Energy in SHM

Energy sloshes between spring and motion while the total stays fixed.

9 minUpdated 21 October 2026energy in SHMpotential energykinetic energytotal energy

Lesson at a glanceShow the key points

Potential energy U = ½kx² = ½mω²x² (zero at the centre) (12.13).
Kinetic energy K = ½mω²(A² − x²); total E = U + K = ½mω²A² = ½kA², independent of time (12.14).
All kinetic at the centre, all potential at the turning points.
U and K each oscillate at twice the frequency of x.

An oscillator is an energy exchange. At the ends of the swing everything is stored; in the middle everything is motion. The total never changes.

Potential energy of a linear restoring force

The force F = −kx does work $\int_0^x(-kx)\,dx = -\tfrac12kx^2$ as the displacement grows from 0 to x. Potential energy rises by the negative of that work. Choosing U = 0 at the centre:

Potential energy in SHM(12.13)

U = \tfrac12 kx^2 = \tfrac12 m\omega^2 x^2

the same as a spring’s stored energy

Kinetic and total energy

With x = A sin(ωt + δ) and v = Aω cos(ωt + δ):

U = \tfrac12m\omega^2A^2\sin^2(\omega t + \delta), \qquad K = \tfrac12m\omega^2A^2\cos^2(\omega t + \delta)

Total energy(12.14)

E = U + K = \tfrac12m\omega^2A^2 = \tfrac12kA^2

constant: sin² + cos² = 1

At the centre U = 0 and all of E is kinetic: $\tfrac12mv_{\max}^2 = \tfrac12m\omega^2A^2$ . At x = ±A the body is momentarily still and all of E is potential.

Interactive

Energy sloshing back and forth

A 1 kg body on a spring. The bowl is the potential energy ½kx²; the flat line is the total energy ½kA². The body can only be where the bowl is below the line, and the gap between them is its kinetic energy. The total never changes.

0.0

3.2

Total

U = \tfrac12kx^2

3.20J

K = \tfrac12mv^2

0.00J

E = \tfrac12kA^2

3.20J

Amplitude

A

0.40m

Spring constant

k

40N/m

Worked example

Energy of a vibrating mass

A 60 g mass oscillates in SHM with amplitude 5.0 cm and period 0.50 s. Find the total mechanical energy.

What to remember

1 $U = \tfrac12kx^2$ , $K = \tfrac12k(A^2 - x^2)$ .
2 $E = \tfrac12kA^2 = \tfrac12m\omega^2A^2$ , constant in time.
3All kinetic at the centre, all potential at the extremes.
4Energy goes as the square of the amplitude.

Check yourself

At what displacement are K and U equal?

Doubling the amplitude multiplies the energy by…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

PlayMasses and SpringsPhET · CU Boulder ReadEnergy in simple harmonic motionOpenStax University Physics §15.2 ReadSimple harmonic motionHyperPhysics

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Energy in SHM

Energy sloshes between spring and motion while the total stays fixed.

9 minUpdated 21 October 2026energy in SHMpotential energykinetic energytotal energy

Lesson at a glanceShow the key points

Potential energy U = ½kx² = ½mω²x² (zero at the centre) (12.13).
Kinetic energy K = ½mω²(A² − x²); total E = U + K = ½mω²A² = ½kA², independent of time (12.14).
All kinetic at the centre, all potential at the turning points.
U and K each oscillate at twice the frequency of x.

Potential energy of a linear restoring force

The force F = −kx does work

\int_0^x(-kx)\,dx = -\tfrac12kx^2

as the displacement grows from 0 to x. Potential energy rises by the negative of that work. Choosing U = 0 at the centre:

Potential energy in SHM(12.13)

U = \tfrac12 kx^2 = \tfrac12 m\omega^2 x^2

the same as a spring’s stored energy

Kinetic and total energy

With x = A sin(ωt + δ) and v = Aω cos(ωt + δ):

U = \tfrac12m\omega^2A^2\sin^2(\omega t + \delta), \qquad K = \tfrac12m\omega^2A^2\cos^2(\omega t + \delta)

Total energy(12.14)

E = U + K = \tfrac12m\omega^2A^2 = \tfrac12kA^2

constant: sin² + cos² = 1

At the centre U = 0 and all of E is kinetic:

\tfrac12mv_{\max}^2 = \tfrac12m\omega^2A^2

. At x = ±A the body is momentarily still and all of E is potential.

Interactive

Energy sloshing back and forth

0.0

3.2

Total

U = \tfrac12kx^2

3.20J

K = \tfrac12mv^2

0.00J

E = \tfrac12kA^2

3.20J

Amplitude

A

0.40m

Spring constant

k

40N/m

Worked example

Energy of a vibrating mass

A 60 g mass oscillates in SHM with amplitude 5.0 cm and period 0.50 s. Find the total mechanical energy.

What to remember

1 $U = \tfrac12kx^2$ , $K = \tfrac12k(A^2 - x^2)$ .
2 $E = \tfrac12kA^2 = \tfrac12m\omega^2A^2$ , constant in time.
3All kinetic at the centre, all potential at the extremes.
4Energy goes as the square of the amplitude.

Check yourself

At what displacement are K and U equal?

Doubling the amplitude multiplies the energy by…

Potential energy of a linear restoring force#

Kinetic and total energy#

Check yourself

Keep exploring

Potential energy of a linear restoring force#

Kinetic and total energy#

Check yourself

Keep exploring

Potential energy of a linear restoring force

Kinetic and total energy

Potential energy of a linear restoring force

Kinetic and total energy