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A particle is performing linear S.H.M. at a point A, on the path, its potential energy is three times kinetic energy. At another point B on the same path, its kinetic energy is 3 times the potential energy. The ratio of the potential energy at A to its potential energy at B isA. `9 : 1`B. `1 : 9`C. `1 : 3`D. `3 : 1` |
Answer» Correct Answer - D `T.E._(A)=K.E_(A)+P.E_(A)` `=KE_(A)+3KE_(A)` `=4KE_(A)` `T.E_(B)=KE_(B)+KE_(B)=KE_(B)+(1)/(3)KE_(B)` `T.E_(B)=(4KE_(B))/(3)` |
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