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In S.H.M. path length is 4 cm and maximu...

In S.H.M. path length is 4 cm and maximum accelertion is `2pi^(2) cm//s^(2)`. Time period of motion is

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The correct Answer is:
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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