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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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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A mass of 2.0 kg is put on a flat plan attached to a vertical spring fixed on the ground
- ₉₂U²³⁴ has 92 protons and 238 nucleons. It decays by emitting an alpha particle
- The displacement of a particle varies according to the relation x=4 (cos π t + sin π t)
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Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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