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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
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Two particles of masses m₁ m₂ move with initial velocities u₁ and u₂
- If n₁, n₂ and n₃ are the fundamental frequencies of three segments into which a string
- A body covers 200 cm in the first 2 seconds and 220 cm in the next 4 seconds.
- Which of the following physical quantities has neither dimensions nor unit
- Three capacitors 3μF,6μF and 6μF are connected in series to a source of 120V.
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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