JEE Main 2024 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 27 January 2024, Shift 1, Physics Q54

Question 54 of 90 in this shift, Physics question 24 of 30, Section B.

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is

\sqrt{\alpha}

cm, where

\alpha

= ________.

Official answer

NTA final key.

Chapter: Oscillations
Topic: Energy in SHM
Idea tested: Velocity-Displacement from Energy Conservation

Same topic in other shifts

All Energy in SHM questions

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.