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Discuss in detail the energy in simple harmonic motion.

Answer» <html><body><p></p>Solution :(a) Expression for Potential Energy <br/> For the simple harmonic motion, the force and the displacement are related by Hooke.s <a href="https://interviewquestions.tuteehub.com/tag/law-184" style="font-weight:bold;" target="_blank" title="Click to know more about LAW">LAW</a> <br/> `vecF=-kvecr` <br/> Since force is a vector quantity, in three dimensions it has three components. Further, the force in the above <a href="https://interviewquestions.tuteehub.com/tag/equation-974081" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATION">EQUATION</a> is a conservative force field, such a force can be derived from a scalar function which has only component. In one <a href="https://interviewquestions.tuteehub.com/tag/dimensional-2582799" style="font-weight:bold;" target="_blank" title="Click to know more about DIMENSIONAL">DIMENSIONAL</a> case <br/> `F=-kx""...(1)` <br/> The work done by the conservative force fieldis independent of path. The potential energy U can be <a href="https://interviewquestions.tuteehub.com/tag/calculated-907694" style="font-weight:bold;" target="_blank" title="Click to know more about CALCULATED">CALCULATED</a> from the following expression. <br/> `F=-(dU)/(dx)""...(2)` <br/> Comparing (1) and (2), we get <br/> `-(dU)/(dx)=-kx` <br/> `dU=-kx dx` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_V02_C10_E01_039_S01.png" width="80%"/> <br/> This work done by the force F during a small displacement dx stores as potential energy <br/> `U(x)=int_(0)^(x)kx.dx.=(1)/(2)k(x.)^(2)|_(0)^(x)=(1)/(2)kx^(2)" "...(3)` <br/> From equation `omega=sqrt((k)/(m))`, we can substitute the value of force constant `k=momega^(2)` in equation (3), <br/> `U(x)=(1)/(2)momega^(2)x^(2)""...(4)` <br/> where `omega` is the natural frequency of the oscillating system. For the particle executing simple harmonic motion from equation `y=Asinomegat`, we get <br/> `x=Asinomegat` <br/> `U(t)=(1)/(2)momega^(2)A^(2)sin^(2)omegat""...(5)` <br/> This variation of U is shown in figure. <br/> (b) Expression for Kinetic Energy <br/> Kinetic energy <br/> `KE=(1)/(2)mv_(x)^(2)=(1)/(2)m((dx)/(dt))^(2)""...(6)` <br/> Since the particle is executing simple harmonic motion, from equation <br/> `y=Asinomegat` <br/> `x=Asinomegat` <br/> Therefore, velocity is <br/> `v_(x)=(dx)/(dt)=Aomegacosomegat""...(7)` <br/> `=Aomegasqrt(1-((x)/(A))^(2))` <br/> `v_(x)=omegasqrt(A^(2)-x^(2))""...(8)` <br/> Hence, `KE=(1)/(2)mv_(x)^(2)=(1)/(2)momega^(2)(A^(2)-x^(2))""...(9)` <br/> `KE=(1)/(2)momega^(2)A^(2)cos^(2)omegat""...(10)` <br/> This variation with time is shown in figure. <br/> (c ) Expression for Total Energy <br/> Total energy is the sum of kinetic energy and potential energy <br/> `E=KE+U""...(11)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_V02_C10_E01_039_S02.png" width="80%"/> <br/> `E=(1)/(2)momega^(2)(A^(2)-x^(2))+(1)/(2)momega^(2)x^(2)` <br/> Hence, cancelling `x^(2)` term, <br/> `E=(1)/(2)momega^(2)A^(2)=` constant `""...(12)` <br/> Alternatively, from equation (5) and equation (10), we get the total energy as <br/> `E=(1)/(2)momega^(2)A^(2)sin^(2)omegat+(1)/(2)momega^(2)A^(2)cos^(2)omegat` <br/> `=(1)/(2)momega^(2)A^(2)(sin^(2)omegat+cos^(2)omegat)` <br/> From trigonometry identify, `(sin^(2)omegat+cos^(2)omegat)=1` <br/> `E=(1)/(2)momega^(2)A^(2)` = constant <br/> which gives the law of conservation of total energy <br/> Thus the <a href="https://interviewquestions.tuteehub.com/tag/amplitude-859568" style="font-weight:bold;" target="_blank" title="Click to know more about AMPLITUDE">AMPLITUDE</a> of simple harmonic oscillator, can be expressed in terms of total energy. <br/> `A=sqrt((2E)/(momega^(2)))=sqrt((2E)/(k))""...(13)` <br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_PHY_XI_V02_C10_E01_039_S03.png" width="80%"/></body></html>


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