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A particle executes S.H.M., according to the displacement equation x = 6 sin `(3pit+pi//6)`m. Then the magnitude of its acceleration at t = 2 s isA. `3pi^(2)m//s^(2)`B. `9pi^(2)m//s^(2)`C. `18pi^(2)m//s^(2)`D. `27pi^(2)m//s^(2)` |
Answer» Correct Answer - D `x=6sin(3pit+(pi)/(6))` `x=6xxsin(3pixx2+(pi)/(6))` `=6sin((pi)/(6))=6xx(1)/(2)=3` `a=omega^(2)x` `=9pi^(2)xx3=27pi^(2)`. |
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