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Saturday, 31 October

JEE Main 2022 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 27 June 2022, Shift 2, Physics Q54

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

A particle executes simple harmonic motion. Its amplitude is 8 cm and time period is 6 s. The time it will take to travel from its position of maximum displacement to the point corresponding to half of its amplitude, is ___________ s.

Official answer

1

NTA final key.

Same topic in other shifts

All SHM Equations questions
  1. 2 Apr 2026, Shift 2 · Q45The equation of motion of a particle is given by x=asin⁡(50t+π3)x = a\sin\left(50t + \frac{\pi}{3}\right) cm. The particle will come to rest at time…MediumSingle correct
  2. 4 Apr 2026, Shift 1 · Q48The velocity of a particle executing simple harmonic motion along xx-axis is described as v2=50−x2v^2=50-x^2, where xx represents…EasyNumerical value
  3. 5 Apr 2026, Shift 2 · Q35Match List - I with List - II. List - I A. sin⁡2ωt\sin^2\omega t B. sin⁡3(2ωt)\sin^3(2\omega t) C. sin⁡(ωt)+cos⁡(πωt)\sin(\omega t)+\cos(\pi\omega t) D.…HardMatch the list
  4. 6 Apr 2026, Shift 1 · Q37A particle is executing simple harmonic motion. Its amplitude is AA and time period is 5 sec. The time required by it to move from x=Ax=A…EasySingle correct
  5. 8 Apr 2026, Shift 2 · Q37The frequency of oscillation of a mass mm suspended by a spring is v1v_1. If the length of the spring is cut to half, the same mass…EasySingle correct
  6. 2 Apr 2025, Shift 1 · Q33A particle is subjected to two simple harmonic motions as : x1=7sin⁡5tx_1=\sqrt{7}\sin5t cm and x2=27sin⁡(5t+π3)x_2=2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right)…MediumSingle correct

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.