A particle performing S.H.M. has an acceleration of 0.2 m//s^2 at a distance of 0.05 m from its mean position then what is it's time period ?

A particle performing S.H.M. has an acceleration of 0.2 m//s^2 at a di

1.	A particle performing S.H.M. has an acceleration of 0.2 m//s^2 at a distance of 0.05 m from its mean position then what is it's time period ?
Answer» SOLUTION :`AC c^n=omega^2x` ` 0.2=((2PI)/T)^2xx0.05` T^2=(4pi^2xx0.05)/0.2` `T^2=pi^2` `T=pi SEC.`

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A particle performing S.H.M. has an acceleration of 0.2 m//s^2 at a distance of 0.05 m from its mean position then what is it's time period ?

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