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The avergae kinetic energy of particle of mass `m` undergoing `S.H.M.` with angular frequency `omega` and amplitude `A,` over half of one time period isA. `(1)/(2) m omega^(2) A^(2)`B. `(1)/(4) m omega^(2)A^(2)`C. `m omega^(2)A^(2)`D. `2 m omega^(2)A^(2)` |
Answer» Correct Answer - B average kinetic energy of a particle is `K.E._(avg) = (1)/(2)m omega^(2) A^(2) (int_(0)^(T//2)cos^(2)omegat.dt)/(int_(0)^(T//2)dt)` by solving we get `K.E_(avg) = (1)/(4)m omega^(2) A^(2)` |
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