MHT CET · Physics · Oscillations
A particle is performing S.H.M. with maximum velocity ' \(\mathrm{v}\) '. If the amplitude is tripled and periodic time is doubled then maximum velocity will be
- A \(1.5 \mathrm{v}\)
- B \(3 \mathrm{v}\)
- C \(2 \mathrm{v}\)
- D \(\mathrm{v}\)
Answer & Solution
Correct Answer
(A) \(1.5 \mathrm{v}\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{V}=\mathrm{V}_{\max }=\omega \mathrm{A}=\frac{2 \pi}{\mathrm{T}} \cdot \mathrm{A}
\)
If \(\mathrm{A}\) is tripled and \(\mathrm{T}\) is doubled, \(\mathrm{V}_{\max }\) will become \(\frac{3}{2}\) times or 1.5 times.
\mathrm{V}=\mathrm{V}_{\max }=\omega \mathrm{A}=\frac{2 \pi}{\mathrm{T}} \cdot \mathrm{A}
\)
If \(\mathrm{A}\) is tripled and \(\mathrm{T}\) is doubled, \(\mathrm{V}_{\max }\) will become \(\frac{3}{2}\) times or 1.5 times.
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