Physics · Simple Harmonic Motion

JEE Main 2024 — 6 April, Shift 1 — Question 58

A particle is doing simple harmonic motion of amplitude 0.06 m and time period 3.14 s . The maximum velocity of the particle is $\qquad$ $\mathrm{cm} / \mathrm{s}$ .

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

We know

$\mathrm{v}_{\max }=\omega \mathrm{A} \quad$

at mean position

\begin{aligned} & =\frac{2 \pi}{\mathrm{~T}} \mathrm{~A}=\frac{2 \pi}{\pi} \times 0.06=0.12 \mathrm{~m} / \mathrm{sec} \mathrm{v}_{\max } & =12 \mathrm{~cm} / \mathrm{sec} \end{aligned}

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Physics
Chapter: Simple Harmonic Motion
Topic: Kinematics of SHM, Phase and Energy in SHM