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Hooke’s Law & Spring Calculator

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Understanding the Hooke's law and spring calculator — concept and principle

Hooke's law describes how an ideal spring pushes back in proportion to how far it is stretched or compressed. This calculator gives the spring force, the elastic energy stored, and the period and frequency with which an attached mass would bounce. It is used for physics labs, choosing springs for mechanisms, suspension and vibration estimates, and spring scales.

Spring stretched 5 cm by 10 NWith k = 200 N/m, a 5 cm stretch needs 10 N and stores ½ × 10 × 0.05 = 0.25 J. A 0.5 kg mass on it bobs with a period of 0.314 s, 3.18 times a second.Stretch it twice as far, pull twice as hardrelaxed5 cm10 Nforce F = k × x200 × 0.05 = 10 Nx 0.05 m10 Nstored energy½ × 10 × 0.05= 0.25 J0.50.5 kg bobbing on it:T = 2π√(0.5 ÷ 200) = 0.314 s≈ 3.18 bounces per second
Force grows in step with stretch; the stored energy is the triangle under it.

How it is calculated

The principle

For small deformations, the atomic bonds in a material behave like tiny springs, so the restoring force is proportional to displacement; Robert Hooke summed this up in 1678 as "ut tensio, sic vis" (as the extension, so the force). The stored energy is the area under the force–extension line, a triangle with height kx and base x, hence ½kx². A mass on such a spring feels a force that always points back to equilibrium and grows with distance, which produces simple harmonic motion. A stiffer spring pulls back harder and speeds the motion up; a larger mass has more inertia and slows it down, which is why T depends on √(m ÷ k) and not on the amplitude.

Worked example

  1. Inputs: k = 200 N/m, x = 0.05 m, m = 0.5 kg.
  2. Force: 200 × 0.05 = 10 N.
  3. Energy: ½ × 200 × 0.05² = 0.25 J.
  4. Period: 2π × √(0.5 ÷ 200) = 2π × 0.05 ≈ 0.31416 s.
  5. Frequency: 1 ÷ 0.31416 ≈ 3.1831 Hz.
How Force (N) changes with Spring constant k (N/m)How Force changes with Spring constant k0510152050100150200250300400Example: 10Spring constant k (N/m)
Each point is the calculator re-run with only “Spring constant k (N/m)” changed; the gold point is the default example. Change the inputs above and this chart follows your numbers.
Spring constant k (N/m)Force (N)
502.5
1005
1507.5
20010
25012.5
30015
40020

How to read the result

A k of 200 N/m means each centimetre of stretch adds 2 N. The period does not change with how far you pull the mass, as long as the spring stays in its linear range. If you hang the mass vertically, it simply sits lower by mg ÷ k (here about 2.45 cm) and oscillates about that point with the same period. To halve the period, you need four times the stiffness or a quarter of the mass.

Common mistakes

Keep in mind: Real springs are linear only within a limited range and have their own mass and damping. Spring selection for safety-related equipment should follow the manufacturer's ratings and qualified engineering advice.

Frequently asked questions

What units does the spring constant have?

Newtons per metre (N/m). Catalogues often use N/mm; 1 N/mm = 1,000 N/m.

How do I measure k?

Hang a known mass, measure the extension, and divide the weight mg by that extension.

Does gravity change the period of a vertical spring?

No. Gravity shifts the equilibrium position, but the period is still 2π√(m ÷ k).

Sources: Robert Hooke, Lectures de Potentia Restitutiva, or of Spring, London, 1678 · Halliday, Resnick and Walker, Fundamentals of Physics, Wiley

Checked 8 Nov 2026. Standard textbook formulas in SI units, computed in your browser. For learning and estimates — real designs must follow your local codes and be checked by a qualified engineer.

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