Oscillations

JEE Main level, single correct, medium

55 of 194

Question 55 of 194

A body is in simple harmonic motion with time period T=0.5sT=0.5 \text{s} and amplitude A=1cmA=1 \text{cm}. Find the average velocity in the interval in which it moves from equilibrium position to half of its amplitude.

Show answer and solution

Answer: (D) 12cm/s12 \text{cm/s}

∴x=Asin⁡⁡ωt∴x=A\sin ⁡\omega t

A2=Asin⁡⁡ωt\dfrac{A}{2}=A\sin ⁡\omega t

ωt=π6\omega t=\dfrac{\pi }{6}

t=π6ωt=\dfrac{\pi }{6\omega }

=π6×2π=\dfrac{\pi }{6\times 2\pi }

t=T12t=\dfrac{T}{12}

∴Timetakentoreachfromx=0tox=A2isT12∴Timetakentoreachfromx=0tox=\dfrac{A}{2}is\dfrac{T}{12}

∴Averagevelocity=DisplacementTime∴Averagevelocity=\dfrac{Displacement}{Time}

=A/2T/12=\dfrac{A/2}{T/12}

=AT×6=\dfrac{A}{T}\times 6

=6×10.5=\dfrac{6\times 1}{0.5}

=12cm/s=12cm/s

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Simple pendulum simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Oscillations (Simple Harmonic Motion), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.