A particle is executing a simple harmonic motion. Its maximum acceleration

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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)

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