Physics
Mass–spring period
Find period, mass or spring constant for an ideal oscillator.
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Number formats & units
Use decimal dots in datasets. Single values also accept decimal commas and scientific notation such as 1.2e-3 or 1.2 × 10⁻³.
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Result & working
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Interpretation & assumptions
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Formula & example
The relationship
What the variables mean
| Symbol | Meaning |
|---|---|
| Oscillation period in seconds. | |
| Oscillating mass in kilograms. | |
| Spring constant in N m⁻¹. |
Worked example
- Solve for
- Period T
- Mass m (kg)
- 0.250
- Spring constant k (N/m)
- 10
0.9934588 s
Follow the example steps
- T = 2π√(m/k)
- = 2π√(0.25 / 10) = 0.9934588 s
Common mistakes
Use kilograms in the SI mass–spring equation.
Incorrect approach
m = 100 g, k = 10 N/m: T = 2π√(100/10)
Correct approach
m = 0.1 kg: T = 2π√(0.1/10) ≈ 0.6283 s
More checks before you submit
- Use kilograms, not grams, in the SI form.
- The static extension mg/k is not the oscillation amplitude.
- Do not assume Hooke’s law remains valid beyond the spring’s linear range.
Method notes
Use this for an ideal horizontal mass–spring system, or for vertical oscillations about equilibrium when the spring remains in its linear range.
For real springs, the spring’s own effective mass, damping and non-linear behaviour can shift the period. Measure several oscillations for a more precise experimental period.