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In S.H.M. the ratio of kinetic energy at mean position to the potential energy when the displacement is half of the amplitude is |
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Answer» `(4)/(1)` At mean position, `y=0:.K.E.=(1)/(2)m omega^(2)a^(2)` At, `y=(a)/(2),P.E.=(1)/(2)m omega^(2)(a^(2))/(4)=(1)/(8)m omega^(2)a^(2)` `:.` RATIO `=((1//2)m omega^(2)a^(2))/((1//8)m omega^(2)a^(2))=("K.E. at mean position")/("PE at "a//2)=(4)/(1)` |
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