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Mechanics · Physics

Hooke's Law Simulation

Pull and push a spring and read F = kx. Join two springs in series or parallel to see the effective spring constant, and watch the stored energy ½kx² grow.

Interactive virtual lab · no download · works on phones and laptops

Chapter
Mechanical Properties of Solids · Oscillations
For
Class 11 · NEET
Time
~10 min

Hooke's Law is a Pro simulation

Pro opens all 125 virtual labs — 483 simulations with guides and quizzes.

How to use it

  1. 1Drag the hand or use the applied-force slider.
  2. 2Change the spring constant.
  3. 3Systems: choose series or parallel and set each spring.
  4. 4Energy: read the force–displacement graph and its area.

What to notice

  • Spring force is proportional to displacement and opposite to it.
  • Stiffer spring (bigger k) → more force for the same stretch.
  • Parallel: k = k₁ + k₂. Series: 1/k = 1/k₁ + 1/k₂.
  • Stored energy ½kx² grows with the square of the stretch.

For teachers

A ready-made teaching kit comes with this lab

  • Lesson plan

    19 min, 4 steps

  • Ask the class

    3 questions with answers

  • Quick quiz

    6 questions, printable

  • Key formulas

    4 formulas

About the Hooke's Law simulation

Hooke's law says that the force needed to stretch or compress a spring is proportional to how far you stretch it: F = kx. The constant k, the spring constant, measures stiffness in newtons per metre. The law holds until the spring is stretched past its elastic limit.

This Hooke's law simulation lets you pull a single spring, or two springs joined in series or in parallel, and read the applied force, displacement and stored energy. You can see why two identical springs in parallel are twice as stiff and in series half as stiff. It supports Mechanical Properties of Solids and Oscillations in Class 11 and the spring-combination problems common in JEE.

Also covers: spring constant, restoring force, elastic potential energy, springs in series, springs in parallel.

Frequently asked questions

What is the spring constant k?

The force needed per metre of stretch, in N/m. A spring with k = 200 N/m needs 200 N to stretch it by 1 m (or 2 N for 1 cm).

Why are springs in parallel stiffer?

Each spring stretches by the same amount and they share the load, so their forces add: k = k₁ + k₂.

How much energy does a stretched spring store?

U = ½kx². Doubling the stretch stores four times the energy.