In short

A spring exerts a restoring force F = -kx, where x is the displacement from the natural length and k is the spring constant (measured in N/m). Stiffer springs have larger k. When springs are connected in parallel, the effective spring constant is k_{\text{eff}} = k_1 + k_2. When they are in series, \frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2}. A block on a spring settles at equilibrium where the spring force balances gravity, and oscillates about that point if disturbed.

Click the top of a ballpoint pen. Feel the spring resist your thumb, then snap back the moment you let go. That small coil inside the pen is doing something precise: the harder you push, the harder it pushes back. Push twice as far, it pushes back twice as hard. This proportionality — force scales linearly with displacement — is Hooke's law, and it governs everything from the suspension of a car bouncing over a speed breaker to the needle of a bathroom weighing scale settling on your weight.

The remarkable thing is not that springs push back. Of course they do — you can feel that. The remarkable thing is that the relationship is linear. Double the stretch, double the force. Triple the stretch, triple the force. This clean proportionality is rare in nature, and it makes springs one of the most useful objects in physics.

The restoring force — why springs always fight back

Take a spring and lay it on a smooth table. Its natural length — the length when nothing is pulling or pushing it — is the reference point. Call any displacement from this natural length x. Positive x means stretched; negative x means compressed.

Now pull the spring by a distance x. The spring pulls back toward its natural length. Compress it by x, and it pushes outward toward its natural length. In both cases, the force points opposite to the displacement. This is why it is called a restoring force — the spring is always trying to restore itself to its natural length.

Three states of a spring: compressed, natural, and stretched A wall on the left, with a spring in three states stacked vertically. Top: compressed, with a force arrow pointing right. Middle: natural length, no force. Bottom: stretched, with a force arrow pointing left. Compressed (x < 0) m F (→) Natural length (x = 0) m F = 0 Stretched (x > 0) m F (←)
The spring always pushes or pulls toward its natural length. Compressed: force points outward. Stretched: force points inward. At the natural length: no force at all.

Hooke discovered that the magnitude of this restoring force is directly proportional to the displacement:

F = -kx

Why the minus sign: the force is always opposite to the displacement. If you stretch the spring in the positive x direction (x > 0), the force is negative (pointing back). If you compress the spring (x < 0), the force is positive (pushing outward). The minus sign encodes the restoring nature of the force.

The constant k is called the spring constant — it tells you how stiff the spring is. A ballpoint pen spring might have k \approx 500 N/m. A car suspension spring might have k \approx 30{,}000 N/m. The bathroom weighing scale spring sits somewhere in between. The units of k are newtons per metre (N/m) — force per unit displacement.

What the spring constant means physically

The spring constant k is the slope of the F-vs-x graph. Since |F| = kx, a stiffer spring (larger k) produces a steeper line. A soft spring (small k) produces a shallower line.

The interactive figure below lets you drag the displacement x and watch how the restoring force changes. The slope of the line is the spring constant.

Interactive: Force vs displacement graph for a spring Two lines from the origin showing F = kx for k = 100 N/m and a second spring whose stiffness you control. Drag the red point along the x-axis to change the displacement and watch the force readout update. displacement x (m) restoring force |F| (N) 0.1 0.2 0.3 0.4 0.5 10 20 30 40 50 drag the red point along the axis
Drag the red point along the horizontal axis to change the displacement. The dashed line rises to the force curve, and the readout updates live. At $x = 0.1$ m the force is 10 N, at $x = 0.3$ m the force is 30 N. The line's slope *is* the spring constant.

This linearity has a dimensional check built in. The spring constant k has units of N/m. Multiplying by x (in metres) gives a force in newtons. The equation is dimensionally consistent, and the constant k carries the information about how stiff this particular spring is.

Think of it this way: k is the answer to the question "how many newtons does it take to stretch this spring by one metre?" A bathroom weighing scale spring with k = 2000 N/m needs 2000 N (about the weight of 200 kg) to stretch by a full metre — which is why the scale does not flatten when you stand on it. A toy slinky with k \approx 1 N/m stretches a metre under the weight of about 100 g.

Deriving the force law — where does the minus sign come from?

The linearity of Hooke's law is an experimental observation — you measure it in the lab by hanging different weights on a spring and recording the extension. But the minus sign is pure logic.

Choose a coordinate system: let x = 0 be the natural length of the spring, and let positive x point in the direction of stretching.

When x > 0 (spring stretched): the spring pulls back toward x = 0. The force points in the negative direction. So F < 0 when x > 0. That gives F = -(\text{positive number}) \times (\text{positive } x), which is negative. Correct.

When x < 0 (spring compressed): the spring pushes outward toward x = 0. The force points in the positive direction. So F > 0 when x < 0. With F = -k \times (\text{negative } x) = +k|x|, which is positive. Correct.

When x = 0 (natural length): no displacement, no force. F = -k \times 0 = 0. Correct.

Why: the single formula F = -kx handles all three cases — stretched, compressed, and natural — because the minus sign automatically flips the direction of the force relative to the displacement. This is the elegance of using a signed variable for displacement.

The spring constant k itself is always positive. The minus sign in F = -kx is not about k being negative — it is about the force opposing the displacement.

Springs in series — sharing the load

Connect two springs end-to-end, one after the other: k_1 first, then k_2. Attach a weight at the free end. Both springs feel the same force (the weight pulls on the bottom spring, which pulls on the top spring — the tension passes through).

Two springs in series supporting a weight A rigid support at the top. Spring k1 hangs from the support, spring k2 hangs from k1, and a mass m hangs from k2. Both springs stretch, and the total extension is x1 + x2. k₁ k₂ m x₁ x₂ mg x₁ + x₂
Two springs in series: both carry the same force $mg$, but each stretches by a different amount. The total extension is $x_1 + x_2$.

Here is the key insight: both springs carry the same force F (the weight mg hanging at the bottom pulls on k_2, and k_2 in turn pulls on k_1 with the same force — Newton's third law). But each spring stretches by a different amount depending on its own stiffness.

Step 1. Write the extension of each spring.

x_1 = \frac{F}{k_1}, \qquad x_2 = \frac{F}{k_2}

Why: from F = kx, solving for x gives x = F/k. The softer spring (smaller k) stretches more for the same force.

Step 2. The total extension is the sum.

x_{\text{total}} = x_1 + x_2 = \frac{F}{k_1} + \frac{F}{k_2}

Why: the springs are connected end-to-end, so their extensions add up. The top spring stretches by x_1, and the bottom spring stretches by x_2 below that.

Step 3. Define the effective spring constant k_{\text{eff}} as the single spring that would produce the same total extension under the same force.

x_{\text{total}} = \frac{F}{k_{\text{eff}}}

Step 4. Equate and simplify.

\frac{F}{k_{\text{eff}}} = \frac{F}{k_1} + \frac{F}{k_2}

Divide both sides by F (which is nonzero):

\boxed{\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2}}

Why: the effective spring constant of springs in series is always less than the smallest individual spring constant. Adding a second spring in series makes the combination softer, not stiffer — because the total extension gets longer while the force stays the same. This is the opposite of what happens in parallel.

For n springs in series:

\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2} + \cdots + \frac{1}{k_n}

Notice the similarity to resistors in parallel in electric circuits — mathematically identical structure.

Springs in parallel — combining the stiffness

Now place two springs side by side, both attached to the same block and to the same rigid support. When the block moves by x, both springs are stretched by the same amount x — but each contributes its own force.

Two springs in parallel supporting a weight A rigid support at the top. Two springs, k1 and k2, hang side by side. Both connect to the same horizontal bar, from which a mass m hangs. Both springs stretch by the same amount x. k₁ k₂ m x k₁x k₂x mg
Two springs in parallel: both stretch by the same amount $x$, but each contributes its own restoring force. The total upward force is $k_1 x + k_2 x$.

Step 1. Write the force from each spring.

F_1 = k_1 x, \qquad F_2 = k_2 x

Why: each spring obeys Hooke's law independently. Since both are stretched by the same displacement x, each produces a force proportional to its own stiffness.

Step 2. The total restoring force is the sum.

F_{\text{total}} = F_1 + F_2 = k_1 x + k_2 x = (k_1 + k_2)\,x

Why: both springs pull upward on the same block, so their forces add. The block feels a combined restoring force.

Step 3. The effective spring constant is the coefficient of x.

\boxed{k_{\text{eff}} = k_1 + k_2}

Why: springs in parallel add their stiffnesses directly. More springs in parallel means a stiffer combination — each spring contributes to fighting the displacement. This is the opposite of the series result.

For n springs in parallel:

k_{\text{eff}} = k_1 + k_2 + \cdots + k_n

Here is a useful mnemonic: in parallel, the springs share the displacement but their forces add, so stiffnesses add. In series, the springs share the force but their extensions add, so compliances (1/k) add.

Ideal vs real springs — the elastic limit

Everything above assumes the spring is ideal: the restoring force is perfectly proportional to displacement, no matter how far you stretch. Real springs have a limit.

Stretch a ballpoint pen spring gently, and it snaps back to its original length. Stretch it too far — really wrench it open — and it stays deformed. The coils have been pulled past the point where the metal can recover, and the spring is permanently bent. This threshold is called the elastic limit.

Below the elastic limit, the material behaves elastically: remove the force, and the spring returns to its natural length with no permanent deformation. Hooke's law holds in this region.

Beyond the elastic limit, the material undergoes plastic deformation: the internal structure of the metal changes permanently, and Hooke's law no longer applies. The force-displacement relationship stops being linear — the graph curves, and the spring does not return to its original length.

Force vs extension for a real spring showing elastic limit A graph showing force on the y-axis and extension on the x-axis. A straight line from the origin represents the Hooke's law region. At the elastic limit, the curve bends and becomes nonlinear, eventually reaching the breaking point. extension x force F elastic limit Hooke's law holds here plastic deformation break
The straight portion is where Hooke's law applies — force scales linearly with extension. Beyond the elastic limit (red dot), the material deforms permanently and the graph curves. Push even further and the spring breaks.

For a bathroom weighing scale, the elastic limit is designed to be well above the maximum weight it will ever measure. The scale relies on Hooke's law being valid for the entire measurement range. A pogo stick spring, a car suspension coil, a trampoline spring — all are engineered to operate safely below their elastic limit.

Assumptions for the rest of this article: Unless stated otherwise, all springs are ideal — meaning Hooke's law F = -kx holds for all displacements, the spring has zero mass, and there is no internal friction or damping.

A block on a spring — equilibrium and oscillation

Hang a block of mass m from a spring of constant k (attached to a rigid ceiling). The block stretches the spring until the spring force balances gravity. That resting position is the equilibrium.

Step 1. At equilibrium, the net force on the block is zero.

k x_0 = mg

Why: the spring pulls upward with force kx_0 (where x_0 is the extension at equilibrium), and gravity pulls downward with force mg. At equilibrium, these two are equal.

Step 2. The equilibrium extension is:

x_0 = \frac{mg}{k}

Why: a heavier block (m larger) stretches the spring more. A stiffer spring (k larger) stretches less for the same weight. This is exactly what you observe when you step on a bathroom weighing scale — your weight compresses the spring by an amount mg/k, and the scale reads the compression.

Now here is where it gets interesting. Pull the block a small distance below the equilibrium and release. The spring is now stretched more than x_0, so the spring force exceeds mg, and the net force pulls the block back upward. The block overshoots the equilibrium (because of its inertia), and then the spring is compressed relative to x_0, so gravity exceeds the spring force, and the block falls back down. This repeats — the block oscillates up and down around the equilibrium position.

Step 3. Measure displacement y from the equilibrium position (y = x - x_0). The net force at displacement y is:

F_{\text{net}} = mg - k(x_0 + y) = mg - kx_0 - ky = -ky

Why: the mg and -kx_0 terms cancel (they are equal at equilibrium). What remains is -ky — a restoring force proportional to displacement from equilibrium. This is Hooke's law all over again, but now about the equilibrium position. This is the seed of simple harmonic motion.

The result F_{\text{net}} = -ky means that a block on a spring, displaced from equilibrium, experiences a restoring force proportional to the displacement. By Newton's second law, ma = -ky, which gives:

a = -\frac{k}{m}\,y

This is the defining equation of simple harmonic motion (SHM). The block oscillates with a time period:

T = 2\pi\sqrt{\frac{m}{k}}

The derivation of this formula from a = -(k/m)y involves solving a differential equation, which you will find in Simple Harmonic Motion. But the physical content is clear from the formula: a heavier block (m larger) oscillates more slowly, and a stiffer spring (k larger) oscillates faster. A bathroom weighing scale needle that bounces before settling is doing exactly this — it is a mass-on-a-spring system undergoing damped simple harmonic motion.

Worked examples

Example 1: Finding the spring constant from a hanging mass

A 200 g mass is hung from a vertical spring, and the spring stretches by 4 cm from its natural length. Find the spring constant k. Then predict the extension when a 500 g mass is hung from the same spring.

Before and after diagram: 200 g stretches spring by 4 cm Left: spring at natural length with no mass. Right: 200 g mass hanging, spring stretched by 4 cm. A dashed line marks the natural length for comparison. No load L₀ 200 g hung 200 g natural end 4 cm mg kx
Left: the spring at its natural length. Right: a 200 g mass stretches the spring by 4 cm. At equilibrium, the spring force $kx$ balances the weight $mg$.

Step 1. Convert to SI units.

m = 200 g = 0.2 kg, x = 4 cm = 0.04 m, g = 9.8 m/s².

Why: Hooke's law uses SI units — force in newtons, displacement in metres. Always convert before substituting.

Step 2. At equilibrium, the spring force equals the weight.

kx = mg
k = \frac{mg}{x} = \frac{0.2 \times 9.8}{0.04} = \frac{1.96}{0.04} = 49 \text{ N/m}

Why: the spring is in equilibrium — it is not accelerating. The upward spring force kx balances the downward gravitational force mg. Solving for k gives the spring constant.

Step 3. Predict the extension for 500 g.

x_{\text{new}} = \frac{m_{\text{new}} \cdot g}{k} = \frac{0.5 \times 9.8}{49} = \frac{4.9}{49} = 0.1 \text{ m} = 10 \text{ cm}

Why: the spring constant k is a property of the spring — it does not change when you change the mass. The new mass is 2.5 times the original, so the new extension is 2.5 times the original 4 cm, giving 10 cm. Hooke's law is linear: double the weight, double the stretch.

Result: The spring constant is k = 49 N/m. With a 500 g mass, the spring stretches 10 cm.

What this shows: Measuring k from one load lets you predict the extension for any other load, as long as you stay within the elastic limit. This is exactly how a bathroom weighing scale works — the manufacturer calibrates the scale by measuring k, then prints a mass scale based on how far the spring compresses under different weights.

Example 2: Two springs in series supporting a load

Two springs with k_1 = 100 N/m and k_2 = 200 N/m are connected end-to-end (in series) and hung from a rigid ceiling. A 3 kg mass hangs from the free end. Find the effective spring constant and the total extension of the combination.

Two springs in series with a 3 kg mass Spring k1 (100 N/m) connects to the ceiling, then spring k2 (200 N/m) connects to k1, with a 3 kg block at the bottom. k1 stretches by 29.4 cm, k2 stretches by 14.7 cm, total 44.1 cm. k₁ = 100 N/m k₂ = 200 N/m 3 kg x₁ = 29.4 cm x₂ = 14.7 cm 29.4 N
Spring $k_1$ (red, softer) stretches nearly twice as much as spring $k_2$ (dark, stiffer), even though both carry the same 29.4 N load.

Step 1. Find the effective spring constant using the series formula.

\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{100} + \frac{1}{200} = \frac{2}{200} + \frac{1}{200} = \frac{3}{200}
k_{\text{eff}} = \frac{200}{3} \approx 66.7 \text{ N/m}

Why: the series formula adds the reciprocals. The effective spring constant (66.7 N/m) is less than both individual constants (100 N/m and 200 N/m) — a series combination is always softer than the softest spring in it.

Step 2. Find the total extension.

x_{\text{total}} = \frac{mg}{k_{\text{eff}}} = \frac{3 \times 9.8}{66.7} = \frac{29.4}{66.7} \approx 0.441 \text{ m} = 44.1 \text{ cm}

Why: from Hooke's law, x = F/k. The effective spring constant treats the entire series combination as a single spring.

Step 3. Verify by computing each extension individually.

The force through both springs is F = mg = 3 \times 9.8 = 29.4 N.

x_1 = \frac{F}{k_1} = \frac{29.4}{100} = 0.294 \text{ m} = 29.4 \text{ cm}
x_2 = \frac{F}{k_2} = \frac{29.4}{200} = 0.147 \text{ m} = 14.7 \text{ cm}
x_1 + x_2 = 29.4 + 14.7 = 44.1 \text{ cm} \checkmark

Why: the individual extensions add up to the total — confirming the series formula is correct. Notice that the softer spring (k_1 = 100 N/m) stretches twice as much as the stiffer one (k_2 = 200 N/m), even though both carry the same force.

Result: The effective spring constant is \frac{200}{3} \approx 66.7 N/m. The total extension is 44.1 cm: the softer spring contributes 29.4 cm and the stiffer spring contributes 14.7 cm.

What this shows: In a series combination, the softer spring dominates the total extension. The effective k is always less than the smallest individual k — adding more springs in series makes the system more flexible, not stiffer. This is why a car suspension with multiple spring stages can be made very soft even if each individual spring is relatively stiff.

Common confusions

If you came here to understand Hooke's law, use the spring constant, and combine springs in series and parallel, you have what you need. What follows is for readers preparing for JEE Advanced who want the energy perspective, the microscopic origin, and the formal series/parallel analogy.

Energy stored in a spring

When you stretch a spring by x, you do work against the restoring force. Since the force varies with displacement (F = kx, increasing as you stretch further), the work is not simply F \times x — you need to integrate.

The work done in stretching the spring from 0 to x:

W = \int_0^x kx'\,dx' = \frac{1}{2}kx^2

Why: the force increases linearly from 0 to kx as the displacement goes from 0 to x. The work is the area under the F-vs-x graph — a triangle with base x and height kx, giving area \frac{1}{2}kx^2. This is stored as elastic potential energy.

This energy is stored as elastic potential energy:

U = \frac{1}{2}kx^2

The energy scales as x^2, not x. Doubling the stretch stores four times the energy. This is why a compressed spring in a toy gun can launch a projectile — the energy stored in the spring is converted to kinetic energy of the projectile.

Why Hooke's law works — the microscopic view

At the atomic level, atoms in a solid sit in potential energy wells created by their interactions with neighbouring atoms. Near the equilibrium position (the bottom of the well), any smooth potential energy curve looks approximately parabolic — you can verify this by expanding U(x) in a Taylor series about the equilibrium position x_0:

U(x) = U(x_0) + U'(x_0)(x - x_0) + \frac{1}{2}U''(x_0)(x - x_0)^2 + \cdots

At equilibrium, U'(x_0) = 0 (that is the definition of equilibrium — no net force). The first nonzero correction is the \frac{1}{2}U''(x_0)(x-x_0)^2 term, which gives a parabolic potential. The force is:

F = -\frac{dU}{dx} = -U''(x_0)(x - x_0)

This is exactly Hooke's law, with k = U''(x_0).

Why: near equilibrium, every potential looks parabolic. This is why Hooke's law applies not just to metal springs, but to rubber bands (at small stretches), vibrating molecules, and even the oscillations of atoms in a crystal lattice. The spring is the universal model for small oscillations about equilibrium.

The higher-order terms in the Taylor expansion become important at large displacements — that is precisely the elastic limit. When x - x_0 is no longer small, the cubic and quartic terms matter, the force is no longer linear, and Hooke's law breaks down.

The series-parallel analogy with resistors

The mathematical structure of spring combinations is identical to that of resistor combinations — but with the series and parallel roles swapped.

Configuration Springs Resistors
Series \frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2} R_{\text{eff}} = R_1 + R_2
Parallel k_{\text{eff}} = k_1 + k_2 \frac{1}{R_{\text{eff}}} = \frac{1}{R_1} + \frac{1}{R_2}

For springs, the "flow-through" quantity is force (same force through series springs), and the "across" quantity is extension. For resistors, the "flow-through" quantity is current (same current through series resistors), and the "across" quantity is voltage. The mathematical inversion arises because Hooke's law (F = kx) is analogous to Ohm's law (V = IR), but k maps to 1/R (conductance), not to R.

This analogy is worth remembering. When you encounter capacitors in series and parallel, you will find the same pattern repeats yet again — and knowing why the formulas flip will save you from memorising them blindly.

Where this leads next