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Exam level3 past questions

Simple Pendulum and Spring Systems

Study the time period formulas, energy considerations, and behavior of simple pendulums and spring-mass systems, including effects of non-inertial frames and spring combinations.

Why this shows up in the exam

You need to apply these formulas and concepts to solve for periods and analyze pendulum and spring problems in NEET.

How NEET tests this

Application · 4 QsNumerical

Learn the idea

For a simple pendulum (small angles) the period T = 2π√(L/g) – it depends only on length and gravity, not on mass or amplitude. For a spring‑mass system T = 2π√(m/k) – only mass and spring constant matter.

🧠 Memory hook: A pendulum’s swing cares only about how long the rope is – like a child’s laughter that doesn’t get louder with a heavier swing.

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

Formulas & facts to keep ready

  • Tpend = 2π√(L/g) for small oscillations
  • Period of a simple pendulum is independent of bob mass
  • Period is independent of amplitude only when θ is small (sinθ≈θ)
  • Tspring = 2π√(m/k) for a mass‑spring system
  • For springs in parallel k_eq = k1+k2; in series 1/k_eq = 1/k1+1/k2
  • Energy in pendulum: E = mgL(1‑cosθ) ≈ ½mgLθ² for small θ

How to approach it

  1. 1Identify which system (pendulum or spring) the question deals with
  2. 2Write the appropriate period formula and note which quantities affect it
  3. 3Apply any changes given (length, mass, k, series/parallel) keeping in mind the independence rules
  4. 4Compute the new period by simple algebraic scaling
  5. 5Check the small‑angle condition if amplitude is mentioned

Worked example — watch it click

A simple pendulum executing small oscillations has period T. If now the length of the pendulum is made 4 times, mass of the bob doubled and amplitude made one-fourth, the new period becomes:

  • A)√2T 3
  • B)T 2√2
  • C)2√2 3 T
  • ✅2T

The concept behind this problem

The worked example checks whether you remember that, for small oscillations, the period of a simple pendulum is governed solely by its length (and g), so changing mass or amplitude does not alter T.

Step by step

  1. 1Period of simple pendulum T = 2π√(L/g).
  2. 2Period is independent of mass and amplitude (for small oscillations).
  3. 3If length becomes 4L: T' = 2π√(4L/g) = 2·2π√(L/g) = 2T.
  4. 4Mass doubling and amplitude change do not affect period.

Watch out

Students often mistakenly think halving the amplitude will change the period, but for small angles the period stays the same.

Common slip-ups that cost marks

  • •Treating a large amplitude as if the small‑angle formula still holds
  • •Assuming mass changes the pendulum period
  • •Confusing series and parallel combinations of springs – remember series reduces k, parallel adds k

🌟 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 3

A simple pendulum executing small oscillations has period T. If now the length of the pendulum is made 4 times, mass of the bob doubled and amplitude made one-fourth, the new period becomes:

Push further

More challenging

6 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 6

A simple pendulum of length L has a time period T. If its length is increased by 44%, what will be the new time period?

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