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A particle is executing a simple harmonic motion. Its maximum acceleration is α and maximum velocity is β. Then, its time period of vibration will be
Options
(a) β²/α
(b) 2πβ/α
(c) β²/α²
(d) α/β
Correct Answer:
2πβ/α
Explanation:
As, we know, in Simple Harmonic Motion
Maximum acceleration of the particle, α = Aω²
Maximum velocity, β = Aω
⇒ ω = α / β
⇒ T = 2π / ω = 2πβ / α [Since, ω = 2π / T].
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Topics: Oscillations
(58)
Subject: Physics
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N
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- A source is moving towards an observer with a speed of 20 m/s and having frequency
- If voltage across a bulb rated 220 volt-100 Watt drops by 2.5% of its rated value
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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