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The particle executing simple harmonic motion has a kinetic energy `K_(0) cos^(2) omega t`. The maximum values of the potential energy and the energy are respectivelyA. `K_(0) and K_(0)`B. `0 and 2K_(0)`C. `(K_(0))/(2) and K_(0)`D. `K_(0) and 2K_(0)` |
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Answer» Correct Answer - A `:. K.E. =K_(0) cos^(2) omegat` ltbtgt `:.` Maximum P.E.= Maximum K.E. = Total energy =`K_(0)` |
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