Library / Foundations: Physics / FPH-54
Foundations: Physics
Potential energy of a simple harmonic oscillator
Energy stored in the restoring force when the oscillator is displaced by x, with zero taken at equilibrium.
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- FPH-54
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- Foundations: Physics
- Result
- Ep [J]
The potential energy stored in the restoring force at displacement x is Ep = ½kx² = ½mω²x², with zero taken at equilibrium. It rises with the square of the displacement and equals the total energy at the ends of the motion, where x = ±A. The sign of x does not matter.
Energy stored in the restoring force when the oscillator is displaced by x, with zero taken at equilibrium. It rises as the square of the displacement.
Calculator
- Stiffness implied, k = mω²50 N/m
- Restoring force at this displacement, kx1.5 N
- Energy at twice this displacement0.09 J
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 from equilibrium the energy stored in the restoring force is 0.0225 J, for a stiffness of 50 N/m.
- Compressing and stretching by the same amount stores the same energy, since only x² matters: the energy curve is a parabola with its minimum at equilibrium.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J) | Local step | Change in result |
|---|---|---|---|---|---|---|
| x Displacement from equilibrium | −1 to 1 m | 0 | 25 | 25 | +0.04 m | +444 % |
| ω Angular frequency | 1 to 100 rad/s | 2.25e-4 | 2.25 | 2.25 | +10% | +21 % |
| m Oscillating mass | 0.01 to 10 kg | 4.5e-4 | 0.45 | 0.4495 | +10% | +10 % |
Sweep of Displacement from equilibrium
| Displacement from equilibrium (m) | Potential energy (J) |
|---|---|
| −1 | 25 |
| −0.75 | 14.06 |
| −0.5 | 6.25 |
| −0.25 | 1.563 |
| 0 | 0 |
| 0.25 | 1.563 |
| 0.5 | 6.25 |
| 0.75 | 14.06 |
| 1 | 25 |
Sweep one variable
Variables and how each one contributes
Potential energy
What the equation gives you. Everything else feeds it.
Oscillating mass
The oscillating mass, in kg. Raises the stored energy in direct proportion for a given ω and x. Raising it by 0.1 kg (to 0.6 kg) moves the result by +20 %, from 0.0225 to 0.027 J.
Angular frequency
The angular frequency, in rad/s. Raises the stored energy with its square, because the stiffness is mω². Raising it by 5 rad/s (to 15 rad/s) moves the result by +125 %, from 0.0225 to 0.05062 J.
Displacement from equilibrium
The displacement from equilibrium, in m. Positive and negative values give the same energy. Raises the stored energy with its square: doubling the displacement gives four times the energy. Raising it by 0.01 m (to 0.04 m) moves the result by +77.8 %, from 0.0225 to 0.04 J.
Worked example with the values above
- Stiffness: k = mω² = 0.5 × 100 = 50 N/m.
- Square the displacement: x² = 9×10−4 m².
- Ep = ½ × 50 × 9×10−4 = 0.0225 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
- Using the amplitude in place of the displacement at the instant considered.
- Expecting a negative energy for a negative displacement.
Statistics and uncertainty
The energy depends on x², so a 1 % error in the displacement gives about a 2 % error in the energy. The stiffness mω² adds the same sensitivity to ω.
Where it is used
- Energy stored in a spring or elastic mount at a given position.
- Splitting the total energy into kinetic and potential parts.
Maths you need
- Angular frequency: ω = 2π times the frequency, in rad/s
- Amplitude: the largest displacement from equilibrium