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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`
`therefore` Kinetic energy , `E-U=E-(1)/(4)E=(3)/(4)E=75%` of E
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