JEE Main 2021 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 18 March 2021, Shift 1, Physics Q25

Question 25 of 90 in this shift, Physics question 25 of 30, Section B.

A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is

\frac{1}{a}

s. The value of 'a' to the nearest integer is __________.

Official answer

NTA final key.

Chapter: Oscillations
Topic: SHM Equations
Idea tested: Phasor Diagram Technique

Same idea in other shifts

Asked 5× in all

6 Apr 2026, Shift 1 · Q37A particle is executing simple harmonic motion. Its amplitude is $A$ and time period is 5 sec. The time required by it to move from $x=A$ …EasySingle correct
8 Apr 2023, Shift 2 · Q32For particle $P$ revolving round the centre $O$ with radius of circular path $r$ and angular velocity $\omega$ , as shown in below figure,…MediumSingle correctHas a figure
3 Aug 2021, Shift 2 · Q1A particle is executing simple harmonic motion with time period $T$ . What is the time taken by the particle to go directly to…EasySingle correct
3 Aug 2021, Shift 2 · Q2A block of mass m rigidly attached with a spring of spring constant $K$ , is compressed through a small distance A. If the block is now…MediumSingle correctHas a figure

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 42 of the 174 JEE Main shifts held from 2017 to 2026.