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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 identical cells of the same e.m.f. and same internal resistance
- A body of length 1 m having cross-sectional area 0.75 m² has heat flow through it
- Pick out the correct statements from the following
- In an electrical circuit R,L,C and an AC voltage source are all connected in series.
- A spherical drop of mercury having a potential of 2.5 v is obtained as a result
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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