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In S.H.M. path length is 4 cm and maximum accelertion is `2pi^(2) cm//s^(2)`. Time period of motion is |
Answer» Correct Answer - A `A=2cm` `a_(m)=Aomega^(2)` `omega^(2)=sqrt((a_(m))/(A))=sqrt((2pi^(2))/(2))` `therefore omega=pi` `T=(2pi)/(omega)=(2pi)/(T_(1))=2s` |
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