A particle is doing simple harmonic motion of amplitude $0.06 \mathrm{~m}$ and time period $3.14 \mathrm{~s}$. The maximum velocity of the particle is. . . . .. . $\mathrm{cm} / \mathrm{s}$.
JEE MAIN 2024, Diffcult
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We know
$\mathrm{V}_{\max } =\omega \mathrm{A} \quad \text { at mean position }$
$=\frac{2 \pi}{\mathrm{T}} \mathrm{A}=\frac{2 \pi}{\pi} \times 0.06=0.12 \mathrm{~m} / \mathrm{sec}$
$\mathrm{V}_{\max } =12 \mathrm{~cm} / \mathrm{sec}$
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