question_answer 22)

Show that for a particle in linear S.H.M., the average kinetic energy over a period of oscillation equals the average potential energy over the same period.

Answer:

Consider a particle of mass m executing S.H.M. with period T. The displacement of the particle at an instant t, when time period is noted from the mean position is given by \[y=-a\sin \omega t\] \[\frac{\text{1}}{\text{2}}{{\text{x}}_{\text{0}}}\text{=}{{\text{x}}_{\text{0}}}{{\text{e}}^{\text{-bt/2m}}}\text{,}\] Velocity, \[\text{2=}{{\text{e}}^{\text{bt/2m}}}\text{,}\] \[\text{lo}{{\text{g}}_{\text{e}}}\text{2=}\frac{\text{bt}}{\text{2m}}\text{lo}{{\text{g}}_{\text{e}}}\text{e=}\frac{\text{bt}}{\text{2m}}\] \[\text{b=}\frac{\text{2mlo}{{\text{g}}_{\text{e}}}\text{2}}{\text{t}}\text{ }....\text{(i)}\] \[\text{t=2 }\!\!\pi\!\!\text{ }\sqrt{\frac{\text{M}}{\text{K}}}\text{=2 }\!\!\times\!\!\text{ }\frac{\text{22}}{\text{7}}\text{ }\!\!\times\!\!\text{ }\sqrt{\frac{\text{3000}}{\text{4 }\!\!\times\!\!\text{ 5 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{4}}}}}\text{=}\frac{\text{44}}{\text{70}}\sqrt{\frac{\text{3}}{\text{2}}}\text{s}\] \[\text{b=}\frac{\text{2mlo}{{\text{g}}_{\text{e}}}\text{2}}{\text{t}}\text{=}\frac{\text{2 }\!\!\times\!\!\text{ 750 }\!\!\times\!\!\text{ 0 }\!\!\times\!\!\text{ 6931}}{\frac{\text{44}}{\text{70}}\sqrt{\frac{\text{3}}{\text{2}}}}\text{=1350}\cdot 4kg/s\] Average K.E. over one cycle \[\omega \] \[\therefore \] \[\text{V=}\frac{\text{dy}}{\text{dt}}\text{=a }\!\!\omega\!\!\text{ cos }\!\!\omega\!\!\text{ t}\text{.}\] \[\text{K}\text{.E}\text{.,}{{\text{E}}_{\text{K}}}\text{=}\frac{\text{1}}{\text{2}}\text{m}{{\text{V}}^{\text{2}}}\text{=}\frac{\text{1}}{\text{2}}\text{m}{{\text{a}}^{\text{2}}}{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\omega\!\!\text{ t}\text{.}\] .(i) Average P.E. over one cycle is \[\text{P}\text{.E,}{{\text{E}}_{\text{P}}}\text{=}\frac{\text{1}}{\text{2}}\text{k}{{\text{y}}^{\text{2}}}\text{=}\frac{\text{1}}{\text{2}}\text{m}{{\text{a}}^{\text{2}}}{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\text{si}{{\text{n}}^{\text{2}}}\text{ }\!\!\omega\!\!\text{ t}\text{.}\] \[\left( \because k=m{{\omega }^{2}} \right)\] \[\therefore \] \[{{\text{E}}_{{{\text{K}}_{\text{a }\!\!\upsilon\!\!\text{ }}}}}\text{=}\frac{\text{1}}{\text{T}}\int\limits_{\text{0}}^{\text{T}}{{{\text{E}}_{\text{K}}}\text{dt=}\frac{\text{1}}{\text{T}}}\int\limits_{\text{0}}^{\text{T}}{\frac{\text{1}}{\text{2}}}\text{m}{{\text{a}}^{\text{2}}}{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\omega\!\!\text{ tdt=}\frac{\text{1}}{\text{2T}}\] .(ii) From (i) and (ii), \[\text{m}{{\text{a}}^{\text{2}}}{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\int\limits_{\text{0}}^{\text{T}}{\frac{\left( \text{1+cos2 }\!\!\omega\!\!\text{ t} \right)}{\text{2}}}\text{dt}\]