The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of 4 m , 2 ms -1 and 16 ms -2 at a certain instant. The amplitude of the motion is sqrt x m where x is .

Physics · Simple Harmonic Motion

JEE Main 2024 — 9 April, Shift 1 — Question 58

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

Answer: 17

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{x}=4 \mathrm{~m}, \mathrm{~V}=2 \mathrm{~m} / \mathrm{s}, \mathrm{a}=16 \mathrm{~m} / \mathrm{s}^{2}$

\begin{aligned} & |a|=\omega^{2} x & \Rightarrow 16=\omega^{2}…(2) & \omega=2 \mathrm{rad} / \mathrm{s} \end{aligned}

\begin{aligned} & v=\omega \sqrt{A^{2}-x^{2}} & A=\sqrt{\frac{v^{2}}{\omega^{2}}+x^{2}} \Rightarrow A=\sqrt{\frac{4}{4}+16} & A=\sqrt{17} m \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 1
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Kinematics of SHM, Phase and Energy in SHM