Expression for K.E of a simple harmonic oscillator: The displacement of the body in S.H.M, `X=A sin omegat`
where `A=` amplitude and `omegat=` Angular displacement.
Velocity at any instant, `V=(dx)/(dt)=Aomega cos omegat`
So its K.E `=(1)/(2)mv^(2)=(1)/(2)mA^(2)omega^(2)cos^(2)omegat`
But `cos^(2)omegat=(1-sin^(2)omegat)`
`thereforeK.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 a 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 in S.H.M is given by, `x=A sin omegat`
For a body in S.H.M acceleration, `a=-omega^(2)Y` ltbgt 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 displacement through a small distance dx
`rArr` Work done, `dW=F.dx`
This work done `=P.E` in the body
`thereforeP.E=momega^(2)x.dx`(where x is its displacement)
Total work done, `W=intdW=int_(0)^(x)momega^(2)x.dx`
work done, `W=(momega^(2)x^(2))/(2).`
This work is stored as potential energy.
`therefore` P.E at any point `=(1)/(2)momega^(2)x^(2)`
For conservative force total Mechanical Energy at any point `=E=P.E+K.E`
`therefore` Total energy.
`E=(1)/(2)momega^(2)(A^(2)-x^(2))+(1)/(2)momega^(2)x^(2)`
`therefore E=(1)/(2)momega^(2){A^(2)-x^(2)+x^(2)}=(1)/(2)momega^(2)A^(2)`
So far a body in S.H.M total energy at any point of its motion is constant and equals to `(1)/(2)momega^(2)A^(2)`.
