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In SHM, if the maximum acceleration of the particle is α = Aω² and the maximum velocity is β = Aω, find the time period of the motion.

1. 2π/ω
2. 2πβ/α
3. 2π/α
4. 2πT/ω

Correct answer: 2πβ/α

Solution

In SHM, maximum acceleration is α = Aω² and maximum velocity is β = Aω. Dividing α by β gives ω = α/β. The time period T is related to ω by T = 2π/ω. Substituting ω = α/β, we get T = 2πβ/α.

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