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The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is : |
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Answer» `4:1` and `K.E.=(1)/(2)m omega^(2)(a^(2)-x^(2))` For `x=(a)/(2)` then the RATIO is `(K.E.)/(P.E.)=(1//2m omega^(2)(a^(2)-a^(2)//4))/(1//2m omega^(2)(a^(2)//4))` `(K.E.)/(P.E.)=(3a^(2)//4)/(a^(2)//4)=3:1` So CORRECT CHOICE is ( c ). |
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