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If a simple harmonic motion is represented by `(d^(2)x)/(dt^(2)) + alphax = 0`, its time period is :A. `2pialpha`B. `2pisqrt(alpha`C. `(2pi)/(alpha)`D. `(2pi)/(sqrt(alpha))` |
Answer» Correct Answer - D Comparing it with the standard equation of S.H.M. `(d^(2)x)/(dt^(2))=-omega^(2)x` We get, `omega^(2)=alpha` `therefore omega=sqrt(alpha)` `T=(2pi)/(omega)=(2pi)/(sqrt(alpha))` |
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