InterviewSolution
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InterviewSolution
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An external harmonic force F whose frequency can be varied, with amplitude maintained constant, acts in a vertical direction on a ball suspended by a weightless spring. The damping coefficient is eta times less than the natural oscillation frequency omega_(0) of the ball. How much, in per cent, does the mean power ( :P: ) developed differ from themaximum mean power ( :P: )_(max)? Averaging is performed over one oscillation period. |
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Answer» Solution :Given ` beta = omega_(0)//eta` . Then from the previous problem `lt Pgt =(F_(0)^(2)omega_(0))/( etam). (1)/( ((omega_(0)^(2))/(OMEGA)-omega)^(2)+ 4 ( omega_(0)^(2))/( eta^(2)))` At displacement resonance ` omega=sqrt(omega_(0)^(2)-2 beta^(2))` `lt P gt_(res)=(F_(0)^(2)omega_(0))/( etam)(1)/((4 beta^(4))/( omega_(0)^(2)- 2 beta^(2))+(4 omega_(0)^(2))/( eta^(2)))=(F_(0)^(2)omega_(0))/( eta m)(1)/((4 omega_(0)^(4)//eta^(4))/( omega_(0)^(2)(1-(2)/( eta^(2))))+4( omega_(0)^(2))/( eta^(2)))` `=(F_(0)^(2))/( 4 eta m omega_(0))(n^(2))/((1)/( n^(2)-2)+1)=(F_(0)^(2)eta)/( 4 m omega_(0))(n^(2)-2)/( n^(2)-1)` while `lt P gt_(MAX)=(F_(0)^(2)eta)/(4 m omega_(0))`. THUS `(lt P gt_(max)- lt P gt _(res))/(lt P gt_(max))=(100)/(eta^(2)-1)%` |
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