Where is kinetic energy maximum in SHM?

At the equilibrium position, where speed is greatest and potential energy is zero.

Why does energy depend on amplitude squared?

Maximum speed is proportional to amplitude, and kinetic energy depends on speed squared.

Why do energy–time graphs have double frequency?

Kinetic energy peaks each time the object passes equilibrium, which happens twice per cycle, once in each direction.

What happens to total energy with damping?

It decreases over time as energy is transferred to the surroundings, and the amplitude shrinks.

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188OscillationsStretch

Energy in SHM

A swinging pendulum is constantly converting energy. At the top of each swing it is momentarily still, with all its energy as potential energy; at the bottom it moves fastest, with all its energy kinetic. In an undamped $\textbf{simple harmonic oscillator}$ , the total never changes.

30 min Topic 13 · Oscillations

What you'll be able to do

Describe the interchange of kinetic and potential energy during SHM
Use $E_{\text{total}} = \tfrac12 m\omega^{2}A^{2}$
Calculate kinetic and potential energy at any displacement
Sketch energy–displacement and energy–time graphs for an oscillator

Energy exchange in an oscillator

In a mass–spring system, energy transfers between kinetic energy of the mass and elastic potential energy of the spring. In a pendulum, it transfers between kinetic and gravitational potential energy.

At maximum displacement, the object is momentarily at rest: kinetic energy is zero and potential energy is maximum. At equilibrium, potential energy is at its minimum (taken as zero) and kinetic energy is maximum.

With no damping, no energy is lost, so the sum stays constant throughout the motion.

Tip — For a vertical spring, the equilibrium position already includes some stretch. Measure the SHM potential energy from equilibrium, not from the spring’s natural length.

Total energy

At equilibrium all energy is kinetic, and the speed is $v_{\max} = A\omega$ . So the total energy is $\tfrac12 m(A\omega)^{2}$ .

Total energy is proportional to $A^{2}$ . Halving the amplitude of an oscillation removes three-quarters of its energy.

For a mass on a spring, $\omega^{2} = k/m$ , so this is equivalent to $\tfrac12 kA^{2}$ — exactly the elastic energy stored at maximum extension.

E_{\text{total}} = \tfrac12 m\omega^{2}A^{2}

Total energy of an SHM oscillator.

E_{\text{total}} = \tfrac12 kA^{2}

For a mass on a spring.

Example 1A 0.25 kg mass oscillates on a spring with amplitude 0.080 m and period 0.50 s. Find its total energy.

\omega = 2\pi/0.50 = 12.57\ \mathrm{rad\,s^{-1}}

E = \tfrac12 \times 0.25 \times 12.57^{2} \times 0.080^{2}

Answer

0.126\ \mathrm{J}

Energy at any displacement

Potential energy at displacement $x$ is $E_{p} = \tfrac12 m\omega^{2}x^{2}$ . Kinetic energy is the remainder: $E_{k} = \tfrac12 m\omega^{2}(A^{2} - x^{2})$ .

Plotted against displacement, $E_{p}$ is an upward parabola from zero at the centre, $E_{k}$ is an inverted parabola, and the total is a horizontal line.

Kinetic and potential energy are equal when $x^{2} = A^{2}/2$ , that is, at $x = A/\sqrt{2} \approx 0.71A$ — not at half the amplitude.

E_{p} = \tfrac12 m\omega^{2}x^{2}

Potential energy.

E_{k} = \tfrac12 m\omega^{2}(A^{2} - x^{2})

Kinetic energy.

Example 1For the oscillator above, find the kinetic energy when the displacement is 0.040 m.

E_{p} = \tfrac14 E_{\text{total}}

, since

x = A/2

gives

x^{2} = A^{2}/4

E_{k} = \tfrac34 \times 0.126

Answer

0.095\ \mathrm{J}

Energy against time

Starting from maximum displacement, $E_{p} = E_{\text{total}}\cos^{2}\omega t$ and $E_{k} = E_{\text{total}}\sin^{2}\omega t$ .

Both are always positive and each reaches its maximum twice per oscillation, so the energy graphs have $\textbf{twice the frequency}$ of the displacement.

At every instant the two curves add to the constant total — they are mirror images about half the total energy.

Tip — A common exam trap: if the oscillation has period $T$ , the energy graphs have period $T/2$ .

Equation recap

E = \tfrac12 m\omega^{2}A^{2}

Total energy.

E_{p} = \tfrac12 m\omega^{2}x^{2}

Potential energy.

E_{k} = \tfrac12 m\omega^{2}(A^{2} - x^{2})

Kinetic energy.

Common mistakes to avoid

Saying energy is shared equally at half amplitude.

Equal sharing is at

x = A/\sqrt{2}

Drawing energy–time graphs with the same period as displacement.

Energy oscillates at twice the frequency.

Saying total energy is proportional to amplitude.

It is proportional to

A^{2}

Drawing kinetic energy as negative for part of the cycle.

Kinetic energy is never negative.

Key takeaways

Energy interchanges between kinetic and potential forms during SHM.
Total energy $\tfrac12 m\omega^{2}A^{2}$ is constant without damping.
Energy is proportional to amplitude squared.
$E_{k} = E_{p}$ at $x = A/\sqrt{2}$ .
Energy–time graphs have twice the frequency of the oscillation.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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