InterviewSolution
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InterviewSolution
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Derivative expressions for the kinetic energy and potential energy of a simple harmonic oscillator. |
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Answer» Solution :Expression for K.E of a SIMPLE harmonic oscillator: The displacement of the body in S.H.M., `X=Asinomegat` where A = amplitude, `omegat=` Angular displacement. Velocity at any instant, `y=dy/dx=Aomegacosomegat` `therefore K.E=1/2mv^(2)=1/2mA^(2)omega^(2)cos^(2)omegat` `therefore K.E=(1)/(2)mA^(2)omega^(2)(1-sin^(2)omegat)` `=(1)/(2)mA^(2)omega^(2)[1-(x^(2))/(A^(2))]` `=(1)/(2)MOMEGA^(2)[A^(2)-x^(2)]` `therefore K.E =(1)/(2)momega^(2)[A^(2)-x^(2)]` At mean position velocity is maximum and displacement `x=0` `therefore K.E_(max)=(1)/(2)mA^(2)omega^(2)` Expression for P.E of simple harmonic oscillator: Let a body of mass 'm' is in S.H.M with an amplitude A.  Let O is the mean position. EQUATION of a body inS.H.M is given by `x=Asinomegat` For body in S.H.M acceleration, `a=-omega^(2)Y` Force, `F=ma=-momega^(2)x` `therefore` Restoring force, `F=momega^(2)x` Potential energy of the body at any point say 'x': Let the body is displaced through a small DISTANCE dx Work done, `dW=F.dx=P.E.` This work done. `therefore P.E=momega^(2)x.dx`(where x is its displacement) Total work done, `W=intdW=int_(0)^(x)momega^(2)x.dx` `rArr` Work done, `W=(momega^(2)x^(2))/(2)`. This work is stored as potential enerygy. `therefore` P.E. at any point `=1/2momega^(2)x^(2)` |
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