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A particle is executing a simple harmonic motion. Its maximum acceleration is a and maximum velocity is b. Then its time period of vibration will be:

1. α / β
2. β² / α
3. 2πβ / α
4. β² / α²

Correct answer: β² / α

Solution

In SHM, maximum acceleration is given by a = ω²A and maximum velocity is given by b = ωA, where ω is the angular frequency and A is the amplitude. Dividing these equations, we get ω = a / b. The time period T is related to ω by T = 2π / ω. Substituting ω = a / b, we get T = 2πb / a. Thus, the correct answer is B.

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