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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2024
  • Hard
  • Numerical

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 _______ \(\mathrm{cm} / \mathrm{s}\).

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Answer & Solution

Correct Answer

(A) \(12\)

Step-by-step Solution

Detailed explanation

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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