Hooke’s Law & Spring Calculator
Spring force, spring constant, stored energy, combined springs and oscillation period.
Try before you buy.
- Free preview: the value each mode solves for in full, with the first rows of its table, a marked chart and the first step of the working; every other figure is hidden.
- Locked until you unlock it: download and copy.
- Unlock: Pro pass, ₹179 for 30 days or ₹79 for 1 week, a one-time payment that never renews. Unlock: Pro pass, $5.99 for 30 days or $2.99 for 1 week, a one-time payment that never renews.
Ways to unlock shows how to get the full result.
Printing this result is locked in the free preview.
Result
Force and extension
Along the way
How it was worked out
Locked in the free preview. Opens the ways to unlock this result.
Locked in the free preview. Batch runs unlock with a pass.
Locked in the free preview. Query results unlock with a pass.
About the Hooke’s Law & Spring Calculator
Hooke’s law says a spring pulls or pushes back with a force in proportion to how far it is stretched or squeezed: F = k·x, where k is the spring constant. This calculator works with it three ways:
- Hooke’s law: give two of the force, the spring constant and the extension, and get the third, the elastic energy stored (½·k·x²), a force–extension chart and the force and energy along the way.
- Springs together: the equivalent spring constant of up to eight springs in series or in parallel, and, with a load, how the force, the stretch and the energy are shared between them.
- A mass on a spring: the period, frequency and angular frequency, or the mass or spring constant that gives a period, with the largest speed, acceleration and energy for an amplitude, and a chart of the motion.
Every unit box works both ways: N, kN, lbf or kgf; N/m, N/mm or lbf/in; m, cm, mm or inches. With a Pro pass, copy the result or download it as a CSV, or unlock this one result. Without a pass the free preview shows the value each mode solves for in full, with the first rows of its table, a marked chart and the first step of the working; every other figure is hidden.
How to use it
- Choose what to calculate: Hooke’s law, springs together or a mass on a spring.
- For Hooke’s law or a mass on a spring, pick what to solve for and type the other values with their units; the unit box of the value solved for sets the unit of the answer.
- For springs together, choose series or parallel, enter each spring constant (add up to eight) and, if you like, the load on them.
- Read the answer, the table, the chart and the working. Copy the result or download the CSV with a Pro pass, or after unlocking this result.
Examples
k = 200 N/m, x = 5 cm
F = 10 N, and it stores ½ × 200 × 0.05² = 0.25 J; on Earth a mass of 1.02 kg stretches it this far
A 10 N force stretches the spring 2 cm
k = F ÷ x = 500 N/m (0.5 N/mm)
100 N/m and 300 N/m end to end, a 15 N load
1/k = 1/100 + 1/300, so k = 75 N/m; the load stretches them 15 cm and 5 cm, 20 cm in all
100 N/m and 300 N/m in parallel, a 20 N load
k = 400 N/m; both stretch 5 cm and carry 5 N and 15 N
Amplitude 5 cm
T = 2π·√(0.5 ÷ 200) = 0.3142 s (3.183 Hz, ω = 20 rad/s); fastest 1 m/s, largest acceleration 20 m/s², energy 0.25 J
Common uses
- Physics homework and lab reports on springs and simple harmonic motion
- Working out the spring constant from a measured stretch or a timed oscillation
- Choosing two springs in series or parallel to get a stiffness you need
- Finding how far a spring will stretch or how much energy it stores under a load
- Checking the bounce frequency of a mass on a spring, a toy or a simple suspension
The formulas
- Hooke’s law F = −k·x: the force is in proportion to the extension x and points back to the spring’s natural length (the minus sign); the calculator shows its size, F = k·x
- Elastic potential energy U = ½·k·x², zero at the natural length; the same as ½·F·x, the area under the force–extension line
- In series (end to end): 1/k = 1/k₁ + 1/k₂ + …; the same force runs through each spring and the extensions add up
- In parallel (side by side): k = k₁ + k₂ + …; every spring stretches the same and the forces add up
- A mass on a spring: ω = √(k ÷ m), T = 2π·√(m ÷ k), f = 1 ÷ T, the same for every amplitude
- The motion with an amplitude A: x(t) = A·cos(ω·t); the largest speed A·ω, the largest acceleration A·ω², the energy ½·k·A²
- Hung vertically, a mass rests lower by Δy = m·g ÷ k and oscillates around that point with the same period
Measuring a spring constant
Hang known masses from the spring and measure how far each one stretches it from where it hung with nothing on it. The force is the weight, m·g, so plot the force against the extension: the points lie on a straight line through zero, and its slope is k. With one mass, k = m·g ÷ x; with several, the slope of the best line averages out the measuring errors.
Or time the oscillation: hang a mass m, set it bouncing gently, time 10 or 20 full oscillations and divide to get the period T; then k = m·(2π ÷ T)². Choose “A mass on a spring” and solve for the spring constant.
Series or parallel?
Springs in series are softer than the softest one: two equal springs end to end stretch twice as far under the same load, so the pair has half the constant. Springs in parallel are stiffer than the stiffest: two equal springs side by side share the load and stretch half as far, so the pair has twice the constant. A car’s four suspension springs and the coils of a mattress share their load in parallel. A long, even spring cut in half gives two springs, each twice as stiff as the whole, because the halves in series make the original.
Sources
- OpenStax, University Physics Volume 1, §5.6 Common Forces: Hooke’s law F = −k·x (CC BY 4.0)
- OpenStax, University Physics Volume 1, §8.1 Potential Energy of a System: the elastic potential energy ½·k·x² (CC BY 4.0)
- OpenStax, University Physics Volume 1, §15.1 Simple Harmonic Motion: ω = √(k ÷ m), T = 2π·√(m ÷ k), f = 1 ÷ T and the vertical spring (CC BY 4.0)
- Series and parallel springs: the equivalent constants and how the force, extension and energy are shared
Limitations
- Hooke’s law holds only up to the spring’s elastic limit; stretched further it bends for good, and real springs may also be stiffer or softer at the ends of their travel.
- Springs are ideal: massless, with no friction or damping, so an oscillation never dies down here. A spring that is heavy compared with its load oscillates more slowly than this.
- Compression is treated like extension; a compressed coil spring stops when its coils touch.
- For designing springs for a product (wire size, coils, stress) use a spring design method; this is a physics calculator.
Privacy
On this device: nothing is sent
Everything happens in your browser. What you enter or open here is not uploaded or stored by MySmartCoPilot.
Frequently asked questions
What do I get without a pass?
Without a pass, Hooke’s Law & Spring Calculator shows the value each mode solves for in full, with the first rows of its table, a marked chart and the first step of the working; every other figure is hidden. Until you unlock it, the result can’t be downloaded or copied. A Pro, Premium or Ultimate pass, a one-time payment that never renews, unlocks the full result. The pricing page lists the passes and their prices.
How do I calculate the spring constant?
Divide the force by the extension it causes: k = F ÷ x. If 10 N stretches a spring 2 cm (0.02 m), k = 10 ÷ 0.02 = 500 N/m. Hanging a mass m gives F = m·g: 1 kg (9.807 N) stretching it 5 cm gives 196 N/m.
What are the units of the spring constant?
Force per length: newtons per metre (N/m) in SI. Spring makers often use N/mm (1 N/mm = 1,000 N/m) or pounds-force per inch (1 lbf/in = 175.1 N/m).
How much energy does a stretched spring store?
U = ½·k·x². A 200 N/m spring stretched 5 cm stores ½ × 200 × 0.05² = 0.25 J. Doubling the stretch quadruples the energy.
How do I combine springs in series and in parallel?
In series add the reciprocals: 1/k = 1/k₁ + 1/k₂ (100 and 300 N/m give 75 N/m). In parallel add the constants: k = k₁ + k₂ (100 and 300 N/m give 400 N/m).
What is the period of a mass on a spring?
T = 2π·√(m ÷ k). A 0.5 kg mass on a 200 N/m spring has T = 2π × √0.0025 = 0.3142 s, about 3.18 oscillations a second. It does not depend on how far you pull it, and four times the mass doubles the period.
Why is there a minus sign in F = −kx?
It gives the direction: the spring’s force always points back towards its natural length, opposite to the stretch or squeeze. The size of the force is k·x, which is what this calculator shows.