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Discuss in detail the energy in simple harmonic motion. |
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Answer» Solution :(a) Expression for Potential Energy For the simple harmonic motion, the force and the displacement are related by Hooke.s LAW `vecF=-kvecr` Since force is a vector quantity, in three dimensions it has three components. Further, the force in the above EQUATION is a conservative force field, such a force can be derived from a scalar function which has only component. In one DIMENSIONAL case `F=-kx""...(1)` The work done by the conservative force fieldis independent of path. The potential energy U can be CALCULATED from the following expression. `F=-(dU)/(dx)""...(2)` Comparing (1) and (2), we get `-(dU)/(dx)=-kx` `dU=-kx dx`  This work done by the force F during a small displacement dx stores as potential energy `U(x)=int_(0)^(x)kx.dx.=(1)/(2)k(x.)^(2)|_(0)^(x)=(1)/(2)kx^(2)" "...(3)` From equation `omega=sqrt((k)/(m))`, we can substitute the value of force constant `k=momega^(2)` in equation (3), `U(x)=(1)/(2)momega^(2)x^(2)""...(4)` where `omega` is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from equation `y=Asinomegat`, we get `x=Asinomegat` `U(t)=(1)/(2)momega^(2)A^(2)sin^(2)omegat""...(5)` This variation of U is shown in figure. (b) Expression for Kinetic Energy Kinetic energy `KE=(1)/(2)mv_(x)^(2)=(1)/(2)m((dx)/(dt))^(2)""...(6)` Since the particle is executing simple harmonic motion, from equation `y=Asinomegat` `x=Asinomegat` Therefore, velocity is `v_(x)=(dx)/(dt)=Aomegacosomegat""...(7)` `=Aomegasqrt(1-((x)/(A))^(2))` `v_(x)=omegasqrt(A^(2)-x^(2))""...(8)` Hence, `KE=(1)/(2)mv_(x)^(2)=(1)/(2)momega^(2)(A^(2)-x^(2))""...(9)` `KE=(1)/(2)momega^(2)A^(2)cos^(2)omegat""...(10)` This variation with time is shown in figure. (c ) Expression for Total Energy Total energy is the sum of kinetic energy and potential energy `E=KE+U""...(11)`  `E=(1)/(2)momega^(2)(A^(2)-x^(2))+(1)/(2)momega^(2)x^(2)` Hence, cancelling `x^(2)` term, `E=(1)/(2)momega^(2)A^(2)=` constant `""...(12)` Alternatively, from equation (5) and equation (10), we get the total energy as `E=(1)/(2)momega^(2)A^(2)sin^(2)omegat+(1)/(2)momega^(2)A^(2)cos^(2)omegat` `=(1)/(2)momega^(2)A^(2)(sin^(2)omegat+cos^(2)omegat)` From trigonometry identify, `(sin^(2)omegat+cos^(2)omegat)=1` `E=(1)/(2)momega^(2)A^(2)` = constant which gives the law of conservation of total energy Thus the AMPLITUDE of simple harmonic oscillator, can be expressed in terms of total energy. `A=sqrt((2E)/(momega^(2)))=sqrt((2E)/(k))""...(13)` 
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