InterviewSolution
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InterviewSolution
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In SHM , potential energy of a particle at mean position is `E_(1)` and kinetic enregy is `E_(2)` , thenA. `E_(1)=E_(2)`B. total potential energy at `x=(sqrt(3)A)/(2)` is `E_(1)+(3E_(2))/(4)`C. total kinetic energy at `x=(sqrt(3)A)/(2)` is `(3E_(2))/(4)`D. total kinetic energy at `x=(A)/(sqrt(2))` is `(E_(2))/(4)` |
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Answer» Correct Answer - B (b) `U=U_("mean")+(1)/(2)kX^(2)` Given , `U_("mean")=E_(1)` and `(1)/(2)kA^(2)=E_(2)` = maximum kinetic energy at mean position `therefore X=(sqrt(3)A)/(2), U=E_(1)+(1)/(2)k((sqrt(3)A)/(2))^(2)=E_(1)+(3)/(4)E_(2)` |
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