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Foundation2 past questions

Energy in Simple Harmonic Motion

Learn how kinetic and potential energy are distributed and conserved in SHM, including the concept of energy equipartition.

Why this shows up in the exam

Questions frequently ask about energy transformations and conservation in oscillatory systems.

How NEET tests this

Numerical · 2 QsApplication

Learn the idea

In SHM the total mechanical energy stays constant and is shared between kinetic (T) and potential (V) energies as the particle moves. The key insight is that T and V are simple quadratic functions of the displacement x, so setting them equal leads to x = a/√2.

🧠 Memory hook: Energy in SHM is like a balanced seesaw: at the halfway height (a/√2) the two sides – kinetic and potential – are exactly equal.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Total energy E = (1/2) m ω² a² (constant)
  • Kinetic energy T = (1/2) m ω² (a² - x²)
  • Potential energy V = (1/2) m ω² x²
  • At mean position (x=0) T = E, V = 0; at extreme (x=±a) T = 0, V = E
  • Equality T = V occurs at |x| = a/√2
  • For a mass‑spring ω = √(k/m); for a simple pendulum (small angle) ω = √(g/L)

How to approach it

  1. 1Write the standard SHM expressions for T and V in terms of a, x and ω
  2. 2Apply the condition given in the question (e.g., T = V, T/V = …) to form an algebraic equation
  3. 3Solve the equation for the unknown (usually x or a ratio) using simple algebra
  4. 4Verify that the solution satisfies |x| ≤ a and matches the physical situation

Worked example — watch it click

The kinetic energy and potential energy of a particle executing SHM will be equal, when displacement is: (amplitude = a)

  • A)(a)/(2)
  • B)a√(2)
  • ✅(a)/(√(2))
  • D)(a)/(2√(3))

The concept behind this problem

The worked example asks for the displacement where kinetic and potential energies are equal, which directly tests the quadratic dependence of T and V on x and the use of the equality condition.

Step by step

  1. 1KE = (1/2)mω²(a² - x²), PE = (1/2)mω²x².
  2. 2For KE = PE: a² - x² = x², so 2x² = a², giving x = a/√2.

Watch out

Students often invert the root and write x = a√2 instead of the correct x = a/√2.

Common slip-ups that cost marks

  • •Confusing displacement x with amplitude a – x is measured from the mean position
  • •Dropping the square on a or x when forming the equation, leading to a√2 instead of a/√2
  • •Assuming energy equality at the extreme points where actually one form is zero

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 2NEET 1991

The angular velocity and the amplitude of a simple pendulum is ω and a, respectively. At a displacement x from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is :

Push further

More challenging

3 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 3

A particle undergoes simple harmonic motion with an amplitude of 10 cm. At what displacement from the equilibrium position will its kinetic energy be equal to its potential energy?

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