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Exam level5 past questions

Velocity and Acceleration in SHM

Explore how velocity and acceleration vary with displacement and time in SHM, including maximum values and their physical significance.

Why this shows up in the exam

NEET tests your understanding of the mathematical relationships and physical meaning of velocity and acceleration in oscillators.

How NEET tests this

Numerical · 3 QsMulti-conceptStatement analysis

Learn the idea

In simple harmonic motion the velocity and acceleration are directly linked to the displacement through the angular frequency ω; knowing any two of v, a, x or ω lets you find the others, and the maximum values occur at the extreme positions (a_max) and equilibrium (v_max).

🧠 Memory hook: V_max = 2π f A – think of a point on a circle of radius A travelling one full revolution each period; distance per second = circumference (2πA) times cycles per second (f).

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

Formulas & facts to keep ready

  • Displacement: x = A cos(ωt+φ) or A sin(ωt+φ) (NCERT)
  • Velocity: v = dx/dt = ±ω √(A²‑x²) = A ω sin(ωt+φ) with |v|_max = A ω
  • Acceleration: a = d²x/dt² = –ω² x with |a|_max = A ω²
  • Relation: v² + ω² x² = (A ω)² (energy‑like)
  • Frequency and angular frequency: ω = 2π f and T = 1/f

How to approach it

  1. 1Write the given maximum speed as v_max = A ω
  2. 2Find ω = v_max / A
  3. 3Convert ω to ordinary frequency using f = ω / (2π)
  4. 4If the question asks for period, use T = 1/f

Worked example — watch it click

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cms⁻¹. The frequency of its oscillation is:

  • ✅1 Hz
  • B)3 Hz
  • C)2 Hz
  • D)4 Hz

The concept behind this problem

The example uses the definition v_max = A ω and the relation ω = 2π f to turn the given amplitude and maximum speed into the oscillation frequency, testing the core link between v_max, A and ω.

Step by step

  1. 1Maximum speed v_max = Aω = A(2πf).
  2. 2Given A = 5 cm, v_max = 31.4 cm/s: 31.4 = 5 × 2π × f, so f = 31.4/(10π) ≈ 31.4/31.4 = 1 Hz.

Watch out

A common mistake is to write f = v_max / A, forgetting the essential 2π factor.

Common slip-ups that cost marks

  • •Do not confuse ω (rad s⁻¹) with f (Hz); ω = 2π f, not the other way round
  • •Never forget the factor 2π when changing between ω and f
  • •Maximum speed occurs at the mean position, not at the extreme displacement

🌟 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 5NEET 2005

A particle executing simple harmonic motion of amplitude 5 cm has maximum speed of 31.4 cms⁻¹. The frequency of its oscillation is:

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

Consider a particle undergoing simple harmonic motion. Which of the following statements is always true?

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