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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
- In a primary coil 5 A current is flowing on 220 V. In the secondary coil 2200 V
- Two vessels separately contain two ideal gases A and B at the same temperature,
- A plane electromagnetic wave of frequency 20 MHz travels through a space along
- Two bodies of mass 10 kg and 5 kg moving in concentric orbits of radius
- A galvanometer having resistance of 50 Ω requires a current of 100μA to give full
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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