Energy in Simple Harmonic Motion

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) for elastic potential energy (when it is momentarily stationary at the extremes). A pendulum swaps kinetic energy for gravitational potential energy. In every case, the total mechanical energy — absent friction — remains constant.

This lesson explores the energy of a simple harmonic oscillator: formulae, graphs, and exam-style worked examples. It is part of OCR A-Level Physics A Module 5.3 (Oscillations).

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Total Energy of an SHM Oscillator

From the last lesson, the velocity of a simple harmonic oscillator satisfies

v² = ω²(A² − x²)

Its kinetic energy is therefore

E_k = ½mv² = ½mω²(A² − x²)

At the maximum displacement, x = ±A and v = 0, so E_k = 0. At equilibrium, x = 0 and the speed is maximum, so

E_k,max = ½mω²A² = ½mv_max²

Because energy is conserved (no friction), the total mechanical energy of the oscillator must be the value at equilibrium, where the entire energy is kinetic:

E_total = ½mω²A²

This is a beautiful result: for a given oscillator, the total energy is proportional to the square of the amplitude. Doubling the amplitude quadruples the total energy.

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Potential Energy

Since total energy is conserved,

E_p = E_total − E_k = ½mω²A² − ½mω²(A² − x²) = ½mω²x²

E_p = ½mω²x²

So potential energy grows as the square of the displacement from equilibrium. This parabolic "energy well" is the signature of any harmonic system — whether it is a mass on a spring, a pendulum at small angles, or an atom in a crystal lattice.

At x = 0, E_p = 0 (all energy is kinetic). At x = ±A, E_p = ½mω²A² (all energy is potential). And at any intermediate x, E_k + E_p = ½mω²A² exactly.

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Graphs of Energy in SHM

Plotting E_k, E_p and E_total against displacement x is a standard OCR exam skill.

• E_p = ½mω²x² — upward parabola through the origin.
• E_k = ½mω²(A² − x²) — downward parabola, zero at x = ±A, maximum at x = 0.
• E_total — a horizontal line at height ½mω²A².

graph TD
    T["E_total<br/>(constant, = ½mω²A²)"]
    E["E (energy axis)"]
    X["x (displacement axis, from −A to +A)"]
    P["E_p = ½mω²x² — parabola, minimum at x = 0"]
    K["E_k = ½mω²(A² − x²) — inverted parabola, maximum at x = 0"]
    T --> P
    T --> K

The two parabolae mirror each other, and at every x their values add up to the flat total-energy line.

Against Time

Against time the picture is different but equally instructive. Using x = A cos(ωt):

• E_p(t) = ½mω²A² cos²(ωt)
• E_k(t) = ½mω²A² sin²(ωt)
• E_total = ½mω²A² (constant)

Because sin² + cos² = 1, the sum is still a flat line. Both E_k and E_p oscillate with a period of T/2 — exactly half the period of the motion itself. (Every cycle the oscillator passes through equilibrium twice and through each extreme twice, so the kinetic energy reaches its maximum twice and the potential energy reaches its maximum twice.)

Exam Tip: A classic OCR trap: "What is the frequency of the kinetic energy variation?" The answer is 2f, twice the frequency of the motion, because KE reaches its maximum twice per cycle.

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Worked Example 1 — Energy of a Mass on a Spring

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

(a)

E_total = ½mω²A² = ½ × 0.40 × 12² × 0.050²
        = 0.5 × 0.40 × 144 × 0.0025
        = 0.072 J

(b)

E_k = ½mω²(A² − x²) = 0.5 × 0.40 × 144 × (0.050² − 0.030²)
    = 28.8 × (0.0025 − 0.0009)
    = 28.8 × 0.0016
    = 0.046 J