A particle executes simple harmonic motion with an amplitude of 4 cm . At the mean position, velocity of the particle is 10 cm / s . The distance of the particle from the mean position when its speed becomes 5 cm / s is √(α) cm , where α=

Physics · Simple Harmonic Motion

JEE Main 2024 — 27 January, Shift 1 — Question 54

A particle executes simple harmonic motion with an amplitude of 4 cm . At the mean position, velocity of the particle is $10 \mathrm{~cm} / \mathrm{s}$ . The distance of the particle from the mean position when its speed becomes $5 \mathrm{~cm} / \mathrm{s}$ is $\sqrt{\alpha} \mathrm{cm}$ , where $\alpha=$ $\qquad$

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{V}_{\mathrm{at} \text { mean position }} \mathrm{A} \omega \Rightarrow 10=4 \omega$

\begin{aligned} & \omega=\frac{5}{2} & \mathrm{v}= \omega \sqrt{\mathrm{A}^{2}-\mathrm{x}^{2}} & 5= \frac{5}{2} \sqrt{4^{2}-\mathrm{x}^{2}} \Rightarrow \mathrm{x}^{2}=16-4 & \mathrm{x}=\sqrt{12} \mathrm{~cm} \end{aligned}

Answer key and solution verified before publishing.

Practise Simple Harmonic Motion

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2024
Paper: 27 January, Shift 1
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Kinematics of SHM, Phase and Energy in SHM