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A particle is executing a simple harmonic motion. Its maximum acceleration is `alpha` and maximum velocity is `beta`. Then, its time period of vibration will beA. `(2pibeta)/(alpha)`B. `(beta^(2))/(alpha^(2))`C. `(alpha)/(beta)`D. `(beta^(2))/(alpha)` |
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Answer» Correct Answer - A Maximum acceleration `alpha=Aomega^(2)` . . (1) Maximum velocity `beta=A omega` `(Eq^(n)(1))/(Eq^(n)(2))implies(alpha)/(beta)implies(alpha)/(beta)=(2pi)/(T)` `T=2pi(beta)/(alpha)` |
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