Library / Foundations: Physics / FPH-55
Foundations: Physics
Kinetic energy of a simple harmonic oscillator
Energy of motion of the oscillator at displacement x.
- Ref
- FPH-55
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- v1
- Field
- Foundations: Physics
- Result
- Ek [J]
The kinetic energy of the oscillator is the total energy ½mω²A² minus the potential energy ½mω²x², which gives Ek = ½mω²(A² − x²). It is greatest at the centre and zero at the ends. At every displacement the sum of kinetic and potential energy equals the same constant. The displacement must not exceed the amplitude.
Energy of motion of the oscillator at displacement x. It is the total energy minus the potential energy, so it is greatest at the centre and zero at the ends.
Calculator
- Total energy ½mω²A²0.0625 J
- Potential energy ½mω²x²0.0225 J
- Speed at this displacement0.4 m/s
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- At 3 cm the oscillator has 0.04 J of kinetic energy and 0.0225 J of potential energy, which add to the total of 0.0625 J.
- At the centre all of the 0.0625 J is kinetic; at the end, none: the energy swaps back and forth twice every cycle.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J) | Local step | Change in result |
|---|---|---|---|---|---|---|
| A Amplitude | 0.001 to 1 m | 0.0025 | 24.98 | 24.97 | +10% | +32.8 % |
| ω Angular frequency | 1 to 100 rad/s | 4e-4 | 4 | 4 | +10% | +21 % |
| m Oscillating mass | 0.01 to 10 kg | 8e-4 | 0.8 | 0.7992 | +10% | +10 % |
| x Displacement from equilibrium | −1 to 1 m | — | — | — | +0.04 m | — |
Sweep of Angular frequency
| Angular frequency (rad/s) | Kinetic energy (J) |
|---|---|
| 1 | 4e-4 |
| 13.38 | 0.07156 |
| 25.75 | 0.2652 |
| 38.13 | 0.5814 |
| 50.5 | 1.02 |
| 62.88 | 1.581 |
| 75.25 | 2.265 |
| 87.63 | 3.071 |
| 100 | 4 |
Sweep one variable
Variables and how each one contributes
Kinetic energy
What the equation gives you. Everything else feeds it.
Oscillating mass
The oscillating mass, in kg. Raises the kinetic energy in direct proportion for a given ω, A and x. Raising it by 0.1 kg (to 0.6 kg) moves the result by +20 %, from 0.04 to 0.048 J.
Angular frequency
The angular frequency, in rad/s. Raises the kinetic energy with its square: a faster oscillation moves the mass faster. Raising it by 5 rad/s (to 15 rad/s) moves the result by +125 %, from 0.04 to 0.09 J.
Amplitude
The amplitude, in m. Raises the kinetic energy, strongly, because it raises the total energy for a given displacement. Raising it by 0.01 m (to 0.06 m) moves the result by +68.8 %, from 0.04 to 0.0675 J.
Displacement from equilibrium
The displacement from equilibrium, in m. Its size cannot exceed A. Lowers the kinetic energy as the mass approaches the end of its range, where it stops. Raising it by 0.01 m (to 0.04 m) moves the result by −43.8 %, from 0.04 to 0.0225 J.
Worked example with the values above
- Squares: A² − x² = 0.0025 − 9×10−4 = 0.0016 m².
- Multiply: m ω² (A² − x²) = 0.5 × 100 × 0.0016 = 0.08 J.
- Ek = ½ × 0.08 = 0.04 J.
Assumptions and limits
- An undamped simple harmonic oscillator: a restoring force proportional to the displacement.
- Constant amplitude and angular frequency; displacement is measured from the equilibrium position.
Common mistakes
- Entering a displacement larger than the amplitude.
- Forgetting that the kinetic and potential parts must sum to a constant.
Statistics and uncertainty
The kinetic energy is a difference of two squares, so it is sensitive to errors when x is close to A, where the two terms are nearly equal. Take care to measure A and x consistently.
Where it is used
- The kinetic energy of a vibrating mass at a given position.
- Checking that energy is conserved in a vibration model.
Maths you need
- Angular frequency: ω = 2π times the frequency, in rad/s
- Amplitude: the largest displacement from equilibrium