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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
- A thin wire of length L and mass M is bent to form a semicircle.
- The moment of inertia of a rod about an axis through its centre and perpendicular
- An air bubble in a glass slab with refractive index 1.5 (near normal incidence)
- In R-L-C series circuit, the potential differences across each element is 20 V.
- A boat taken 2h to travel 8km and back in still water.If the velocity of water 4 km/h
Topics: Oscillations (58)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
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