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Friday, 9 October

NEET UG 2019 · Physics · Question 10

NEET 2019 Physics, question 10

Question 10 of the NEET 2019 paper (code P1) is a Physics question from Oscillations. Its idea, Superposition of Parallel SHMs, was asked only this once in NEET 2017–2026. The NTA final answer key gives option 2.

Updated 2026-10-08

Q10 · Physics

The displacement of a particle executing simple harmonic motion is given by

y=A0+Asin⁡ωt+Bcos⁡ωty = A_0 + A\sin\omega t + B\cos\omega t.

Then the amplitude of its oscillation is given by :

  1. 1A0+A2+B2A_0 + \sqrt{A^2 + B^2}
  2. 2A2+B2\sqrt{A^2 + B^2}Official answer
  3. 3A02+(A+B)2\sqrt{A_0^2 + (A + B)^2}
  4. 4A+B

Official answer: option 2 · NTA final answer key

Same idea

The only question on Superposition of Parallel SHMs

No other NEET paper from 2017–2026 tested this exact idea.

Superposition of Parallel SHMs: every question and how it was framed

The chapter

Oscillations in ten years of NEET

17 questions from 2017–2026, asked in 10 of 10 papers, about 1.6 a paper. Simple Harmonic Motion alone had 16.

Chapter analysisStudy this chapterPractise this chapter

Oscillations: questions per paper

0242’171’183’191’202’211’221’232’242’252’26