JEE Main 2022 · PhysicsNumerical answerNumerical valueMediumCalculation

JEE Main 28 July 2022, Shift 2, Physics Q25

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

The potential energy of a particle of mass 4 kg in motion along the

x

-axis is given by U = 4(1 – cos 4

x

) J. The time period of the particle for small oscillation (

\sin\theta=\theta

) is

\left(\frac{\pi}{\mathrm{K}}\right)

s. The value of K is _____.

Official answer

NTA final key.

Chapter: Oscillations
Topic: SHM Equations
Idea tested: Effective Spring Constant from Potential Energy

Same topic in other shifts

All SHM Equations questions

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5 Apr 2026, Shift 2 · Q35Match List - I with List - II. List - I A. $\sin^2\omega t$ B. $\sin^3(2\omega t)$ C. $\sin(\omega t)+\cos(\pi\omega t)$ D.…HardMatch the list
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 2026, Shift 2 · Q37The frequency of oscillation of a mass $m$ suspended by a spring is $v_1$ . If the length of the spring is cut to half, the same mass…EasySingle correct
2 Apr 2025, Shift 1 · Q33A particle is subjected to two simple harmonic motions as : $x_1=\sqrt{7}\sin5t$ cm and $x_2=2\sqrt{7}\sin\left(5t+\frac{\pi}{3}\right)$ …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.