When the displacement is one half the amplitude in SHM The fraction of the total energy that is potential is?

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Solution : Fraction of total energy which is kinetic `=(K.E.)/(T.E.)` <br> `=((1)/(2)momega^(2)(a^(2)-y^(2)))/((1)/(2)m omega^(2)a^(2))=(a^(2)-y^(2))/(a^(2))=(a^(2)-(a^(2))/(4))/(a^(2))=(3)/(4)` <br> Fraction of total energy which is potential <br> `=(P.E.)/(T.E.)=((1)/(2)momega^(2)y^(2))/((1)/(2)momega^(2)a^(2))=(y^(2))/(a^(2))=(a^(2)//4)/(a^(2))=(1)/(4)` <br> KE `=` PE <br> `(1)/(2)momega^(2)(a^(2)-y^(2))=(1)/(2)momega^(2)y^(2)` <br> On solving , `y=a//sqrt(2)`