JEE Main 2026 · PhysicsNumerical answerNumerical valueEasyCalculation

JEE Main 4 April 2026, Shift 1, Physics Q48

Question 48 of 75 in this shift, Physics question 23 of 25, Section B.

The velocity of a particle executing simple harmonic motion along

x

-axis is described as

v^2=50-x^2

, where

x

represents displacement. If the time period of motion is

\frac{x}{7}

s, the value of

x

is ______.

Official answer

NTA final key.

Chapter: Oscillations
Topic: SHM Equations
Idea tested: Velocity and Accel vs Position

Same idea in other shifts

Asked 3× in all

9 Apr 2024, Shift 1 · Q54The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4$ m,…MediumNumerical value
31 Jan 2024, Shift 1 · Q54A particle performs simple harmonic motion with amplitude $A$ . Its speed is increased to three times at an instant when its displacement…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 42 of the 174 JEE Main shifts held from 2017 to 2026.