Spring Constant Calculator with Steps | k = F/x, Any Units

A spring constant calculator finds the stiffness k of a spring from Hooke’s law: divide the force by the extension, k = F ÷ x. A force of 2.00 N that stretches a spring by 4.00 cm (0.0400 m) gives k = 2.00 ÷ 0.0400 = 50.0 N/m.

Rule of thumb: always use the change in length, in metres. For a hanging mass, the force is its weight, so k = m g ÷ x.

A spring constant calculator works out how stiff a spring is. The spring constant (k), also called the force constant or spring rate, is the force needed to stretch or compress a spring by one metre. It is measured in newtons per metre (N/m). A soft spring in a ballpoint pen might be a few hundred N/m; a car suspension spring is tens of thousands of N/m.

To calculate the spring constant, use Hooke’s law, F = k x, where F is the force on the spring and x is the extension or compression, the change in length from its natural length. Rearranging gives k = F ÷ x. If a mass m hangs at rest on the spring, the force is its weight, so k = m g ÷ x. If you know how fast it bounces instead, use k = 4π²m ÷ T².

A worked one-liner: a force of 2.00 N stretches a spring by 4.00 cm. Converting to metres, x = 0.0400 m, so k = 2.00 N ÷ 0.0400 m = 50.0 N/m. On a graph of force against extension, this is the gradient of the straight line.

Who uses it: students in CBSE and ICSE Class 11, IGCSE and GCSE (the required practical on springs), A level, AP Physics 1, JEE and NEET, and first-year engineering. Engineers use spring rates to design car suspensions, valve springs, mattresses, weighing scales and vibration mounts. Robert Hooke first published the law in 1678 as the Latin anagram ut tensio, sic vis, “as the extension, so the force”.

Pick what you know, choose what to solve for, then type your values in any units.

4.1 What k means

The spring constant is the stiffness of a spring. A spring with k = 50.0 N/m needs 50.0 N to stretch it 1 m, 0.500 N to stretch it 1 cm, and 2.00 N to stretch it 4.00 cm. A bigger k means a stiffer spring that stretches less for the same force. The unit N/m is the same as kg/s².

4.2 Extension, not length

The most common mistake is dividing by the total length of the spring. Hooke’s law uses x, the change in length from the natural (unloaded) length. A spring that is 12.0 cm long unloaded and 15.5 cm long with 1.40 N on it has x = 3.5 cm, so k = 1.40 ÷ 0.035 = 40.0 N/m.

4.3 Is it a sensible size?

School lab springs are usually 5 to 100 N/m. A retractable pen spring is around 100 to 500 N/m, a mattress spring a few thousand N/m, and a car suspension spring 20 000 to 100 000 N/m (20 to 100 N/mm). If you get 0.4 N/m or 4000 N/m for a lab spring, check whether the extension was in cm or mm instead of metres.

4.4 Three independent ways to measure k

You can find k statically (hang a mass and measure the stretch), dynamically (time the oscillations and use k = 4π²m ÷ T²) or from a graph of several readings. The two methods should agree within a few percent. If the dynamic value is lower, the spring’s own mass is the usual reason: about one third of it joins the oscillating mass.

4.5 Combining springs

Springs combine the opposite way to resistors. Side by side (parallel), they share the load and k = k₁ + k₂. End to end (series), each carries the full force and stretches by F ÷ kᵢ, so 1/k = 1/k₁ + 1/k₂. Cutting a spring in half doubles its spring constant, because each half is one of two equal springs in series.

4.6 What the charts show

The force–extension chart is the heart of Hooke’s law: a straight line through the origin whose gradient is k. The energy chart is a parabola, because E = ½ k x². For oscillations, a graph of T² against m is a straight line of gradient 4π² ÷ k. The point marked on each chart is your spring.

4.7 Where Hooke’s law breaks down

Hooke’s law holds only up to the limit of proportionality. Past the elastic limit, the spring is permanently deformed and does not return to its original length. Tightly wound extension springs also need a small initial tension before they start to stretch, so their force–extension line does not pass through the origin; use the gradient, not one reading.

4.8 Where this leads next

The spring constant sets the natural frequency of anything that vibrates: a car on its suspension, a building in an earthquake, atoms in a crystal and the quartz in a watch. The same idea, stiffness, appears as Young’s modulus for materials and as the torsion constant for twisting springs.

▶ Press Calculate above to fill all five write-ups with your own numbers: the full GRESS working, every unit, an uncertainty, a mark scheme and a plain-text version.

This flowchart shows how to calculate the spring constant step by step, from a word problem to a checked answer. Every box uses the same worked example as the Excel, R and Python sections: a force of 2.00 N stretches a spring by 4.00 cm. Find the spring constant.

Check any box yourself: the Excel sheet in Section 9, the R script in Section 10 and the Python script in Section 11 all return 50.0 N/m for this spring. Section 14 shows the method written out by hand.

1. Choose what you know. Pick force and extension, a hanging mass, the oscillation period, stored energy, springs in series or parallel, coil dimensions, or the Graph tab for lab data.
2. Choose what to solve for. Click a Solve for chip, for example the spring constant, the force, the extension, the mass or the period. That field turns into a dashed answer box.
3. Type the known values. Plain numbers, decimals and scientific notation all work: 2.00, 0.0400 or 4e-2.
4. Pick the unit beside each value. Grams, pounds, mm, cm, inches, lbf, N/mm, lbf/in, GPa and psi are converted for you.
5. Use the change in length. Enter the extension or compression, or use the two-lengths option and let the calculator subtract.
6. Set g for a hanging mass. Use 9.81 for IGCSE, A level and most US courses, 9.8 for CBSE and AP, or 10 for quick estimates.
7. Press Calculate. Nothing is worked out until you do, and changing any input clears the old answer so you never read stale numbers.
8. Read the answer box and the badges. They show k with its units, other units, Units check ✓ and Verified.
9. Follow the working, the diagram and the charts. The GRESS table, numbered steps, results table, diagram and four charts explain the answer.
10. Copy or download the working. Section 5 gives five ready-to-paste write-ups, and Download Doc or Download PDF saves the whole solution.

Excel has no spring constant function, and it does not need one. k = F ÷ x is one division, so a sheet with one spring per row handles a whole class set. The two habits that matter are converting every extension to metres in its own column, and using SLOPE on a set of readings rather than a single division.

9.1 The functions you need

9.2 Step by step

9.3 The awkward cases

• Initial tension. A closely wound extension spring needs some force before it stretches at all. Its line does not pass through zero, so use SLOPE, not a single F ÷ x.
• Two lengths. If you recorded lengths, not extensions, make a column =C2-$C$2 to subtract the unloaded length from every row.
• Units of spring rate. Divide N/m by 1000 for N/mm, or use =CONVERT(E2,"N","lbf")*0.0254 for lbf/in.
• Precision: Excel keeps about 15 significant digits. Round only in the final display with =ROUND(E2,1) or the number format.

9.4 Why Excel might disagree with R, Python or this page

For force-and-extension problems g does not appear, so every tool gives exactly the same k. When a mass is involved, a different last digit is usually a different g. A graph gradient also depends on whether you force the line through zero: LINEST(C2:C7,D2:D7,FALSE) gives 40.0 N/m for the sample data, and the free line gives 40.0 N/m too, because the intercept is tiny.

9.5 Charting it

1. Put the extension (m) in one column and the force (N) in the next.
2. Insert → Charts → Scatter. Do not use the Line chart type, which spaces x values evenly.
3. Click a point → Add Trendline → Linear → tick Display Equation on chart and Display R-squared.
4. The number in front of x in the equation is k. An intercept far from zero points to a zero error or initial tension.
5. Label the axes with units: Extension x (m) and Force F (N).

9.6 Errors you will hit

R works out the spring constant in one line, k <- F / x. The script below does more: it finds k from a hanging mass, the period, stored energy, series and parallel springs and coil dimensions, fits a straight line to lab readings with lm() for the gradient and its standard error, runs a substitution check, and saves one 300 dpi figure of force against extension. It uses base R only.

10.1 The complete script

R, base only, saves spring_constant_plot.png

# ============================================================
# Spring constant calculator in R, with one 300 dpi figure
# StatsUnlock - https://statsunlock.com/spring-constant-calculator/
# Base R only. No packages to install.
# ============================================================

# ---- 1. Your values (SI units) -----------------------------
F <- 2.00          # force on the spring in N
x <- 4.00 / 100    # extension in m (4.00 cm)
g <- 9.81          # m/s^2; use 9.80665, 9.8 or 10 to match your book

# ---- 2. Hooke's law: k = F / x -----------------------------
k <- F / x                                   # N/m

# ---- 3. Other ways to find k -------------------------------
k_mass   <- 0.200 * g / (4.90 / 100)         # hanging mass 0.200 kg, stretch 4.90 cm
k_period <- 4 * pi^2 * 0.500 / 0.800^2       # 0.500 kg bouncing with T = 0.800 s
k_energy <- 2 * 0.90 / (6.0 / 100)^2         # 0.90 J stored at 6.0 cm
k_series <- 1 / (1 / 200 + 1 / 300)          # 200 N/m and 300 N/m end to end
k_par    <- 200 + 300                        # the same springs side by side
k_coil   <- 79.3e9 * (2.00e-3)^4 / (8 * (20.0e-3)^3 * 10)   # G d^4 / (8 D^3 n)

# ---- 4. From a graph: least-squares gradient ---------------
ext  <- c(0, 2.4, 5.0, 7.3, 9.9, 12.2) / 100        # extension, m
load <- c(0, 100, 200, 300, 400, 500) / 1000 * g    # weight of the masses, N
fit  <- lm(load ~ ext)
k_graph <- unname(coef(fit)[2])
se      <- summary(fit)$coefficients[2, 2]

# ---- 5. Print the answers with units -----------------------
cat(sprintf("Hooke's law k     = %.1f N/m\n", k))
cat(sprintf("From hanging mass = %.1f N/m\n", k_mass))
cat(sprintf("From the period   = %.3g N/m\n", k_period))
cat(sprintf("From the energy   = %.0f N/m\n", k_energy))
cat(sprintf("Series / parallel = %.3g / %.3g N/m\n", k_series, k_par))
cat(sprintf("Coil spring rate  = %.3g N/mm\n", k_coil / 1000))
cat(sprintf("Graph gradient    = %.1f +/- %.2f N/m\n", k_graph, se))
stopifnot(abs(k * x - F) < 1e-12)             # substitution check
cat("Units: N / m = kg/s^2 = N/m\n")

# ---- 6. One figure: force against extension ----------------
png("spring_constant_plot.png", width = 7, height = 4.5, units = "in", res = 300)
plot(ext * 100, load, pch = 19, col = "#2563eb", cex = 1.3,
     xlab = "Extension x (cm)", ylab = "Force F (N)",
     main = sprintf("Gradient k = %.1f N/m", k_graph))
abline(coef(fit)[1], coef(fit)[2] / 100, lwd = 3, col = "#9333ea")
legend("topleft", c("Measured", "Best-fit line"), col = c("#2563eb", "#9333ea"),
       pch = c(19, NA), lwd = c(NA, 3), bty = "n")
invisible(dev.off())

Output when run: Hooke's law k = 50.0 N/m, From hanging mass = 40.0 N/m, From the period = 30.8 N/m, From the energy = 500 N/m, Series / parallel = 120 / 500 N/m, Coil spring rate = 1.98 N/mm, Graph gradient = 40.0 +/- 0.32 N/m. These match Sections 3, 7, 9 and 11.

10.2 What each part does

R traps: lm(ext ~ load) swaps the axes and returns 1/k; -2^2 is −4 in R; F is also R’s shortcut for FALSE, so the script overwrites it on purpose and you should write FALSE in full afterwards.

10.3 What the figure shows

The blue points are the six readings of force against extension, and the purple line is the least-squares fit. The points lie almost exactly on a straight line through the origin, which is Hooke’s law. The gradient, 40.0 N/m, is the spring constant. If the top points bent away from the line, the spring would be past its limit of proportionality and those points should be left out of the fit.

10.4 Common problems and fixes

10.5 Useful one-liners

# k for several springs at once (force in N, extension in cm)
c(2.00, 3.0, 5.0) / (c(4.00, 0.80, 2.5) / 100)
# k from 20 oscillations of 0.250 kg in 15.8 s
4 * pi^2 * 0.250 / (15.8 / 20)^2
# line forced through the origin
coef(lm(load ~ 0 + ext))
# 95% confidence interval for k
confint(fit)[2, ]

Python does the same job with NumPy and Matplotlib. The script finds k from force and extension, a hanging mass, the period, stored energy, series and parallel springs and coil dimensions, fits the lab readings with np.polyfit, prints each answer with its unit, and saves one 300 dpi figure of force against extension.

Install once with pip install numpy matplotlib. Google Colab already has both.

11.1 The complete script

Python 3 with NumPy, saves spring_constant_plot.png

# ============================================================
# Spring constant calculator in Python, with one 300 dpi figure
# StatsUnlock - https://statsunlock.com/spring-constant-calculator/
# pip install numpy matplotlib
# ============================================================
import numpy as np
import matplotlib.pyplot as plt

# ---- 1. Your values (SI units) -----------------------------
F = 2.00            # force on the spring, N
x = 4.00 / 100      # extension, m (4.00 cm)
g = 9.81            # m/s^2 (scipy.constants.g would give 9.80665)

# ---- 2. Hooke's law: k = F / x -----------------------------
k = F / x           # N/m

# ---- 3. Other ways to find k -------------------------------
k_mass = 0.200 * g / (4.90 / 100)
k_period = 4 * np.pi**2 * 0.500 / 0.800**2
k_energy = 2 * 0.90 / (6.0 / 100)**2
k_series = 1 / (1 / 200 + 1 / 300)
k_par = 200 + 300
k_coil = 79.3e9 * (2.00e-3)**4 / (8 * (20.0e-3)**3 * 10)

# ---- 4. From a graph: least-squares gradient ---------------
ext = np.array([0, 2.4, 5.0, 7.3, 9.9, 12.2]) / 100
load = np.array([0, 100, 200, 300, 400, 500]) / 1000 * g
k_graph, c = np.polyfit(ext, load, 1)                 # gradient and intercept
res = load - (k_graph * ext + c)
se = np.sqrt((res**2).sum() / (len(ext) - 2) / ((ext - ext.mean())**2).sum())   # standard error of the gradient

# ---- 5. Print the answers with units -----------------------
print(f"Hooke's law k     = {k:.1f} N/m")
print(f"From hanging mass = {k_mass:.1f} N/m")
print(f"From the period   = {k_period:.3g} N/m")
print(f"From the energy   = {k_energy:.0f} N/m")
print(f"Series / parallel = {k_series:.3g} / {k_par:.3g} N/m")
print(f"Coil spring rate  = {k_coil / 1000:.3g} N/mm")
print(f"Graph gradient    = {k_graph:.1f} +/- {se:.2f} N/m")
assert abs(k * x - F) < 1e-12          # substitution check

# ---- 6. One figure: force against extension ----------------
xs = np.linspace(0, ext.max(), 50)
fig, ax = plt.subplots(figsize=(7, 4.5))
ax.plot(ext * 100, load, "o", color="#2563eb", ms=7, label="Measured")
ax.plot(xs * 100, k_graph * xs + c, lw=3, color="#9333ea", label="Best-fit line")
ax.set_xlabel("Extension x (cm)")
ax.set_ylabel("Force F (N)")
ax.set_title(f"Gradient k = {k_graph:.1f} N/m")
ax.legend(frameon=False)
fig.savefig("spring_constant_plot.png", dpi=300, bbox_inches="tight")

Output when run: Hooke's law k = 50.0 N/m, From hanging mass = 40.0 N/m, From the period = 30.8 N/m, From the energy = 500 N/m, Series / parallel = 120 / 500 N/m, Coil spring rate = 1.98 N/mm, Graph gradient = 40.0 +/- 0.32 N/m.

11.2 What each part does

Python traps: ^ is not a power; f"{50.0:.3g}" prints 50, dropping the significant zero; np.polyfit(load, ext, 1) swaps the axes and returns 1/k; scipy.constants.g is 9.80665.

11.3 What the figure shows

The same force–extension graph as in R: six readings on a straight line through the origin, with a gradient of 40.0 N/m. Change the two arrays at the top of the script to your own readings and the title updates with your spring constant.

11.4 Common problems and fixes

11.5 Useful one-liners

# k for several springs at once (force in N, extension in cm)
import numpy as np; np.array([2.00, 3.0, 5.0]) / (np.array([4.00, 0.80, 2.5]) / 100)
# k from 20 oscillations of 0.250 kg in 15.8 s
4 * np.pi**2 * 0.250 / (15.8 / 20)**2
# line forced through the origin
(ext @ load) / (ext @ ext)
# three springs in series
1 / sum(1 / np.array([200, 300, 600]))

Run it in a terminal with python sk.py, in a Jupyter notebook cell, or in Google Colab without installing anything.

12.1 Spring constant formulas at a glance

12.2 Typical spring constants

Typical published ranges; real springs vary widely. Measure your own spring when accuracy matters.

12.3 Spring wire materials

Standard spring-design values for wire about 1 to 3 mm in diameter. G is used for compression and extension springs; E is used for torsion springs.

12.4 Spring rate unit conversions

12.5 Constants

Source: NIST, CODATA 2022 recommended values (Mohr et al., 2025, reference 1).

This is the written method examiners expect, worked on the same example as Sections 7, 9, 10 and 11: a force of 2.00 N stretches a spring by 4.00 cm. Find the spring constant, then the force needed to stretch it by 7.00 cm.

14.1 The written method (GRESS)

1. Given. F = 2.00 N, x = 4.00 cm = 0.0400 m.
2. Required. Spring constant k (N/m), then the force for x = 7.00 cm.
3. Equation. Hooke’s law, F = k x, so k = F ÷ x.
4. Substitute, with units. k = 2.00 N ÷ 0.0400 m. Then F = k × 0.0700 m.
5. Solve. k = 50.0 N/m. F = 50.0 × 0.0700 = 3.50 N.
6. Check the units and the size. N ÷ m = N/m. A longer stretch needs a bigger force, in proportion: 7.00 ÷ 4.00 × 2.00 N = 3.50 N.
7. Write the answer with units. k = 50.0 N/m; F = 3.50 N.

14.2 On a scientific calculator

The statistics mode on most scientific calculators does linear regression: enter extension as x and force as y, and the coefficient B is the spring constant. Make sure the extensions are in metres first.

14.3 Measure it yourself: a school lab practical

Sample class data with g = 9.81 m/s². The single F ÷ x values scatter from 39.2 to 40.9 N/m; the gradient of the graph, 40.0 ± 0.3 N/m, is more reliable. These are the data preloaded in the Graph tab.

14.4 How to calculate the uncertainty of the spring constant

For a single reading, add the percentage uncertainties: if F = 2.00 ± 0.02 N (1%) and x = 4.00 ± 0.05 cm (1.25%), then k = 50.0 N/m ± 2.25%, so 50.0 ± 1.1 N/m. For a graph, draw the steepest and shallowest lines that still pass through every error bar; half the difference of their gradients is the uncertainty in k. The Graph tab gives the statistical version, the standard error of the gradient, directly.

14.5 Practice problems with answers

14.6 Mistakes that lose marks

• Dividing by the total length of the spring instead of the extension.
• Leaving the extension in centimetres or millimetres.
• Dividing the mass by the extension: you need the weight, m g.
• Forgetting to square the period, or using 2π instead of 4π² after rearranging.
• Adding spring constants for springs in series.
• Swapping the axes on the graph, so the gradient is 1/k.
• Giving k the unit N or N·m instead of N/m.

Hooke’s law and the spring constant describe springs very well inside a clear range.

Use them when:

• ✓ The force–extension graph is a straight line through the origin (or with a small offset for initial tension).
• ✓ The spring returns to its original length when the load is removed.
• ✓ The spring is light compared with the mass on it, for the period method.
• ✓ The coils do not touch each other (for compression springs, they are not fully squashed).

Use something else when:

• ✗ The spring is stretched past its limit of proportionality: the graph curves and k changes with the load.
• ✗ The spring is heavy compared with the load: add about one third of the spring’s mass to m in the period formula.
• ✗ The material is rubber or a bungee cord: its force–extension curve is not straight and shows hysteresis.
• ✗ It is a constant-force spring (a coiled flat strip, as in a tape measure): its force hardly changes with extension, so it has no single k. Manufacturers give its rated force instead.
• ✗ It is a conical or variable-pitch spring: k rises as coils bottom out, so use the manufacturer’s rate curve.

Real-world examples:

1. Car suspension: the spring rate and the car’s mass set the ride frequency, f = (1/2π)√(k/m), usually about 1 to 2 Hz.
2. Weighing scales and newton meters: the stretch is proportional to the load, so the scale can be marked evenly.
3. Mattresses and seats: hundreds of springs in parallel share the load.
4. Engine valves and door closers: a stiff compression spring returns the part quickly.
5. Torsion springs in clothes pegs and mousetraps: the torque grows with the angle wound.
6. Atoms and molecules: chemical bonds act like tiny springs, and their k sets the frequencies seen in infrared spectra.

Assumptions

1. Linear (Hookean) spring. Check: the force–extension readings lie on a straight line. Past the limit of proportionality k = F ÷ x underestimates how much the spring stretches.
2. Massless spring. Check: the spring is much lighter than the load. A heavy spring makes the period method underestimate k unless about a third of its mass is added to m.
3. Static readings at rest. Check: the mass has stopped bouncing before you read the extension.
4. Small, undamped oscillations. Check: the amplitude dies away slowly. Strong damping lengthens the period slightly.
5. Ideal series and parallel springs. In parallel the bar stays level and the load is centred; in series the joints are light.
6. Round wire, uniform coils (dimensions mode). Only active coils count; ground ends are usually inactive. Real rates are typically within about 10% of the formula.
7. Uniform g. g is taken as constant. Over the surface of the Earth it varies from 9.78 to 9.83 m/s², an effect of about 0.5%.

Limitations

• It does not model non-linear springs such as rubber bands, conical springs or progressive-rate car springs.
• It does not model constant-force springs, whose force is set by the strip thickness and coil radius rather than a k.
• It does not include damping or forced vibration.
• The dimensions mode gives the rate, not the stress; check stresses with a spring design standard before building a real spring.
• Typical values in Section 12 are guides only; always measure or check the maker’s rating.

A spring constant calculator rests on one short law, F = k x, yet spring questions still lose marks every year. The division is easy. The hard parts are using the extension rather than the length, converting it to metres, and using the weight rather than the mass.

Three habits fix nearly every error. First, find x as the change in length and convert it to metres before you do anything else. Second, choose the right route to k: F ÷ x for a force, m g ÷ x for a hanging mass, 4π²m ÷ T² for a timed bounce, 2E ÷ x² for stored energy. Third, for real data, plot force against extension and take the gradient instead of trusting one reading.

For several springs, remember that they combine the opposite way to resistors: add k in parallel, add 1/k in series. For spring design, the rate depends on the fourth power of the wire diameter, so small changes in wire size make big changes in stiffness.

Pick your method, type your values in any units, and let this spring constant calculator show every step, a units check and a diagram drawn from your numbers.

These peer-reviewed articles support Hooke’s law, the period method and the spring’s effective mass, springs in series, graph fitting, the lab methods and the common mistakes covered by this spring constant calculator. Every DOI was checked in October 2026.

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2. Moyer, A. E. (1977). Robert Hooke’s ambiguous presentation of “Hooke’s law”. Isis, 68(2), 266–275. https://doi.org/10.1086/351771
3. Fox, J. G., & Mahanty, J. (1970). The effective mass of an oscillating spring. American Journal of Physics, 38(1), 98–100. https://doi.org/10.1119/1.1976240
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7. Gash, P. (2020). A Slinky’s elastic potential energy. The Physics Teacher, 58(3), 198–199. https://doi.org/10.1119/1.5145416
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9. Wulandari, S., Iswanto, B. H., & Sugihartono, I. (2021). Determination of springs constant by Hooke’s law and simple harmonic motion experiment. Journal of Physics: Conference Series, 2019(1), 012053. https://doi.org/10.1088/1742-6596/2019/1/012053
10. Castro-Palacio, J. C., Velázquez-Abad, L., Giménez, M. H., & Monsoriu, J. A. (2013). Using a mobile phone acceleration sensor in physics experiments on free and damped harmonic oscillations. American Journal of Physics, 81(6), 472–475. https://doi.org/10.1119/1.4793438
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13. York, D. (1966). Least-squares fitting of a straight line. Canadian Journal of Physics, 44(5), 1079–1086. https://doi.org/10.1139/p66-090
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