Hooke's Law Calculator
Enter any two of force, spring constant and extension to get the third, along with the elastic energy stored in the spring. Switch modes to combine springs in series or parallel and see how a load divides among them, or to find the period of a mass bouncing on a spring. The working is written out with your numbers. Free, no sign-up.
Positive to stretch, negative to compress. Measured from the spring's natural length.
Results
10.00 N
Force applied
0.05000 m
Extension x
0.2500 J
Elastic energy U = ½kx²
200.0 N/m
Spring constant k
-10.00 N
Restoring force −F
Stretched
The spring is
The working
F = k x
F = 200 × 0.05 = 10.00 N
U = ½ k x²
U = ½ × 200 × (0.05)² = 0.2500 J
Assumes an ideal, massless spring within its elastic limit, where force is exactly proportional to extension. Beyond the limit a real spring stays deformed and the law fails. For a spring hung vertically the period is unchanged; only the equilibrium position shifts.
How to Use the Hooke's Law Calculator
- 1. Pick the mode. Spring force solves F = kx for any one of its terms. Springs in series or parallel finds the equivalent spring constant of up to six springs. Mass on a spring gives the oscillation period, frequency and, if you add an amplitude, the greatest speed and energy.
- 2. Choose the unknown and fill in the rest. The box for the unknown turns into the answer. Each box has a unit menu, from N/m to lbf/in and from millimetres to feet; scientific notation such as 2.5e3 works. The Start from an example menu loads a worked case.
- 3. Mind the sign. A positive extension is a stretch and a negative one a compression. The force you apply has the same sign; the spring pushes back with the opposite force, −F.
- 4. Read the answer and the working. You get the energy stored, a graph that shows why the energy is the area under the line, and the equations with your numbers in SI units.
Hooke's Law and Elastic Energy
Hooke's law
F = k x
Elastic energy
U = ½ k x² = ½ F x
Series
1/kₛ = 1/k₁ + 1/k₂ + …
Parallel
kₚ = k₁ + k₂ + …
Mass on a spring
T = 2π √(m / k)
Robert Hooke published the law in 1678: the force needed to extend or compress a spring by some distance is proportional to that distance. The constant of proportionality, k, is the spring constant, in newtons per metre. A stiff spring has a large k. The spring pulls back with the restoring force −kx, which is why the law is often written with a minus sign; this calculator reports the force you apply, and shows −F separately. OpenStax University Physics covers springs and oscillations in section 15.1.
Because force grows steadily with extension, the work done in stretching a spring is the area of the triangle under the force–extension line, ½ × x × kx = ½kx². Doubling the stretch therefore needs four times the energy. That energy is returned when the spring is released.
Worked Examples
- Force and energy. A 200 N/m spring stretched 5 cm needs F = 200 × 0.05 = 10 N and stores U = ½ × 200 × 0.05² = 0.25 J.
- Compression. A 5 kN/m spring compressed by 2 cm (x = −0.02 m) gives F = −100 N and U = 1 J: the negative sign marks a push rather than a pull.
- Finding k. A 1 kg mass hanging from a spring stretches it 4.9 cm. The weight is 9.80665 N, so k = 9.80665 / 0.049 = 200.1 N/m.
- Two 100 N/m springs, 10 N load. In series, kₛ = 50 N/m and the combination stretches 20 cm, each spring carrying the full 10 N. In parallel, kₚ = 200 N/m and it stretches 5 cm, each spring carrying 5 N.
- A mass on a spring. A 0.25 kg mass on a 100 N/m spring has ω = √(k/m) = 20 rad/s, so T = 2π/ω = 0.3142 s and f = 3.183 Hz. With a 5 cm amplitude the greatest speed is Aω = 1 m/s and the greatest acceleration is Aω² = 20 m/s².
Series makes it softer; parallel makes it stiffer
Two springs in series stretch by the sum of their separate extensions, so the pair is softer than either. Two in parallel share the load but stretch together, so the pair is stiffer than either. For equal springs the combination is k/2 in series and 2k in parallel. This is the reverse of the rule for resistors, where series resistance adds.
Assumptions and Limits
- Elastic limit. Hooke's law holds only up to the limit of proportionality. Past it a spring stays deformed or breaks, and the law fails. Rubber bands and many biological tissues are non-linear well before that point.
- Massless spring. The mass-on-a-spring period ignores the spring's own mass. For a heavy spring, about one third of its mass adds to the oscillating mass.
- Vertical springs. Hanging the mass vertically shifts the equilibrium position by mg/k but does not change the period.
- No damping. Friction and air resistance shrink the amplitude over time and lengthen the period slightly.
Frequently Asked Questions
- What is Hooke's law?
- Hooke's law states that the force needed to stretch or compress a spring is proportional to the extension: F = kx. The constant k, in N/m, measures the spring's stiffness. It holds only within the spring's elastic limit.
- How do you calculate the spring constant?
- Divide the force by the extension: k = F / x. For a 1 kg mass (weight 9.80665 N) that stretches a spring by 4.9 cm, k = 9.80665 / 0.049 ≈ 200 N/m. Using several loads and fitting a line through a force–extension graph gives a more reliable value than a single measurement.
- What are the units of the spring constant?
- The SI unit is newtons per metre (N/m), equivalent to kg/s². Other common units are N/cm, N/mm, kN/m and pounds-force per inch (lbf/in), where 1 lbf/in = 175.13 N/m. The calculator converts between them.
- How do springs add in series and in parallel?
- In parallel the spring constants add: kₚ = k₁ + k₂ + …. In series it is the reciprocals that add: 1/kₛ = 1/k₁ + 1/k₂ + …. So a pair of identical springs of constant k gives k/2 in series and 2k in parallel.
- How much energy is stored in a stretched spring?
- U = ½kx², which is the area under the force–extension line. A 200 N/m spring stretched 5 cm stores 0.25 J, and stretching it 10 cm stores four times as much, 1 J.
- Does the period of a mass on a spring depend on gravity or amplitude?
- Neither. T = 2π√(m/k) depends only on the mass and the spring constant. The same mass-spring system oscillates at the same rate on Earth, on the Moon, or horizontally on a frictionless track, and for small or large amplitudes within the elastic limit.