Introduction to Hooke’s law, its formula, derivation and examples

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Hooke’s law is an elasticity law discovered in 1660 by English scientist Robert Hooke. Hooke’s Law is used to relate a material’s stress (force per unit area) to its strain (change in length over its original length).

It describes a material’s behaviour under load and forecasts the type of deformation that will occur. It depends, as it does with most things in life.

Hooke’s law is simply a first-order linear approximation to the real reaction of springs, as well as maybe other elastic things, under applied forces.

It fails when the forces reach a limit that is unique to each item, because no material can be compressed or stretched beyond a specific minimum size without causing at least minor permanent deformation or change of condition.

Many materials will diverge noticeably from Hooke’s rule long before the elasticity limits are reached.

  • What is Hook’s Law?

According to Wikipedia

“Hooke’s law is a law of physics that states that the force (F) needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance”

Mathematically it is represented as 

Fs = -kx

Where x is distance from mean position, Fs is restoring force and k is proportionality constant. Here negative sign indicates that restoring force is always opposite to the direction of displacement.

  • Limitations of Hooke’s law

Robert Hooke carried out experiments with metal prongs and rods, concluding that the deformation caused by the load supplied to the bar is precisely proportional to the load applied to it. This is true for some assumptions, however:

  • Isotropic material is required ( i.e., it has same material properties in any direction)
  • The material must be uniform (i.e., the density is uniform throughout)
  • The load on the body must be within the linear elastic range.

Hooke’s law does not hold true if any of the above bodily conditions are not met.

  1. Derivation of formula

Consider a spring that can be stretched and compressed while maintaining the same spring constant. Consider a mass attached to the spring and the system positioned on a frictionless table, allowing the mass to glide freely in response to the spring’s force.

Let xo be the displacement of mass m at the mean position or the equilibrium position of spring.

Now if a force is applied on mass towards the right side the spring stretches and induces a restoring force Fs. Now at the stretched position, the mass m covers a new displacement that is x at its extreme position.

When the applied force is released the restoring force allows the mass m to start moving towards its mean position with velocity v.

It gains its maximum velocity at its mean position after reaching the mean position the spring starts compression and the mass m starts moving towards its second extreme position.

It gains its lowest velocity at both of its extremes. Now the restoring force that causes the mass to move to and fro about its mean position is directly proportional to the net displacement covered by mass m. Fs Net displacement

Fs -(x-xo)

Here the minus sign shows that the displacement is always opposite in direction to restoring


Fs = -k (x-xo)

Where k is spring constant or force constant. The SI unit of spring constant is Newton per meter (Nm-1). By using this formula we can calculate the restoring force, spring constant, displacement of spring’s equilibrium position, and displacement from equilibrium position. 

The above equation can also be written as 

Fs = -k Δx

Where Δx is change in displacement. Following are some examples of Hooke’s law.


Example 1:

What force is required to pull a spring with a spring constant of 20 N/m when the distance from equilibrium is 35cm and the spring equilibrium position is 10cm?

Step 1: write given data values

Spring constant = k = 20 N/m

Distance from equilibrium = x = 35cm = 0.35m

Spring equilibrium position = xo = 10cm = 0.10m

Force =?

Step 2: Write general formula

Fs = -k(x-xo)

Step3: Put the given data values

Fs = -20(0.35-0.10)

Fs = -20(0.25)

Fs = -5N

Hence the required force is -5N.

Many times, in scientific and mathematical calculations involving complex operations with intricate numbers, the computation is not practical since it would take a long time and would be prone to errors if done manually.

Hooke’s law calculator helps to reduce the errors and time consumption. You can get result in just one click. The method to use Hooke’s law calculator is mentioned below.

Step 1: Select the value you want to calculate

Step 2: Put the given data values and click on calculate button

Example 2:

A spring is dragged to 5cm from its equilibrium point of 4cm and held in place by a 500N force. What is the spring constant’s value?

Step 1: Write the given data values

Distance from Equilibrium = x = 5cm = 0.5m

Spring Equilibrium Position = xo = 4cm = 0.4m

Force = F =500N

Spring Force Constant =?


Step 2: Rewrite general formula for spring constant

k = Fs/x-xo

Step 3: Put the given data values in above formula

k = 500N/0.5m-0.4m

k = 500N/0.1m

k = 500N/0.1m

k = 5000Nm-1

Hence the spring constant is 5000Nm-1

This problem can also be solved with the help of Hooke’s law calculator.


Hooks law is among the fundamental equations that explains the stress-strain correlation. It allows the calculation of how load affects the body. It was established through experimentation, just like any other constitutional equation. It specifies that strain is proportional to stress.

Hooke’s Law established a foundation of material science strength, which is essential for engineering and building. Because of its simplicity, it enabled the development of a slew of mathematical procedures that form the foundation of elasticity theory.

Engineering requirements resulted in mathematical triumphs, such as a potential theory, complex numbers calculus. 

Hooke’s law calculators allow us to find answers to simple harmonic motion problems without expending too much energy or time, as well as with less effort.

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