JEE Mains · Physics · STD 11 - 13. oscillations
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 _______ \(\mathrm{cm} / \mathrm{s}\).
- A \(12\)
- B \(15\)
- C \(20\)
- D \(22\)
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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