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A particle executes SHM with amplitude A...

A particle executes SHM with amplitude A and time period T. When the displacement from the equilibrium position is half the amplitude , what fractions of the total energy are kinetic and potential energy?

Text Solution

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Total energy `E = (1)/(2) kA^(2)`
when `x= +- (A)/(2)`
The potential energy `U = (1)/(2) kx^(2)`
`=(1)/(2) k. ( +- (A)/(2))^(2) = (1)/(4). (1)/(2) kA^(2) = (1)/(4) E= 25% ` of `E`
`:. `Kinetic energy `= E-U = E=-(1)/(4) E = (3)/(4) E =75%` of `E `
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