Q83 / 90 · JEE Main 2024 (06 Apr Shift 1) · Physics
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 _______ $\mathrm{cm} / \mathrm{s}$.
Correct answer: 12
Solution
We know
$\begin{aligned}
\mathrm{V}_{\max } & =\omega \mathrm{A} \quad \text { at mean position } \\
& =\frac{2 \pi}{\mathrm{T}} \mathrm{A}=\frac{2 \pi}{\pi} \times 0.06=0.12 \mathrm{~m} / \mathrm{sec} \\
\mathrm{V}_{\max } & =12 \mathrm{~cm} / \mathrm{sec}
\end{aligned}$
Hence, the correct answer is 12.