's natural frequency to match the dominant vibration frequency of the environment, maximising A through resonance and therefore maximising the harvested energy. Modern wearable devices and wireless sensors increasingly use this technology.

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A-Level Deep Dive: Energy in SHM

Spec mapping

Edexcel 9PH0 specification, Topic 13 — Oscillations requires students to analyse the interchange of kinetic and potential energy in a simple-harmonic oscillator, to use $E_{\text{total}} = \tfrac{1}{2}m\omega^2 A^2$Etotal​=21​mω2A2, and to interpret displacement-energy and time-energy graphs (refer to the official specification document for exact wording). Energy in SHM appears in Paper 2 (Section B Topics 6–8 carry forward, with Topic 13 sitting in Paper 2's later content) and is regularly cross-examined with Topic 14 (Astrophysics, via stellar pulsations) and Topic 13's own damping/forced-vibration sub-strands. The Edexcel A-Level data and formulae booklet does include the SHM kinematic equations ($x = A\cos\omega t$x=Acosωt and $v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​) but does not tabulate the energy expressions — students must derive $\tfrac{1}{2}m\omega^2 A^2$21​mω2A2 on demand from $\tfrac{1}{2}mv_{\max}^2$21​mvmax2​.

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Worked example with full mark scheme

Question (8 marks):

A trolley of mass $0.40\ \text{kg}$0.40 kg is attached to two horizontal springs and undergoes SHM with amplitude $A = 0.060\ \text{m}$A=0.060 m and period $T = 0.50\ \text{s}$T=0.50 s.

(a) Calculate the maximum kinetic energy of the trolley. (3)

(b) Calculate the potential energy stored when the displacement is $x = 0.030\ \text{m}$x=0.030 m, and verify that $\text{KE} + \text{PE}$KE+PE at this displacement equals the total energy from part (a). (5)

Solution with mark scheme:

(a) Step 1 — find $\omega$ω.

$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{0.50} = 4\pi \approx 12.57\ \text{rad s}^{-1}$ω=T2π​=0.502π​=4π≈12.57 rad s−1

M1 — correct use of $\omega = 2\pi/T$ω=2π/T. Common slip: writing $\omega = 1/T$ω=1/T (confusing angular frequency with frequency).

Step 2 — apply $\text{KE}_{\max} = \tfrac{1}{2}m\omega^2 A^2$KEmax​=21​mω2A2.

$\text{KE}_{\max} = \tfrac{1}{2}(0.40)(12.57)^2 (0.060)^2 = \tfrac{1}{2}(0.40)(157.9)(0.0036) \approx 0.114\ \text{J}$KEmax​=21​(0.40)(12.57)2(0.060)2=21​(0.40)(157.9)(0.0036)≈0.114 J

M1 — substitution into the correct expression. A1 — answer to 2–3 sig fig with unit ($0.11\ \text{J}$0.11 J acceptable).

(b) Step 1 — calculate PE at $x = 0.030\ \text{m}$x=0.030 m.

$\text{PE} = \tfrac{1}{2}m\omega^2 x^2 = \tfrac{1}{2}(0.40)(157.9)(0.030)^2 = \tfrac{1}{2}(0.40)(157.9)(9.0\times 10^{-4}) \approx 0.0284\ \text{J}$PE=21​mω2x2=21​(0.40)(157.9)(0.030)2=21​(0.40)(157.9)(9.0×10−4)≈0.0284 J

M1 — substitution into $\tfrac{1}{2}m\omega^2 x^2$21​mω2x2. A1 — $\text{PE} \approx 0.028\ \text{J}$PE≈0.028 J.

Step 2 — calculate KE at this displacement using $v^2 = \omega^2(A^2 - x^2)$v2=ω2(A2−x2).

$v^2 = (12.57)^2(0.060^2 - 0.030^2) = 157.9 \times (0.0036 - 0.0009) = 157.9 \times 0.0027 = 0.426$v2=(12.57)2(0.0602−0.0302)=157.9×(0.0036−0.0009)=157.9×0.0027=0.426

$\text{KE} = \tfrac{1}{2}(0.40)(0.426) \approx 0.0853\ \text{J}$KE=21​(0.40)(0.426)≈0.0853 J

M1 — correct method (KE from velocity-displacement relation, not by simply subtracting).

Step 3 — verify conservation.

$\text{KE} + \text{PE} = 0.0853 + 0.0284 = 0.1137\ \text{J} \approx \text{KE}_{\max}\ \checkmark$KE+PE=0.0853+0.0284=0.1137 J≈KEmax​ ✓

A1 — explicit verification statement that the sum equals the part (a) value within rounding.

Total: 8 marks (M4 A4).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A pendulum bob of mass $0.20\ \text{kg}$0.20 kg swings with frequency $0.80\ \text{Hz}$0.80 Hz and amplitude $0.040\ \text{m}$0.040 m.

(a) Show that the maximum kinetic energy of the bob is approximately $1.0\ \text{mJ}$1.0 mJ. (3)

(b) State and explain at what displacement the kinetic energy first equals the potential energy. (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO2) — recognising $\omega = 2\pi f = 2\pi(0.80) \approx 5.03\ \text{rad s}^{-1}$ω=2πf=2π(0.80)≈5.03 rad s−1.
• M1 (AO2) — substituting into $\text{KE}_{\max} = \tfrac{1}{2}m\omega^2 A^2 = \tfrac{1}{2}(0.20)(5.03)^2(0.040)^2$KEmax​=21​mω2A2=21​(0.20)(5.03)2(0.040)2.
• A1 (AO1) — evaluating to $\approx 1.01\times 10^{-3}\ \text{J} \approx 1.0\ \text{mJ}$≈1.01×10−3 J≈1.0 mJ as printed.

(b)

• B1 (AO1) — stating $x = \pm A/\sqrt{2} \approx \pm 0.028\ \text{m}$x=±A/2​≈±0.028 m.
• M1 (AO2) — setting $\tfrac{1}{2}m\omega^2 x^2 = \tfrac{1}{2}\cdot\tfrac{1}{2}m\omega^2 A^2$21​mω2x2=21​⋅21​mω2A2 giving $x^2 = A^2/2$x2=A2/2.
• A1 (AO3) — explaining that at this displacement PE is half the total energy, so by conservation KE is also half.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. Edexcel weights energy questions toward AO2 (applying SHM relations) with a small AO3 strand reserved for the conservation reasoning step.

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Synoptic links

Energy in SHM connects across the 9PH0 specification to:

1. SHM kinematics (Topic 13, earlier sub-strand): the velocity-displacement equation $v = \pm\omega\sqrt{A^2 - x^2}$v=±ωA2−x2​ is energy conservation in disguise. Squaring and multiplying by \tfrac{1}{2}m\ yields $\tfrac{1}{2}mv^2 = \tfrac{1}{2}m\omega^2(A^2 - x^2)$21​mv2=21​mω2(A2−x2), i.e. $\text{KE} = E_{\text{total}} - \text{PE}$KE=Etotal​−PE.

2. Hooke's law and elastic PE (Topic 4, Materials): for a mass-spring system, $\text{PE} = \tfrac{1}{2}kx^2$PE=21​kx2 from the area under a force-extension graph. Combined with $\omega^2 = k/m$ω2=k/m, this collapses to $\text{PE} = \tfrac{1}{2}m\omega^2 x^2$PE=21​mω2x2 — the same expression derived independently from SHM. Two routes, one answer.

3. Damping (Topic 13, later sub-strand): light damping causes the amplitude to decay approximately exponentially, $A(t) = A_0 e^{-\gamma t}$A(t)=A0​e−γt. Since $E \propto A^2$E∝A2, the energy decays as $E(t) = E_0 e^{-2\gamma t}$E(t)=E0​e−2γt — twice as fast as the amplitude. This factor of 2 is examined regularly.

4. Resonance (Topic 13): at the resonant driving frequency, the steady-state amplitude (and hence stored energy) is maximised. The width of the resonance peak (the Q-factor) determines how sharply tuned the system is — directly relevant to engineering applications and to MRI / NMR in Topic 12.

5. Gravitational and circular motion (Topic 13's neighbouring topics): orbital energy $E = -\tfrac{GMm}{2r}$E=−2rGMm​ for a circular orbit shows the same "half-and-half" KE-PE structure that emerges in SHM, foreshadowing the virial theorem (see "Going further").

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Mark-scheme literacy

Energy-in-SHM questions on 9PH0 typically split AO marks as follows: