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Show that for a particle executing simple harmonic motion a. the average value of kinetic energy is equal to the average value of potential energy. b. average potential energy = average kinetic energy =1/2 (total energy) (Hint : average kinetic energy = < kinetic energy > =1/Tint_(0)^(T) (Kinetic energy) dt and average Potential energy = =1/Tint_(0)^(T) (potential energy) dt |
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Answer» <P> Solution :a). `(E_(k))"ave"=1/Tint_(0)^(T)E_(k)dt=1/Tint_(0)^(T)(1)/(2)mA^(2)omega^(2)cos^(2)omegatdt``=1/(4T)mA^(2)square^(2).T""...(1)` `(E_(p))"ave"=1/Tint_(0)^(T)E_(p)dt=1/Tint_(0)^(T)(1)/(2)mA^(2)omega^(2)SIN^(2)omegatdt` `=1/(2T)momega^(2)A^(2)int_(0)^(2)((1-cos2omegat))/(2).dt` `=1/(4T).momega^(2)A^(2)[t-(sin2omegat)/(2omega)]_(0)^(T)` `(E_(p))_("ave")=1/(4T)msquare^(2)A^(2)T""..(2)` `=msquare^(2)A^(2)//4` From EQUATIONS (1) and (2) it can show that the average K.E. is equal toaverage P.E. (Total Energy) `TE=1/2msquare^(2)A^(2)` `(E_(k))_("ave")=(E_(p))_("ave")=1/2.((1)/(2)momega^(2)A^(2))=1/2(TE)` b. average kinetic energy = `=1/Tint_(0)^(T)` (Kinetic energy) dt and average Potential energy = `=1/Tint_(0)^(T)` (Potentialenergy) dt |
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