InterviewSolution
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InterviewSolution
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A particle of mass m executing SHM with amplitude A and angular frequency `omega`. The average value of the kinetic energy and potential energy over a period isA. `0,(1)/(2)momega^(2)A^(2)`B. `(1)/(2)momega^(2)A^(2),0`C. `(1)/(2)momega^(2)A^(2),(1)/(2)momega^(2)A^(2)`D. `(1)/(4)momega^(2)A^(2),(1)/(4)momega^(2)A^(2)` |
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Answer» Correct Answer - D Let the displacement of the particle executing SHM at any instant of time t from its equilibrium position is given by `x=Acos(omegat+phi)` velocity, `v=(dx)/(dt)=-omegaAsin(omegat+phi)` Kinetic energy of the particle is `K=(1)/(2)mv^(2)=(1)/(2)momega^(2)A^(2)sin^(2)(omegat+phi)` Potential energy of the particle is `U=(1)/(2)momega^(2)x^(2)=(1)/(2)momega^(2)A^(2)cos^(2)(omegat+phi)` Average value of kinetic energy over a period is `ltKgt=(1)/(T)int_(0)^(T)Kdt=(1)/(T)int_(0)^(T)(1)/(2)momega^(2)A^(2)(omegat+phi)dt` `=(1)/(2T)momega^(@)A^(2)int_(0)^(T)[(1-cos2(omegat+phi))/(2)]dt` Since the average value of both a sine and a cosine function for a complete cycle or over a time period T is 0. `therefore lt K gt =(1)/(4T)momega^(2)A^(2)[t]_(0)^(T)=(1)/(4T)momega^(2)A^(2)T=(1)/(4)momega^(2)A^(2)` Average value of potential enery over a period is `ltUgt=(1)/(T)int_(0)^(T)(1)/(2)momega^(2)A^(2)cos^(2)(omegat+phi)dt` `=(1)/(2T)momega^(2)A^(2)int_(0)^(T)[(1+cos2(omegat+phi))/(2)]dt` Since the average value of both a sine and a cosine function for a complete cycle or over a time period T is zero. `therefore lt U gt =(1)/(4T)momega^(2)A^(2)[t]_(0)^(T)=(1)/(4T)momega^(2)A^(2)T=(1)/(4)momega^(2)A^(2)` |
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