Oscillations

JEE Main level, numerical answer, medium

13 of 194

Question 13 of 194

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of 4 m,2 ms−14 \mathrm{~m}, 2 \mathrm{~ms}^{-1} and 16 ms−216 \mathrm{~ms}^{-2} at a certain instant. The amplitude of the motion is x, m\sqrt{x}, \mathrm{~m} where xx is ________

Show answer and solution

Answer: 17

x=4 m,V=2 m/s,a=16 m/s2∣a∣=ω2x⇒16=ω2(4)ω=2rad/sv=ωA2−x2A=v2ω2+x2⇒A=44+16A=17 m\begin{aligned} & x=4 \mathrm{~m}, V=2 \mathrm{~m} / \mathrm{s}, \mathrm{a}=16 \mathrm{~m} / \mathrm{s}^2 \\ & |\mathrm{a}|=\omega^2 \mathrm{x} \\ & \Rightarrow 16=\omega^2(4) \\ & \omega=2 \mathrm{rad} / \mathrm{s} \\ & \mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\ & A=\sqrt{\frac{\mathrm{v}^2}{\omega^2}+\mathrm{x}^2} \Rightarrow \mathrm{A}=\sqrt{\frac{4}{4}+16} \\ & A=\sqrt{17} \mathrm{~m}\end{aligned}

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.