LEARN WITH
SIMULATIONSBY NANDAX · SCIENCE IN MOTION
हिंदी Launch a Lab

Oscillations · NCERT / JEE · Class 11

Spring–Mass SHM: Hooke’s Law, Period and Energy

Explore spring–mass simple harmonic motion: F = −kx, period, frequency, energy exchange and solved numerical examples with a free lab.

Educational content by NANDAX · Original explanation and interactive experiments

A simple way to see restoring forces

Pull a block attached to a spring to one side and release it. It accelerates toward equilibrium, overshoots because it has momentum, slows at the opposite extreme, and returns. The repeating motion is simple harmonic motion (SHM) when the restoring force is proportional to displacement and directed toward equilibrium.

What you will learn

  • Read the negative sign in Hooke’s law as a direction, not a negative spring strength.
  • Derive the mass and spring-constant dependence of period and frequency.
  • Explain why kinetic and spring potential energy exchange while total mechanical energy stays constant in an ideal oscillator.
  • Make predictions and verify them with the interactive Spring–Mass Lab.

Hooke’s law and Newton’s second law

Let x be displacement from equilibrium. For an ideal spring, F = −kx, where k is the spring constant measured in N/m. With mass m on a frictionless horizontal surface, ma = −kx. Therefore acceleration is opposite to displacement and proportional to its magnitude.

F = −kx
a = −(k/m)x
ω = √(k/m)
T = 2π√(m/k)
f = (1/2π)√(k/m)

The period depends on m and k, not on amplitude A, as long as the spring remains linear and damping is absent. A stiffer spring oscillates faster; a heavier block oscillates more slowly. No individual piece of matter is moving sinusoidally merely because it appears to move back and forth: the proportional restoring-force condition matters.

Displacement, speed and acceleration

For an oscillator released from the positive extreme at t = 0, displacement follows x(t) = A cos(ωt). Different release conditions change the phase but not the frequency. The speed magnitude is largest at the centre and zero at either turning point. Acceleration magnitude is largest at the extremes, where the restoring force is greatest.

x(t) = A cos(ωt)
v(t) = −Aω sin(ωt)
a(t) = −Aω² cos(ωt)
vₘₐₓ = A√(k/m)

Energy: where does the motion go?

A displaced spring stores elastic potential energy U = ½kx². At the extremes, x = ±A, all mechanical energy is spring potential energy in the ideal horizontal model. At equilibrium x = 0, spring potential energy is zero and kinetic energy reaches its maximum. In between, one decreases as the other increases. Their sum stays at E = ½kA². Energy is not “used up” in this frictionless idealisation.

U = ½kx²
K = ½mv²
E = K + U = ½kA²

Worked example: mass 1 kg, stiffness 40 N/m

Let m = 1.00 kg, k = 40 N/m and amplitude A = 0.20 m. Then ω = √40 = 6.325 rad/s, T = 2π/√40 ≈ 0.993 s, and f ≈ 1.007 Hz. The total energy is ½ × 40 × (0.20)² = 0.80 J. At x = 0.10 m, spring potential energy is ½ × 40 × (0.10)² = 0.20 J, leaving 0.60 J as kinetic energy. Speed there is √(2K/m) = √1.2 ≈ 1.095 m/s.

Guided virtual experiment

  1. Open the Spring–Mass Oscillator and set k = 40 N/m and m = 1 kg if those settings are available.
  2. Before starting, predict the natural period. Compare it with the live readout.
  3. Increase m while leaving k unchanged. Note how T changes and state the square-root relationship.
  4. Return to the original mass, then increase k. Explain the shorter period using stronger restoring acceleration.
  5. Observe the energy indicators at x = 0 and at x = ±A.
  6. Where an amplitude control is available, change A and check total energy. Predict why doubling A quadruples E without altering ideal T.

Common misconceptions

  • “The force is largest at the centre.” In this model, force is zero at equilibrium and maximal in magnitude at the extremes.
  • “A larger amplitude always gives a longer period.” Not for a linear undamped spring.
  • “The total energy is constant in every real oscillator.” Real damping converts mechanical energy to other forms.
  • “A spring obeys F = −kx at any stretch.” Real materials have elastic limits; beyond the linear range, Hooke’s law fails.

Practice with worked answers

  1. Doubling mass at fixed k gives T′/T = √2, approximately 1.414.
  2. Increasing k to 4k at fixed m gives T′ = T/2 and frequency 2f.
  3. If amplitude doubles, E′ = 4E in the ideal linear model.
  4. A spring with k = 100 N/m stretched 0.05 m exerts a restoring force of magnitude 5 N.
  5. At an extreme position, why does the block reverse direction? Its speed is momentarily zero but the nonzero restoring force accelerates it toward equilibrium.

Beyond the model

Suspension systems, instrument vibration and molecular motion can resemble oscillators, but damping, forcing and nonlinear stiffness often matter. For NCERT and JEE, always check whether the spring is ideal, whether equilibrium is defined correctly and whether friction can be neglected. Use the worksheet to connect the equations to the moving object rather than memorising them alone.

Frequently asked questions

Where is the spring force greatest?

At the positive and negative extreme displacements, measured by magnitude kA.

Does amplitude control the ideal period?

No. T = 2π√(m/k) depends on mass and spring constant.

When is speed maximum?

At the equilibrium position x = 0.

What happens to total energy if amplitude doubles?

It quadruples for an ideal linear spring because E = ½kA².

Further reading and references

References are for further independent learning; explanations and examples on this page are written for NANDAX.