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If a simple harmonic motion is represented by (d^(2)X)/(dt^(2))+alphax=0 its time period is : |
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Answer» `2PI//alpha` Comparing it with STANDARD equation of SHM. `(d^(2)X)/(dt^(2))+W^(2)X=0` we have `w^(2)=a` Time period `T=(2pi)/(w)=(2pi)/(sqrt(a))` So the CORRECT choice is (b). |
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