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If a simple harmonic motion is represented by `(d^(2)x)/(dt^(2)) + alphax = 0`, its time period is :A. `(2pi)/(alpha)`B. `(2pi)/(sqrt(alpha))`C. `2pialpha`D. `2pisqrt(alpha)` |
Answer» Correct Answer - B `(d^(2)x)/(dt^(2)) = -alphax………(i)` We know `a = (d^(2)x)/(dt^(2)) = -omega^(2)x …….(ii)` From Eq. `(i)` and `(ii)`, we have `omega^(2) = alpha` `omega = sqrt(alpha)` or `(2pi)/(T) = sqrt(alpha) :. T = (2pi)/(sqrt(a))` |
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