Oscillations

JEE Main level, numerical answer, medium

23 of 194

Question 23 of 194

A particle of mass 250g250g executes a simple harmonic motion under a periodic force F=(–25x)NF=(–25x)N. The particle attains a maximum speed of 4ms−14m{s}^{-1} during its oscillation. The amplitude of the motion is ______cmcm.

Show answer and solution

Answer: 40

Given, force F=(–25x)NF=(–25x)N
⇒0.25a=−25x\Rightarrow 0.25a=-25x
⇒a=−100x\Rightarrow a=-100x
Now comparing it with standard equation, a=−ω2xa=-{\omega }^{2}x we get
ω=10\omega =10.
Now, vmax=ωA{v}_{max}=\omega A
⇒ωA=4\Rightarrow \omega A=4
⇒A=410=0.4m\Rightarrow A=\dfrac{4}{10}=0.4m
⇒A=40cm\Rightarrow A=40cm.

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.