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Saturday, 31 October

JEE Main 2023 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 13 April 2023, Shift 1, Physics Q57

Question 57 of 90 in this shift, Physics question 27 of 30, Section B.

At a given point of time the value of displacement of a simple harmonic oscillator is given as y=Acos⁡(30∘)y = A\cos(30^\circ). If amplitude is 40 cm and kinetic energy at that time is 200 J, the value of force constant is 1.0×10x Nm−11.0 \times 10^{x}\ \mathrm{Nm^{-1}}. The value of xx is ___________

Official answer

4

NTA final key.

Chapter
Oscillations
Topic
Energy in SHM
Idea tested
Kinetic Energy in SHM

Same idea in other shifts

Asked 2× in all
  1. 8 Apr 2024, Shift 2 · Q54An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of (25π)\left(\frac{25}{\pi}\right) Hz. At the position…MediumNumerical value

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.